aluminium can: Solution Method 2: Trial and improvement (a) (by substituting the value of r ) V= πr²h πr²h= 1000 h= 1000 πr² ………………(1) A= 2πr² +2πrh = 2πr²+ 2πr( 1000) πr² = 2πr³+2000 r ……………...(2) If r = 1, A= 2π+2000 1 = 2006.2832 r = 2, A= 2π(2)³+2000 2 = 1025.1327 r = 3, A= 2π(3)³+2000 3 = 723.2153 r = 4, A= 2π(4)³+2000 4 = 600.5310 r = 5, A= 2π(5)³+2000 5 = 557.0796 r = 5, A= 2π(5)³+2000 5 = 557.0796 r = 6, A= 2π(6)³+2000 6 = 559.5280 r = 7, A= 2π(7)³+2000 7 = 593.5904 r = 8, A= 2π(8)³+2000 8 = 652.1239 r = 9, A= 2π(9)³+2000 9 = 731.1602 The value of r that falls between 4 and 6 will produce the minimum surface area of the aluminium sheet used. r A 4.5 571.6789 4.6 567.7348 4.7 564.3275 4.8 561.4313 4.9 559.0225 5 557.0796 5.1 555.5825 5.2 554.5127 5.3 553.8532 5.4 553.5881 5.5 553.7027 5.6 554.1835 5.7 555.0179 5.8 556.1939 The lowest surface area is obtained when r=5.4 By substituting r=5.4 into the equation (1) h = 1000 π(5.4)² h = 10.92 h = 10.92 r 5.4 = 2.022,therefore approximately h=2r Surfeca area is minimum,when r=5.4,h=10.92 and h = 2.022 r (b) Alternative method (by substituting the value of h) V= πr²h πr²h = 1000 r² = 1000 πh ………………(1) A = 2πr² + 2πrh = 2π(1000) + 2πh (1000) πh πh = 2000+ 20h#8730; 10πh …………(2) h if h = 1,A = 2000+20 #8730; 10π = 2112.0998 h = 2,A = 2000+40 #8730; 20π 2 = 1158.5331 h = 3,A = 2000+60 #8730; 30π 3 = 860.8293 h = 4,A = 2000+80 #8730; 40π 4 = 724.1996 h = 5,A = 2000+100 #8730; 50π 5 = 650.6628 h = 6,A = 2000+120 #8730; 60π 6 = 607.9207 h = 7,A = 2000+140 #8730; 70π 7 = 582.3025 h = 8,A = 2000+160 #8730; 80π 8 = 567.0662 h = 9,A = 2000+180 #8730; 90π 9 = 558.5217 h = 10,A = 2000+200 #8730; 100π 10 = 554.4908 h = 11, A = 2000+220 #8730; 110π 11 = 553.6112 h = 12, A = 2000+240 #8730; 120π 12 = 554.9918 h = 13, A = 2000+260 #8730; 130π 13 = 558.0278 h = 14, A = 2000+280 #8730; 140π 14 = 562.2963 The value of h that lies between 10 and 12 will produce the least surface area r A 10.1 554.2786 10.2 554.0966 10.3 553.9436 10.4 553.8188 10.5 553.7211 10.6 553.6498 10.7 553.604 10.8 553.5828 10.9 553.5855 11 553.6112 11.1 553.6594 11.2 553.7292 11.3 553.82 11.4 553.9312 Minimum surface area occurs when h= 10.8 By substituting h=10.8 into the equation (1) r² = 1000 ……………(1) πh r = #8730;1000 πh = #8730;1000 10.8π = 5.429 h = 10.8 r 5.429 = 1.989 = 2 Surface area is minimum,when h=10.8, r=5.429 and h=2 r Method 2: Calculus πr² + 2πrh …………… (1)#61472; A = 2 πr²h = 1000 h = 1000 π r² &nbs p; ………………….(2) Substituting equation (2) into (1): A = 2πr² + 2πr (1000) r² #61552; A=2πr² + 2000 r A’(r) = 4πr-2000 r² When A’(r) = 0.4πr = 2000 r² r³ = 500 π r = 500 π r = 5.419 h= 1000 π x 5.419² = 10.84 h =10.84 r 5.419 ~2 A”(r) =4π + 4000 r³ A”(5.419)>0 (because r>0), therefore A is aluminium. The sirface area is minimum, when r = 5.419, h=10.84 and h =2 r Method 3: Graphical Method By plotting the surface area A, against radius, r, for which A=2πr²+200 , the following graph is plotted. r r A 1 2006.283 2 1025.133 3 723.2153 4 600.531 5 557.0796 6 559.528 7 593.5904 8 652.1239 9 731.1602 Graph Area (A) Versus Radius (r) r A 5.1 555.5825 5.2 554.5137 5.3 553.8532 5.4 553.5881 5.5 553.7027 5.6 554.1835 5.7 555.0179 5.8 556.1939 5.9 557.7007 Graph Area (A) Versus Radius (r) r A 5.38 553.6102 5.39 553.5972 5.4 553.5881 5.41 553.5827 5.42 553.5811 5.43 553.5832 5.44 553.5891 5.45 553.5988 5.46 553.6122 5.47 553.6293 5.48 553.6501 From the graph, the value of r = 5.42 h = 1000 πr² = 1000 π(5.42)² = 10.84 h = 10.84 r 5.42 = 2 Surface area is minimum, when r = 5.4, h = 10.84 and h = 2 r To verify the answer , the following method can be used . Method: Construct a Table r A &nbs p; 5 557.0796 &nb sp; The table shows that the minimum surface area occurs 5.05 556.2765 when the value of r falls between 5.4 and 5.5. Therefore 5.1 555.5825 & nbsp; the minimum surface area lines between 553.5881 and 5.15 554.9953 553.7027. 5.2 554.5127 & nbsp; 5.25 554.1327 5.3 553.8532 & nbsp; 5.35 553.6722 5.4 553.5881 & nbsp; 5.45 553.5986 5.5 553.7027 & nbsp; 5.55 553.8982 5.6 554.1835 & nbsp; 5.65 554.5573 r A &nbs p; 5.41 553.5827 To find the more precise value of r and A, the values 5.42 553.5811 of r from 5.41 to 5.54 are used. 5.43 553.5832 5.44 553.5891 From the table, the least value for surface area is 5.45 553.5988 553.5811 and it occurs when r = 5.42. 5.46 553.6122 5.47 553.6293 5.48 553.6501 5.49 553.6746 5.5 553.7027 & nbsp; 5.51 553.7345 5.52 553.77 &n bsp; 5.53 553.8091 5.54 553.8518 r A &nbs p; 5.411 553.5823 ; To find the more precise of value r and A, the values of r 5.412 553.582 From 5.411 to 5.424 are used. 5.413 553.5818 ; 5.414 553.5816 ; From the table, the least value for surface area is 553.581 5.415 553.5814 ; and it occurs when r = 5.419 5.416 553.5812 ; 5.417 553.5811 ; 5.418 553.5811 ; 5.419 553.581 5.42 553.5811 5.421 553.5811 ; 5.422 553.5812 ; 5.423 553.5813 ; 5.424 553.5815 ; Solution for Question 2 Method 1:Calculus Area of aluminium sheet used, A = 2(2r 2r) + 2πrh……….(1) πr2h = 1000………………..(2) A = 8r2 +2πr( ) =8r2 +( ) = 8r2 +2000r -1 16r -2000r -2 When A is minimum, 0. 16r-2000r -2 = 0 16r = r3 = =125 r = 5 h = = =12.73 = =2.546 Minimum area of aluminium sheet used, when r = 5, h = 12.73 and =2.546. Method 2 : Draw a Graph By plotting the surface area A, against radius, r, for which A=8r2 + 2000r -1 , the following graph is produced. r A 1 2008 2 1032 3 738.6667 4 628 5 600 6 621.3333 7 677.7143 8 762 9 870.2222 Graph Area (A) Versus Radius (r) r A 4.7 602.2519 4.8 600.9867 4.9 600.2433 5 600 5.1 600.2369 5.2 600.9354 5.3 602.0785 5.4 603.6504 5.5 605.6364 From the graph, value of r = 5. h = = =12.73 = =2.546 Minimum area of aluminium sheet used, when r = 5, h = 12.73 and =2.546. To verify the answer, the following method can be used. Method: Construct a table r A &nbs p;Table shows that the minimum 1 2008 & nbsp;surface area occurs when the 1.2 1678.187 & nbsp; value of r falls between 4.8 and 1.4 1444.251 & nbsp; 5.2. 1.6 1270.48 &n bsp; 1.8 1137.031 & nbsp; 2 1032 & nbsp; 2.1 947.8109 & nbsp; 2.4 879.4133 & nbsp; 2.6 823.3108 & nbsp; 2.8 777.0057 & nbsp; 3 738.6667 &nb sp; 3.2 706.92 &nb sp; 3.4 680.7153 & nbsp; 3.6 659.2356 & nbsp; 3.8 641.8358 & nbsp; 4 628 &n bsp; 4.2 617.3105 & nbsp; 4.4 609.4255 & nbsp; 4.6 604.0626 & nbsp; 4.8 600.9867 & nbsp; 5 600 &n bsp; 5.2 600.9354 & nbsp; 5.4 603.6504 & nbsp; 5.6 608.0229 & nbsp; r A 4.9 600.2433 4.91 600.1968 4.92 600.1553 4.93 600.1187 4.94 600.0871 4.95 600.0604 4.96 600.0386 4.97 600.0271 4.98 600.0096 4.99 600.0024 5 600 5.01 600.0024 5.02 600.0096 5.03 600.0215 To get a more precise value of A and r, a smaller range of r from 4.8 to 5.2 s used. From the table, the minimum area is 600 and this occurs when r = 5. h = = = 12.73 = =2.546 Solution for Question 3 Consider The Following Shapes (1) Equilateral Triangle Method 1:Trigonometry and Calculus tan 60o = x = r Area of PQR = 3 Area of OPQ = 3 2X r = 3 xr = 3 r2 Total area of Aluminium Sheets Used, A = 6 r2 + 2πr( ) = 6 r2 +2000r -1 12 r – 2000r -2 When A is minimum, 0. 12 r = r3 = 96.225 r = 4.582 h = =15.16 =3.309 12 + 4000r -3 (>0, because r>0) Therefore , A is minimum. Minimum area or aluminium sheet used when r = 4.582, h= 15.16 and = 3.309. Method 2: Graphical Method By using the equation A=6#8730;3r²+2000-1, graph A versus r can be plotted to determine the value of r for the minimum value of A. Graph Area (A) Versus Radius (r) r A 1 2010.392 2 1041.568 3 760.1947 4 666.272 5 659.8 6 707.4453 7 794.9223 8 915.088 9 1063.974 From the graph, minimum area occurs when r falls between 4 cm and 5 cm. The following graph is plotted using the values of r between 4.54 and 4.64 Graph Area (A) Versus Radius (r) r A 4.54 654.7244 4.55 654.7008 4.56 654.6836 4.57 654.6726 4.58 654.668 4.59 654.6695 4.6 654.6773 4.61 654.6913 4.62 654.7114 4.63 654.7377 4.64 654.7701 From the graph, area is minimum when r #8776; 4.58 cm. h = 1000 Πr² = 15.17 h = 15.17 r 4.58 = 3.312 Minimun area of aluminium sheet used when r = 4.58, h = 15.17 and h = 3.312 Method 3: Construct a Table A table can be constructed using A = 6 3 r2 + 2000r-1 r h &nbs p;A h/r 1 318.3099 &nb sp; 2010.392 318.3099 2 79.57747 1041.568 &n bsp; 39.78874 3 35.36777 &nb sp; 760.1947 11.78874 4 19.89437 &nb sp; 666.272 11.78926 5 12.7324 &nbs p; 659.8 4.973592 6 8.841941 &nb sp; 707.4453 2.546479 7 6.49612 &nbs p; 794.9223 0.928017 8 4.973592 &nb sp; 915.088 0.621699 By using a more precise range of value r, we get the following tables. r h &nbs p;A h/r 4.1 18.93575 & nbsp; 662.4944 4.618475 4.2 18.04478 & nbsp; 659.5054 4.296376 4.3 17.21525 & nbsp; 657.2644 4.003545 4.4 16.44163 & nbsp; 655.7346 3.736733 4.5 15.71901 & nbsp; 654.8824 3.493113 4.6 15.043 &nb sp; 654.6773 3.270217 4.7 14.40968 & nbsp; 655.0912 3.06589 4.8 13.81553 & nbsp; 656.0983 2.878236 r h &nbs p;A h/r 4.51 15.64938 654.8333 3.469928 4.52 15.58021 654.7906 3.446949 4.53 15.5115 & nbsp; 654.7543 3.424172 4.54 15.44324 654.7244 3.401595 4.55 15.37543 654.7008 3.379216 4.56 15.30807 654.6836 3.357033 4.57 15.24115 654.6726 3.335044 4.58 15.17467 654.668 3.313246 4.59 15.10862 654.6695 3.291638 4.6 15.043 &nb sp; 654.6773 3.270217 4.61 14.97781 654.6913 3.248982 From the above table, minimum area of aluminium sheet used when r = 4.58, h = 15.17 and the ratio of h = 3.312 ( ii ) Regular Hexagon Method 1: Trionometry and Calculus Tan 300 = x r x = r tan 300 = r 3 Area of OPQ & nbsp;= 1 x 2x x r 2 = 1 x 2( r ) x r 2 = r2 3 = 3 r2 3 Area of Hexagon = 6 x 3r2 3 = 2 #8730; 3 r2 Total Area of Aluminium Sheets Used, A = 2(2 #8730; 3 r2) + 2 π r ( 1000 ) = 4 #8730; 3 r2 + 2000r-1 dA = 8 #8730; 3 r – 2000 r-2 dr For minimum surface area, dA = 0 8 #8730;3 r = 2000 r2 r3 = 2000 8 #8730;3 = 250 #8730; 3 r = 5.246 h = 1000 πr² = 1000 π x ( 5.246 ) = 11.57 h = 11.57 r 5.246 = 2.205 Method 2: Graphical Method By using the equation A=4#8730;3r²+2000-1, graph A versus r can be plotted using the value of r from 1 cm to 11 cm. Graph Area (A) Versus Radius (r) r A 1 2006.928 2 1027.712 3 729.0187 4 610.848 5 573.2 6 582.7413 7 625.1863 8 693.392 9 783.3902 10 892.8 11 1020.106 From the graph, the minimum area occurs when the value of r is between 5 cm and 6 cm. By plotting a graph using values of r from 5.18 to 5.3, a more precise value of r can be determined. Graph Area (A) Versus Radius (r) r A 5.18 571.9953 5.19 571.9698 5.2 571.9485 5.21 571.9315 5.22 571.9187 5.23 571.9101 5.24 571.9056 5.25 571.9054 5.26 571.9093 5.27 571.9173 5.28 571.9294 5.29 571.9457 5.3 571.966 From the graph,the area is minimum when r#8776; 5.25. h =1000 πr² =11.57 h =11.57 r 5.25 =2.204 h Minimum area of aluminium sheet used when r =5.25,h=11.57 and r =2.204. Method 3 : Construct a table A table can be constructed by using equation A = 4#8730;3r²+2000r¯¹ and values of r between 1 and 8. r h A h/r 1 318.3 099 2006.928 & nbsp; 318.3099 2 79.57747 &nbs p; 1027.712 39.78874 3 35.36777 &nbs p; 729.0187 11.78926 4 19.89437 &nbs p; 610.848 4.973592 5 12.7324 ; 573.2 2.546479 6 8.841941 &nbs p; 582.7413 1.473657 7 6.49612 ; 625.1863 0.928017 8 4.973592 &nbs p; 693.392 0.621699 Value of r is around 5cm when the area is minimum.Another two tables can be constructed to obtain a more precise value of r. r h A h/r 4.8 13.81553 576.2878 2.878236 4.9 13.25739& nbsp; 574.5045 ; 2.705589 5 12.7324 ; 573.2 2.546479 5.1 12.23798& nbsp; 572.3541 ; 2.399604 5.2 11.77182& nbsp; 571.9485 ; 2.263811 5.3 11.33179& nbsp; 571.966 2.138073 5.4 10.91598& nbsp; 572.3909 ; 2.021477 5.5 10.52264& nbsp; 573.2084 ; 1.913207 r ; h &nbs p; A &nb sp; h/r 5.2 11.77182 &nbs p; 571.9485 2.263811 5.21 11.72667 571.9315 2.2508 5.22 11.68178 571.9187 2.237889 5.23 11.63715 571.9101 2.225077 5.24 11.59278 571.9056 2.212362 5.25 11.54866 571.9054 2.199744 5.26 11.50479 571.9093 2.187222 5.27 11.46117 571.9173 2.174795 5.28 11.4178 & nbsp; 571.9294 2.162461 Minimum area of aluminium sheet used when r= 5.25,h=11.55 and h =2.2. r Additional Task Considering a regular polygon with n sides . Interior angle,z = 180(n-2) N Tan z = r 2 x x = r tan z ……………………..(1) ² Area of n –sided regular polygon =nxr =nr( r ) tan z ................................... (2) ² Tota; Area of Aluminium Sheets Used, A =2nr( r )+2#12640;rh Tan z ² =2nr² +2πr²(1000) , h=1000 tan z πr² πr² ² = 2nr² + 2000 tan z r ² dA = 4nr - 2000 dr tan z r² ² ; When A is minimum ,dA = 0 Dr 4nr = 2000 tan z r² ² ; z r³ =500 tan 2 n r = 3#8730; 500tan n By varying the values of n ,the corresponding values of z, r, h and h can be found from the following table n Z=((n-2)*180)/n &n bsp; z/n Tanz/2 ; R=((500tan z/2)/n) ^(1/3) h=1000/((22/7)*(r ^2)) h/r 3 60.0 & nbsp;30.0 0.5776 &nb sp; 4.58317552 15.14 754949 3.305034 4 90.0 & nbsp;45.0 1.0006 &nb sp; 5.001053852 12.7 2190936 2.543846 5 108.0 54.0 1.3775 &n bsp; 5.164502417 11. 92939418 2.309883 6 120.0 60.0 1.7337 &n bsp; 5.24727803 11.5 5599132 2.202283 7 128.6 64.3 2.0789 &n bsp; 5.295456734 11. 34667227 2.142718 8 135.0 67.5 2.4175 &n bsp; 5.326142937 11. 21630278 2.105896 9 140.0 70.0 2.7517 &n bsp; 5.34698339 11.1 2903964 2.081368 10 144.0 ; 72.0 3.0830 & nbsp; 5.361841508 11 .06744607 2.044113 11 147.3 ; 73.6 3.4122 & nbsp; 5.372848674 11 .02214555 2.051453 12 150.0 ; 75.0 3.7399 & nbsp; 5.381261993 10 .9877074 2.041846 13 152.3 ; 76.2 4.0665 & nbsp; 5.387863064 10 .96080018 2.03435 14 154.3 ; 77.1 4.3923 & nbsp; 5.393159096 10 .93928395 2.028363 15 156.0 ; 78.0 4.7173 & nbsp; 5.397491134 10 .92173121 2.023483 16 157.5 ; 78.8 5.0419 & nbsp; 5.401095477 10 .90715915 2.019435 17 158.8 ; 79.4 5.3661 & nbsp; 5.404140104 10 .89487269 2.016023 18 160.0 ; 80.0 5.6900 & nbsp; 5.406747113 10 .88436871 2.013109 19 161.1 ; 80.5 6.0136 & nbsp; 5.409006999 10 .87527562 2.010586 20 162.0 ; 81.0 6.3371 & nbsp; 5.410987996 10 .86731406 2.008379 21 162.9 ; 81.4 6.6604 & nbsp; 5.412742349 10 .86027068 2.006427 22 163.6 ; 81.8 6.9836 & nbsp; 5.414310607 10 .85398022 2.004684 23 164.3 ; 82.2 7.3068 & nbsp; 5.41572464 10. 84831306 2.003114 24 165 & nbsp;82.5 7.6299 &nb sp; 5.417009773 10.8 4316636 2.001689 Conclusion The above table shows that as the number of sides of a regular polygon increases,the ratio of h/r approaches 2. 1. Material wastage is reduced if the number of the sides of the aluminium sheet increases (regular polygon shapes) 2. Geometrical shapes that can be fitted into each other without any space in between are a) equilateral triangles b) squares c) regular hexagons 3. Therefore,circles must be cut from the geometrical shapes above to reduce wastage. Among these geometrical shapes,cutting circles from regular hexagons will result in the least wastage of the aluminium sheets.

aluminium can: Solution Method 2: Trial and improvement (a) (by substituting the value of r ) V= πr²h πr²h= 1000 h= 1000 πr² ………………(1) A= 2πr² +2πrh = 2πr²+ 2πr( 1000) πr² = 2πr³+2000 r ……………...(2) If r = 1, A= 2π+2000 1 = 2006.2832 r = 2, A= 2π(2)³+2000 2 = 1025.1327 r = 3, A= 2π(3)³+2000 3 = 723.2153 r = 4, A= 2π(4)³+2000 4 = 600.5310 r = 5, A= 2π(5)³+2000 5 = 557.0796 r = 5, A= 2π(5)³+2000 5 = 557.0796 r = 6, A= 2π(6)³+2000 6 = 559.5280 r = 7, A= 2π(7)³+2000 7 = 593.5904 r = 8, A= 2π(8)³+2000 8 = 652.1239 r = 9, A= 2π(9)³+2000 9 = 731.1602 The value of r that falls between 4 and 6 will produce the minimum surface area of the aluminium sheet used. r A 4.5 571.6789 4.6 567.7348 4.7 564.3275 4.8 561.4313 4.9 559.0225 5 557.0796 5.1 555.5825 5.2 554.5127 5.3 553.8532 5.4 553.5881 5.5 553.7027 5.6 554.1835 5.7 555.0179 5.8 556.1939 The lowest surface area is obtained when r=5.4 By substituting r=5.4 into the equation (1) h = 1000 π(5.4)² h = 10.92 h = 10.92 r 5.4 = 2.022,therefore approximately h=2r Surfeca area is minimum,when r=5.4,h=10.92 and h = 2.022 r (b) Alternative method (by substituting the value of h) V= πr²h πr²h = 1000 r² = 1000 πh ………………(1) A = 2πr² + 2πrh = 2π(1000) + 2πh (1000) πh πh = 2000+ 20h#8730; 10πh …………(2) h if h = 1,A = 2000+20 #8730; 10π = 2112.0998 h = 2,A = 2000+40 #8730; 20π 2 = 1158.5331 h = 3,A = 2000+60 #8730; 30π 3 = 860.8293 h = 4,A = 2000+80 #8730; 40π 4 = 724.1996 h = 5,A = 2000+100 #8730; 50π 5 = 650.6628 h = 6,A = 2000+120 #8730; 60π 6 = 607.9207 h = 7,A = 2000+140 #8730; 70π 7 = 582.3025 h = 8,A = 2000+160 #8730; 80π 8 = 567.0662 h = 9,A = 2000+180 #8730; 90π 9 = 558.5217 h = 10,A = 2000+200 #8730; 100π 10 = 554.4908 h = 11, A = 2000+220 #8730; 110π 11 = 553.6112 h = 12, A = 2000+240 #8730; 120π 12 = 554.9918 h = 13, A = 2000+260 #8730; 130π 13 = 558.0278 h = 14, A = 2000+280 #8730; 140π 14 = 562.2963 The value of h that lies between 10 and 12 will produce the least surface area r A 10.1 554.2786 10.2 554.0966 10.3 553.9436 10.4 553.8188 10.5 553.7211 10.6 553.6498 10.7 553.604 10.8 553.5828 10.9 553.5855 11 553.6112 11.1 553.6594 11.2 553.7292 11.3 553.82 11.4 553.9312 Minimum surface area occurs when h= 10.8 By substituting h=10.8 into the equation (1) r² = 1000 ……………(1) πh r = #8730;1000 πh = #8730;1000 10.8π = 5.429 h = 10.8 r 5.429 = 1.989 = 2 Surface area is minimum,when h=10.8, r=5.429 and h=2 r Method 2: Calculus πr² + 2πrh …………… (1)#61472; A = 2 πr²h = 1000 h = 1000 π r² &nbs p; ………………….(2) Substituting equation (2) into (1): A = 2πr² + 2πr (1000) r² #61552; A=2πr² + 2000 r A’(r) = 4πr-2000 r² When A’(r) = 0.4πr = 2000 r² r³ = 500 π r = 500 π r = 5.419 h= 1000 π x 5.419² = 10.84 h =10.84 r 5.419 ~2 A”(r) =4π + 4000 r³ A”(5.419)>0 (because r>0), therefore A is aluminium. The sirface area is minimum, when r = 5.419, h=10.84 and h =2 r Method 3: Graphical Method By plotting the surface area A, against radius, r, for which A=2πr²+200 , the following graph is plotted. r r A 1 2006.283 2 1025.133 3 723.2153 4 600.531 5 557.0796 6 559.528 7 593.5904 8 652.1239 9 731.1602 Graph Area (A) Versus Radius (r) r A 5.1 555.5825 5.2 554.5137 5.3 553.8532 5.4 553.5881 5.5 553.7027 5.6 554.1835 5.7 555.0179 5.8 556.1939 5.9 557.7007 Graph Area (A) Versus Radius (r) r A 5.38 553.6102 5.39 553.5972 5.4 553.5881 5.41 553.5827 5.42 553.5811 5.43 553.5832 5.44 553.5891 5.45 553.5988 5.46 553.6122 5.47 553.6293 5.48 553.6501 From the graph, the value of r = 5.42 h = 1000 πr² = 1000 π(5.42)² = 10.84 h = 10.84 r 5.42 = 2 Surface area is minimum, when r = 5.4, h = 10.84 and h = 2 r To verify the answer , the following method can be used . Method: Construct a Table r A &nbs p; 5 557.0796 &nb sp; The table shows that the minimum surface area occurs 5.05 556.2765 when the value of r falls between 5.4 and 5.5. Therefore 5.1 555.5825 & nbsp; the minimum surface area lines between 553.5881 and 5.15 554.9953 553.7027. 5.2 554.5127 & nbsp; 5.25 554.1327 5.3 553.8532 & nbsp; 5.35 553.6722 5.4 553.5881 & nbsp; 5.45 553.5986 5.5 553.7027 & nbsp; 5.55 553.8982 5.6 554.1835 & nbsp; 5.65 554.5573 r A &nbs p; 5.41 553.5827 To find the more precise value of r and A, the values 5.42 553.5811 of r from 5.41 to 5.54 are used. 5.43 553.5832 5.44 553.5891 From the table, the least value for surface area is 5.45 553.5988 553.5811 and it occurs when r = 5.42. 5.46 553.6122 5.47 553.6293 5.48 553.6501 5.49 553.6746 5.5 553.7027 & nbsp; 5.51 553.7345 5.52 553.77 &n bsp; 5.53 553.8091 5.54 553.8518 r A &nbs p; 5.411 553.5823 ; To find the more precise of value r and A, the values of r 5.412 553.582 From 5.411 to 5.424 are used. 5.413 553.5818 ; 5.414 553.5816 ; From the table, the least value for surface area is 553.581 5.415 553.5814 ; and it occurs when r = 5.419 5.416 553.5812 ; 5.417 553.5811 ; 5.418 553.5811 ; 5.419 553.581 5.42 553.5811 5.421 553.5811 ; 5.422 553.5812 ; 5.423 553.5813 ; 5.424 553.5815 ; Solution for Question 2 Method 1:Calculus Area of aluminium sheet used, A = 2(2r 2r) + 2πrh……….(1) πr2h = 1000………………..(2) A = 8r2 +2πr( ) =8r2 +( ) = 8r2 +2000r -1 16r -2000r -2 When A is minimum, 0. 16r-2000r -2 = 0 16r = r3 = =125 r = 5 h = = =12.73 = =2.546 Minimum area of aluminium sheet used, when r = 5, h = 12.73 and =2.546. Method 2 : Draw a Graph By plotting the surface area A, against radius, r, for which A=8r2 + 2000r -1 , the following graph is produced. r A 1 2008 2 1032 3 738.6667 4 628 5 600 6 621.3333 7 677.7143 8 762 9 870.2222 Graph Area (A) Versus Radius (r) r A 4.7 602.2519 4.8 600.9867 4.9 600.2433 5 600 5.1 600.2369 5.2 600.9354 5.3 602.0785 5.4 603.6504 5.5 605.6364 From the graph, value of r = 5. h = = =12.73 = =2.546 Minimum area of aluminium sheet used, when r = 5, h = 12.73 and =2.546. To verify the answer, the following method can be used. Method: Construct a table r A &nbs p;Table shows that the minimum 1 2008 & nbsp;surface area occurs when the 1.2 1678.187 & nbsp; value of r falls between 4.8 and 1.4 1444.251 & nbsp; 5.2. 1.6 1270.48 &n bsp; 1.8 1137.031 & nbsp; 2 1032 & nbsp; 2.1 947.8109 & nbsp; 2.4 879.4133 & nbsp; 2.6 823.3108 & nbsp; 2.8 777.0057 & nbsp; 3 738.6667 &nb sp; 3.2 706.92 &nb sp; 3.4 680.7153 & nbsp; 3.6 659.2356 & nbsp; 3.8 641.8358 & nbsp; 4 628 &n bsp; 4.2 617.3105 & nbsp; 4.4 609.4255 & nbsp; 4.6 604.0626 & nbsp; 4.8 600.9867 & nbsp; 5 600 &n bsp; 5.2 600.9354 & nbsp; 5.4 603.6504 & nbsp; 5.6 608.0229 & nbsp; r A 4.9 600.2433 4.91 600.1968 4.92 600.1553 4.93 600.1187 4.94 600.0871 4.95 600.0604 4.96 600.0386 4.97 600.0271 4.98 600.0096 4.99 600.0024 5 600 5.01 600.0024 5.02 600.0096 5.03 600.0215 To get a more precise value of A and r, a smaller range of r from 4.8 to 5.2 s used. From the table, the minimum area is 600 and this occurs when r = 5. h = = = 12.73 = =2.546 Solution for Question 3 Consider The Following Shapes (1) Equilateral Triangle Method 1:Trigonometry and Calculus tan 60o = x = r Area of PQR = 3 Area of OPQ = 3 2X r = 3 xr = 3 r2 Total area of Aluminium Sheets Used, A = 6 r2 + 2πr( ) = 6 r2 +2000r -1 12 r – 2000r -2 When A is minimum, 0. 12 r = r3 = 96.225 r = 4.582 h = =15.16 =3.309 12 + 4000r -3 (>0, because r>0) Therefore , A is minimum. Minimum area or aluminium sheet used when r = 4.582, h= 15.16 and = 3.309. Method 2: Graphical Method By using the equation A=6#8730;3r²+2000-1, graph A versus r can be plotted to determine the value of r for the minimum value of A. Graph Area (A) Versus Radius (r) r A 1 2010.392 2 1041.568 3 760.1947 4 666.272 5 659.8 6 707.4453 7 794.9223 8 915.088 9 1063.974 From the graph, minimum area occurs when r falls between 4 cm and 5 cm. The following graph is plotted using the values of r between 4.54 and 4.64 Graph Area (A) Versus Radius (r) r A 4.54 654.7244 4.55 654.7008 4.56 654.6836 4.57 654.6726 4.58 654.668 4.59 654.6695 4.6 654.6773 4.61 654.6913 4.62 654.7114 4.63 654.7377 4.64 654.7701 From the graph, area is minimum when r #8776; 4.58 cm. h = 1000 Πr² = 15.17 h = 15.17 r 4.58 = 3.312 Minimun area of aluminium sheet used when r = 4.58, h = 15.17 and h = 3.312 Method 3: Construct a Table A table can be constructed using A = 6 3 r2 + 2000r-1 r h &nbs p;A h/r 1 318.3099 &nb sp; 2010.392 318.3099 2 79.57747 1041.568 &n bsp; 39.78874 3 35.36777 &nb sp; 760.1947 11.78874 4 19.89437 &nb sp; 666.272 11.78926 5 12.7324 &nbs p; 659.8 4.973592 6 8.841941 &nb sp; 707.4453 2.546479 7 6.49612 &nbs p; 794.9223 0.928017 8 4.973592 &nb sp; 915.088 0.621699 By using a more precise range of value r, we get the following tables. r h &nbs p;A h/r 4.1 18.93575 & nbsp; 662.4944 4.618475 4.2 18.04478 & nbsp; 659.5054 4.296376 4.3 17.21525 & nbsp; 657.2644 4.003545 4.4 16.44163 & nbsp; 655.7346 3.736733 4.5 15.71901 & nbsp; 654.8824 3.493113 4.6 15.043 &nb sp; 654.6773 3.270217 4.7 14.40968 & nbsp; 655.0912 3.06589 4.8 13.81553 & nbsp; 656.0983 2.878236 r h &nbs p;A h/r 4.51 15.64938 654.8333 3.469928 4.52 15.58021 654.7906 3.446949 4.53 15.5115 & nbsp; 654.7543 3.424172 4.54 15.44324 654.7244 3.401595 4.55 15.37543 654.7008 3.379216 4.56 15.30807 654.6836 3.357033 4.57 15.24115 654.6726 3.335044 4.58 15.17467 654.668 3.313246 4.59 15.10862 654.6695 3.291638 4.6 15.043 &nb sp; 654.6773 3.270217 4.61 14.97781 654.6913 3.248982 From the above table, minimum area of aluminium sheet used when r = 4.58, h = 15.17 and the ratio of h = 3.312 ( ii ) Regular Hexagon Method 1: Trionometry and Calculus Tan 300 = x r x = r tan 300 = r 3 Area of OPQ & nbsp;= 1 x 2x x r 2 = 1 x 2( r ) x r 2 = r2 3 = 3 r2 3 Area of Hexagon = 6 x 3r2 3 = 2 #8730; 3 r2 Total Area of Aluminium Sheets Used, A = 2(2 #8730; 3 r2) + 2 π r ( 1000 ) = 4 #8730; 3 r2 + 2000r-1 dA = 8 #8730; 3 r – 2000 r-2 dr For minimum surface area, dA = 0 8 #8730;3 r = 2000 r2 r3 = 2000 8 #8730;3 = 250 #8730; 3 r = 5.246 h = 1000 πr² = 1000 π x ( 5.246 ) = 11.57 h = 11.57 r 5.246 = 2.205 Method 2: Graphical Method By using the equation A=4#8730;3r²+2000-1, graph A versus r can be plotted using the value of r from 1 cm to 11 cm. Graph Area (A) Versus Radius (r) r A 1 2006.928 2 1027.712 3 729.0187 4 610.848 5 573.2 6 582.7413 7 625.1863 8 693.392 9 783.3902 10 892.8 11 1020.106 From the graph, the minimum area occurs when the value of r is between 5 cm and 6 cm. By plotting a graph using values of r from 5.18 to 5.3, a more precise value of r can be determined. Graph Area (A) Versus Radius (r) r A 5.18 571.9953 5.19 571.9698 5.2 571.9485 5.21 571.9315 5.22 571.9187 5.23 571.9101 5.24 571.9056 5.25 571.9054 5.26 571.9093 5.27 571.9173 5.28 571.9294 5.29 571.9457 5.3 571.966 From the graph,the area is minimum when r#8776; 5.25. h =1000 πr² =11.57 h =11.57 r 5.25 =2.204 h Minimum area of aluminium sheet used when r =5.25,h=11.57 and r =2.204. Method 3 : Construct a table A table can be constructed by using equation A = 4#8730;3r²+2000r¯¹ and values of r between 1 and 8. r h A h/r 1 318.3 099 2006.928 & nbsp; 318.3099 2 79.57747 &nbs p; 1027.712 39.78874 3 35.36777 &nbs p; 729.0187 11.78926 4 19.89437 &nbs p; 610.848 4.973592 5 12.7324 ; 573.2 2.546479 6 8.841941 &nbs p; 582.7413 1.473657 7 6.49612 ; 625.1863 0.928017 8 4.973592 &nbs p; 693.392 0.621699 Value of r is around 5cm when the area is minimum.Another two tables can be constructed to obtain a more precise value of r. r h A h/r 4.8 13.81553 576.2878 2.878236 4.9 13.25739& nbsp; 574.5045 ; 2.705589 5 12.7324 ; 573.2 2.546479 5.1 12.23798& nbsp; 572.3541 ; 2.399604 5.2 11.77182& nbsp; 571.9485 ; 2.263811 5.3 11.33179& nbsp; 571.966 2.138073 5.4 10.91598& nbsp; 572.3909 ; 2.021477 5.5 10.52264& nbsp; 573.2084 ; 1.913207 r ; h &nbs p; A &nb sp; h/r 5.2 11.77182 &nbs p; 571.9485 2.263811 5.21 11.72667 571.9315 2.2508 5.22 11.68178 571.9187 2.237889 5.23 11.63715 571.9101 2.225077 5.24 11.59278 571.9056 2.212362 5.25 11.54866 571.9054 2.199744 5.26 11.50479 571.9093 2.187222 5.27 11.46117 571.9173 2.174795 5.28 11.4178 & nbsp; 571.9294 2.162461 Minimum area of aluminium sheet used when r= 5.25,h=11.55 and h =2.2. r Additional Task Considering a regular polygon with n sides . Interior angle,z = 180(n-2) N Tan z = r 2 x x = r tan z ……………………..(1) ² Area of n –sided regular polygon =nxr =nr( r ) tan z ................................... (2) ² Tota; Area of Aluminium Sheets Used, A =2nr( r )+2#12640;rh Tan z ² =2nr² +2πr²(1000) , h=1000 tan z πr² πr² ² = 2nr² + 2000 tan z r ² dA = 4nr - 2000 dr tan z r² ² ; When A is minimum ,dA = 0 Dr 4nr = 2000 tan z r² ² ; z r³ =500 tan 2 n r = 3#8730; 500tan n By varying the values of n ,the corresponding values of z, r, h and h can be found from the following table n Z=((n-2)*180)/n &n bsp; z/n Tanz/2 ; R=((500tan z/2)/n) ^(1/3) h=1000/((22/7)*(r ^2)) h/r 3 60.0 & nbsp;30.0 0.5776 &nb sp; 4.58317552 15.14 754949 3.305034 4 90.0 & nbsp;45.0 1.0006 &nb sp; 5.001053852 12.7 2190936 2.543846 5 108.0 54.0 1.3775 &n bsp; 5.164502417 11. 92939418 2.309883 6 120.0 60.0 1.7337 &n bsp; 5.24727803 11.5 5599132 2.202283 7 128.6 64.3 2.0789 &n bsp; 5.295456734 11. 34667227 2.142718 8 135.0 67.5 2.4175 &n bsp; 5.326142937 11. 21630278 2.105896 9 140.0 70.0 2.7517 &n bsp; 5.34698339 11.1 2903964 2.081368 10 144.0 ; 72.0 3.0830 & nbsp; 5.361841508 11 .06744607 2.044113 11 147.3 ; 73.6 3.4122 & nbsp; 5.372848674 11 .02214555 2.051453 12 150.0 ; 75.0 3.7399 & nbsp; 5.381261993 10 .9877074 2.041846 13 152.3 ; 76.2 4.0665 & nbsp; 5.387863064 10 .96080018 2.03435 14 154.3 ; 77.1 4.3923 & nbsp; 5.393159096 10 .93928395 2.028363 15 156.0 ; 78.0 4.7173 & nbsp; 5.397491134 10 .92173121 2.023483 16 157.5 ; 78.8 5.0419 & nbsp; 5.401095477 10 .90715915 2.019435 17 158.8 ; 79.4 5.3661 & nbsp; 5.404140104 10 .89487269 2.016023 18 160.0 ; 80.0 5.6900 & nbsp; 5.406747113 10 .88436871 2.013109 19 161.1 ; 80.5 6.0136 & nbsp; 5.409006999 10 .87527562 2.010586 20 162.0 ; 81.0 6.3371 & nbsp; 5.410987996 10 .86731406 2.008379 21 162.9 ; 81.4 6.6604 & nbsp; 5.412742349 10 .86027068 2.006427 22 163.6 ; 81.8 6.9836 & nbsp; 5.414310607 10 .85398022 2.004684 23 164.3 ; 82.2 7.3068 & nbsp; 5.41572464 10. 84831306 2.003114 24 165 & nbsp;82.5 7.6299 &nb sp; 5.417009773 10.8 4316636 2.001689 Conclusion The above table shows that as the number of sides of a regular polygon increases,the ratio of h/r approaches 2. 1. Material wastage is reduced if the number of the sides of the aluminium sheet increases (regular polygon shapes) 2. Geometrical shapes that can be fitted into each other without any space in between are a) equilateral triangles b) squares c) regular hexagons 3. Therefore,circles must be cut from the geometrical shapes above to reduce wastage. Among these geometrical shapes,cutting circles from regular hexagons will result in the least wastage of the aluminium sheets.

aluminium can: Solution Method 2: Trial and improvement (a) (by substituting the value of r ) V= πr²h πr²h= 1000 h= 1000 πr² ………………(1) A= 2πr² +2πrh = 2πr²+ 2πr( 1000) πr² = 2πr³+2000 r ……………...(2) If r = 1, A= 2π+2000 1 = 2006.2832 r = 2, A= 2π(2)³+2000 2 = 1025.1327 r = 3, A= 2π(3)³+2000 3 = 723.2153 r = 4, A= 2π(4)³+2000 4 = 600.5310 r = 5, A= 2π(5)³+2000 5 = 557.0796 r = 5, A= 2π(5)³+2000 5 = 557.0796 r = 6, A= 2π(6)³+2000 6 = 559.5280 r = 7, A= 2π(7)³+2000 7 = 593.5904 r = 8, A= 2π(8)³+2000 8 = 652.1239 r = 9, A= 2π(9)³+2000 9 = 731.1602 The value of r that falls between 4 and 6 will produce the minimum surface area of the aluminium sheet used. r A 4.5 571.6789 4.6 567.7348 4.7 564.3275 4.8 561.4313 4.9 559.0225 5 557.0796 5.1 555.5825 5.2 554.5127 5.3 553.8532 5.4 553.5881 5.5 553.7027 5.6 554.1835 5.7 555.0179 5.8 556.1939 The lowest surface area is obtained when r=5.4 By substituting r=5.4 into the equation (1) h = 1000 π(5.4)² h = 10.92 h = 10.92 r 5.4 = 2.022,therefore approximately h=2r Surfeca area is minimum,when r=5.4,h=10.92 and h = 2.022 r (b) Alternative method (by substituting the value of h) V= πr²h πr²h = 1000 r² = 1000 πh ………………(1) A = 2πr² + 2πrh = 2π(1000) + 2πh (1000) πh πh = 2000+ 20h#8730; 10πh …………(2) h if h = 1,A = 2000+20 #8730; 10π = 2112.0998 h = 2,A = 2000+40 #8730; 20π 2 = 1158.5331 h = 3,A = 2000+60 #8730; 30π 3 = 860.8293 h = 4,A = 2000+80 #8730; 40π 4 = 724.1996 h = 5,A = 2000+100 #8730; 50π 5 = 650.6628 h = 6,A = 2000+120 #8730; 60π 6 = 607.9207 h = 7,A = 2000+140 #8730; 70π 7 = 582.3025 h = 8,A = 2000+160 #8730; 80π 8 = 567.0662 h = 9,A = 2000+180 #8730; 90π 9 = 558.5217 h = 10,A = 2000+200 #8730; 100π 10 = 554.4908 h = 11, A = 2000+220 #8730; 110π 11 = 553.6112 h = 12, A = 2000+240 #8730; 120π 12 = 554.9918 h = 13, A = 2000+260 #8730; 130π 13 = 558.0278 h = 14, A = 2000+280 #8730; 140π 14 = 562.2963 The value of h that lies between 10 and 12 will produce the least surface area r A 10.1 554.2786 10.2 554.0966 10.3 553.9436 10.4 553.8188 10.5 553.7211 10.6 553.6498 10.7 553.604 10.8 553.5828 10.9 553.5855 11 553.6112 11.1 553.6594 11.2 553.7292 11.3 553.82 11.4 553.9312 Minimum surface area occurs when h= 10.8 By substituting h=10.8 into the equation (1) r² = 1000 ……………(1) πh r = #8730;1000 πh = #8730;1000 10.8π = 5.429 h = 10.8 r 5.429 = 1.989 = 2 Surface area is minimum,when h=10.8, r=5.429 and h=2 r Method 2: Calculus πr² + 2πrh …………… (1)#61472; A = 2 πr²h = 1000 h = 1000 π r² &nbs p; ………………….(2) Substituting equation (2) into (1): A = 2πr² + 2πr (1000) r² #61552; A=2πr² + 2000 r A’(r) = 4πr-2000 r² When A’(r) = 0.4πr = 2000 r² r³ = 500 π r = 500 π r = 5.419 h= 1000 π x 5.419² = 10.84 h =10.84 r 5.419 ~2 A”(r) =4π + 4000 r³ A”(5.419)>0 (because r>0), therefore A is aluminium. The sirface area is minimum, when r = 5.419, h=10.84 and h =2 r Method 3: Graphical Method By plotting the surface area A, against radius, r, for which A=2πr²+200 , the following graph is plotted. r r A 1 2006.283 2 1025.133 3 723.2153 4 600.531 5 557.0796 6 559.528 7 593.5904 8 652.1239 9 731.1602 Graph Area (A) Versus Radius (r) r A 5.1 555.5825 5.2 554.5137 5.3 553.8532 5.4 553.5881 5.5 553.7027 5.6 554.1835 5.7 555.0179 5.8 556.1939 5.9 557.7007 Graph Area (A) Versus Radius (r) r A 5.38 553.6102 5.39 553.5972 5.4 553.5881 5.41 553.5827 5.42 553.5811 5.43 553.5832 5.44 553.5891 5.45 553.5988 5.46 553.6122 5.47 553.6293 5.48 553.6501 From the graph, the value of r = 5.42 h = 1000 πr² = 1000 π(5.42)² = 10.84 h = 10.84 r 5.42 = 2 Surface area is minimum, when r = 5.4, h = 10.84 and h = 2 r To verify the answer , the following method can be used . Method: Construct a Table r A &nbs p; 5 557.0796 &nb sp; The table shows that the minimum surface area occurs 5.05 556.2765 when the value of r falls between 5.4 and 5.5. Therefore 5.1 555.5825 & nbsp; the minimum surface area lines between 553.5881 and 5.15 554.9953 553.7027. 5.2 554.5127 & nbsp; 5.25 554.1327 5.3 553.8532 & nbsp; 5.35 553.6722 5.4 553.5881 & nbsp; 5.45 553.5986 5.5 553.7027 & nbsp; 5.55 553.8982 5.6 554.1835 & nbsp; 5.65 554.5573 r A &nbs p; 5.41 553.5827 To find the more precise value of r and A, the values 5.42 553.5811 of r from 5.41 to 5.54 are used. 5.43 553.5832 5.44 553.5891 From the table, the least value for surface area is 5.45 553.5988 553.5811 and it occurs when r = 5.42. 5.46 553.6122 5.47 553.6293 5.48 553.6501 5.49 553.6746 5.5 553.7027 & nbsp; 5.51 553.7345 5.52 553.77 &n bsp; 5.53 553.8091 5.54 553.8518 r A &nbs p; 5.411 553.5823 ; To find the more precise of value r and A, the values of r 5.412 553.582 From 5.411 to 5.424 are used. 5.413 553.5818 ; 5.414 553.5816 ; From the table, the least value for surface area is 553.581 5.415 553.5814 ; and it occurs when r = 5.419 5.416 553.5812 ; 5.417 553.5811 ; 5.418 553.5811 ; 5.419 553.581 5.42 553.5811 5.421 553.5811 ; 5.422 553.5812 ; 5.423 553.5813 ; 5.424 553.5815 ; Solution for Question 2 Method 1:Calculus Area of aluminium sheet used, A = 2(2r 2r) + 2πrh……….(1) πr2h = 1000………………..(2) A = 8r2 +2πr( ) =8r2 +( ) = 8r2 +2000r -1 16r -2000r -2 When A is minimum, 0. 16r-2000r -2 = 0 16r = r3 = =125 r = 5 h = = =12.73 = =2.546 Minimum area of aluminium sheet used, when r = 5, h = 12.73 and =2.546. Method 2 : Draw a Graph By plotting the surface area A, against radius, r, for which A=8r2 + 2000r -1 , the following graph is produced. r A 1 2008 2 1032 3 738.6667 4 628 5 600 6 621.3333 7 677.7143 8 762 9 870.2222 Graph Area (A) Versus Radius (r) r A 4.7 602.2519 4.8 600.9867 4.9 600.2433 5 600 5.1 600.2369 5.2 600.9354 5.3 602.0785 5.4 603.6504 5.5 605.6364 From the graph, value of r = 5. h = = =12.73 = =2.546 Minimum area of aluminium sheet used, when r = 5, h = 12.73 and =2.546. To verify the answer, the following method can be used. Method: Construct a table r A &nbs p;Table shows that the minimum 1 2008 & nbsp;surface area occurs when the 1.2 1678.187 & nbsp; value of r falls between 4.8 and 1.4 1444.251 & nbsp; 5.2. 1.6 1270.48 &n bsp; 1.8 1137.031 & nbsp; 2 1032 & nbsp; 2.1 947.8109 & nbsp; 2.4 879.4133 & nbsp; 2.6 823.3108 & nbsp; 2.8 777.0057 & nbsp; 3 738.6667 &nb sp; 3.2 706.92 &nb sp; 3.4 680.7153 & nbsp; 3.6 659.2356 & nbsp; 3.8 641.8358 & nbsp; 4 628 &n bsp; 4.2 617.3105 & nbsp; 4.4 609.4255 & nbsp; 4.6 604.0626 & nbsp; 4.8 600.9867 & nbsp; 5 600 &n bsp; 5.2 600.9354 & nbsp; 5.4 603.6504 & nbsp; 5.6 608.0229 & nbsp; r A 4.9 600.2433 4.91 600.1968 4.92 600.1553 4.93 600.1187 4.94 600.0871 4.95 600.0604 4.96 600.0386 4.97 600.0271 4.98 600.0096 4.99 600.0024 5 600 5.01 600.0024 5.02 600.0096 5.03 600.0215 To get a more precise value of A and r, a smaller range of r from 4.8 to 5.2 s used. From the table, the minimum area is 600 and this occurs when r = 5. h = = = 12.73 = =2.546 Solution for Question 3 Consider The Following Shapes (1) Equilateral Triangle Method 1:Trigonometry and Calculus tan 60o = x = r Area of PQR = 3 Area of OPQ = 3 2X r = 3 xr = 3 r2 Total area of Aluminium Sheets Used, A = 6 r2 + 2πr( ) = 6 r2 +2000r -1 12 r – 2000r -2 When A is minimum, 0. 12 r = r3 = 96.225 r = 4.582 h = =15.16 =3.309 12 + 4000r -3 (>0, because r>0) Therefore , A is minimum. Minimum area or aluminium sheet used when r = 4.582, h= 15.16 and = 3.309. Method 2: Graphical Method By using the equation A=6#8730;3r²+2000-1, graph A versus r can be plotted to determine the value of r for the minimum value of A. Graph Area (A) Versus Radius (r) r A 1 2010.392 2 1041.568 3 760.1947 4 666.272 5 659.8 6 707.4453 7 794.9223 8 915.088 9 1063.974 From the graph, minimum area occurs when r falls between 4 cm and 5 cm. The following graph is plotted using the values of r between 4.54 and 4.64 Graph Area (A) Versus Radius (r) r A 4.54 654.7244 4.55 654.7008 4.56 654.6836 4.57 654.6726 4.58 654.668 4.59 654.6695 4.6 654.6773 4.61 654.6913 4.62 654.7114 4.63 654.7377 4.64 654.7701 From the graph, area is minimum when r #8776; 4.58 cm. h = 1000 Πr² = 15.17 h = 15.17 r 4.58 = 3.312 Minimun area of aluminium sheet used when r = 4.58, h = 15.17 and h = 3.312 Method 3: Construct a Table A table can be constructed using A = 6 3 r2 + 2000r-1 r h &nbs p;A h/r 1 318.3099 &nb sp; 2010.392 318.3099 2 79.57747 1041.568 &n bsp; 39.78874 3 35.36777 &nb sp; 760.1947 11.78874 4 19.89437 &nb sp; 666.272 11.78926 5 12.7324 &nbs p; 659.8 4.973592 6 8.841941 &nb sp; 707.4453 2.546479 7 6.49612 &nbs p; 794.9223 0.928017 8 4.973592 &nb sp; 915.088 0.621699 By using a more precise range of value r, we get the following tables. r h &nbs p;A h/r 4.1 18.93575 & nbsp; 662.4944 4.618475 4.2 18.04478 & nbsp; 659.5054 4.296376 4.3 17.21525 & nbsp; 657.2644 4.003545 4.4 16.44163 & nbsp; 655.7346 3.736733 4.5 15.71901 & nbsp; 654.8824 3.493113 4.6 15.043 &nb sp; 654.6773 3.270217 4.7 14.40968 & nbsp; 655.0912 3.06589 4.8 13.81553 & nbsp; 656.0983 2.878236 r h &nbs p;A h/r 4.51 15.64938 654.8333 3.469928 4.52 15.58021 654.7906 3.446949 4.53 15.5115 & nbsp; 654.7543 3.424172 4.54 15.44324 654.7244 3.401595 4.55 15.37543 654.7008 3.379216 4.56 15.30807 654.6836 3.357033 4.57 15.24115 654.6726 3.335044 4.58 15.17467 654.668 3.313246 4.59 15.10862 654.6695 3.291638 4.6 15.043 &nb sp; 654.6773 3.270217 4.61 14.97781 654.6913 3.248982 From the above table, minimum area of aluminium sheet used when r = 4.58, h = 15.17 and the ratio of h = 3.312 ( ii ) Regular Hexagon Method 1: Trionometry and Calculus Tan 300 = x r x = r tan 300 = r 3 Area of OPQ & nbsp;= 1 x 2x x r 2 = 1 x 2( r ) x r 2 = r2 3 = 3 r2 3 Area of Hexagon = 6 x 3r2 3 = 2 #8730; 3 r2 Total Area of Aluminium Sheets Used, A = 2(2 #8730; 3 r2) + 2 π r ( 1000 ) = 4 #8730; 3 r2 + 2000r-1 dA = 8 #8730; 3 r – 2000 r-2 dr For minimum surface area, dA = 0 8 #8730;3 r = 2000 r2 r3 = 2000 8 #8730;3 = 250 #8730; 3 r = 5.246 h = 1000 πr² = 1000 π x ( 5.246 ) = 11.57 h = 11.57 r 5.246 = 2.205 Method 2: Graphical Method By using the equation A=4#8730;3r²+2000-1, graph A versus r can be plotted using the value of r from 1 cm to 11 cm. Graph Area (A) Versus Radius (r) r A 1 2006.928 2 1027.712 3 729.0187 4 610.848 5 573.2 6 582.7413 7 625.1863 8 693.392 9 783.3902 10 892.8 11 1020.106 From the graph, the minimum area occurs when the value of r is between 5 cm and 6 cm. By plotting a graph using values of r from 5.18 to 5.3, a more precise value of r can be determined. Graph Area (A) Versus Radius (r) r A 5.18 571.9953 5.19 571.9698 5.2 571.9485 5.21 571.9315 5.22 571.9187 5.23 571.9101 5.24 571.9056 5.25 571.9054 5.26 571.9093 5.27 571.9173 5.28 571.9294 5.29 571.9457 5.3 571.966 From the graph,the area is minimum when r#8776; 5.25. h =1000 πr² =11.57 h =11.57 r 5.25 =2.204 h Minimum area of aluminium sheet used when r =5.25,h=11.57 and r =2.204. Method 3 : Construct a table A table can be constructed by using equation A = 4#8730;3r²+2000r¯¹ and values of r between 1 and 8. r h A h/r 1 318.3 099 2006.928 & nbsp; 318.3099 2 79.57747 &nbs p; 1027.712 39.78874 3 35.36777 &nbs p; 729.0187 11.78926 4 19.89437 &nbs p; 610.848 4.973592 5 12.7324 ; 573.2 2.546479 6 8.841941 &nbs p; 582.7413 1.473657 7 6.49612 ; 625.1863 0.928017 8 4.973592 &nbs p; 693.392 0.621699 Value of r is around 5cm when the area is minimum.Another two tables can be constructed to obtain a more precise value of r. r h A h/r 4.8 13.81553 576.2878 2.878236 4.9 13.25739& nbsp; 574.5045 ; 2.705589 5 12.7324 ; 573.2 2.546479 5.1 12.23798& nbsp; 572.3541 ; 2.399604 5.2 11.77182& nbsp; 571.9485 ; 2.263811 5.3 11.33179& nbsp; 571.966 2.138073 5.4 10.91598& nbsp; 572.3909 ; 2.021477 5.5 10.52264& nbsp; 573.2084 ; 1.913207 r ; h &nbs p; A &nb sp; h/r 5.2 11.77182 &nbs p; 571.9485 2.263811 5.21 11.72667 571.9315 2.2508 5.22 11.68178 571.9187 2.237889 5.23 11.63715 571.9101 2.225077 5.24 11.59278 571.9056 2.212362 5.25 11.54866 571.9054 2.199744 5.26 11.50479 571.9093 2.187222 5.27 11.46117 571.9173 2.174795 5.28 11.4178 & nbsp; 571.9294 2.162461 Minimum area of aluminium sheet used when r= 5.25,h=11.55 and h =2.2. r Additional Task Considering a regular polygon with n sides . Interior angle,z = 180(n-2) N Tan z = r 2 x x = r tan z ……………………..(1) ² Area of n –sided regular polygon =nxr =nr( r ) tan z ................................... (2) ² Tota; Area of Aluminium Sheets Used, A =2nr( r )+2#12640;rh Tan z ² =2nr² +2πr²(1000) , h=1000 tan z πr² πr² ² = 2nr² + 2000 tan z r ² dA = 4nr - 2000 dr tan z r² ² ; When A is minimum ,dA = 0 Dr 4nr = 2000 tan z r² ² ; z r³ =500 tan 2 n r = 3#8730; 500tan n By varying the values of n ,the corresponding values of z, r, h and h can be found from the following table n Z=((n-2)*180)/n &n bsp; z/n Tanz/2 ; R=((500tan z/2)/n) ^(1/3) h=1000/((22/7)*(r ^2)) h/r 3 60.0 & nbsp;30.0 0.5776 &nb sp; 4.58317552 15.14 754949 3.305034 4 90.0 & nbsp;45.0 1.0006 &nb sp; 5.001053852 12.7 2190936 2.543846 5 108.0 54.0 1.3775 &n bsp; 5.164502417 11. 92939418 2.309883 6 120.0 60.0 1.7337 &n bsp; 5.24727803 11.5 5599132 2.202283 7 128.6 64.3 2.0789 &n bsp; 5.295456734 11. 34667227 2.142718 8 135.0 67.5 2.4175 &n bsp; 5.326142937 11. 21630278 2.105896 9 140.0 70.0 2.7517 &n bsp; 5.34698339 11.1 2903964 2.081368 10 144.0 ; 72.0 3.0830 & nbsp; 5.361841508 11 .06744607 2.044113 11 147.3 ; 73.6 3.4122 & nbsp; 5.372848674 11 .02214555 2.051453 12 150.0 ; 75.0 3.7399 & nbsp; 5.381261993 10 .9877074 2.041846 13 152.3 ; 76.2 4.0665 & nbsp; 5.387863064 10 .96080018 2.03435 14 154.3 ; 77.1 4.3923 & nbsp; 5.393159096 10 .93928395 2.028363 15 156.0 ; 78.0 4.7173 & nbsp; 5.397491134 10 .92173121 2.023483 16 157.5 ; 78.8 5.0419 & nbsp; 5.401095477 10 .90715915 2.019435 17 158.8 ; 79.4 5.3661 & nbsp; 5.404140104 10 .89487269 2.016023 18 160.0 ; 80.0 5.6900 & nbsp; 5.406747113 10 .88436871 2.013109 19 161.1 ; 80.5 6.0136 & nbsp; 5.409006999 10 .87527562 2.010586 20 162.0 ; 81.0 6.3371 & nbsp; 5.410987996 10 .86731406 2.008379 21 162.9 ; 81.4 6.6604 & nbsp; 5.412742349 10 .86027068 2.006427 22 163.6 ; 81.8 6.9836 & nbsp; 5.414310607 10 .85398022 2.004684 23 164.3 ; 82.2 7.3068 & nbsp; 5.41572464 10. 84831306 2.003114 24 165 & nbsp;82.5 7.6299 &nb sp; 5.417009773 10.8 4316636 2.001689 Conclusion The above table shows that as the number of sides of a regular polygon increases,the ratio of h/r approaches 2. 1. Material wastage is reduced if the number of the sides of the aluminium sheet increases (regular polygon shapes) 2. Geometrical shapes that can be fitted into each other without any space in between are a) equilateral triangles b) squares c) regular hexagons 3. Therefore,circles must be cut from the geometrical shapes above to reduce wastage. Among these geometrical shapes,cutting circles from regular hexagons will result in the least wastage of the aluminium sheets.

aluminium can: Solution Method 2: Trial and improvement (a) (by substituting the value of r ) V= πr²h πr²h= 1000 h= 1000 πr² ………………(1) A= 2πr² +2πrh = 2πr²+ 2πr( 1000) πr² = 2πr³+2000 r ……………...(2) If r = 1, A= 2π+2000 1 = 2006.2832 r = 2, A= 2π(2)³+2000 2 = 1025.1327 r = 3, A= 2π(3)³+2000 3 = 723.2153 r = 4, A= 2π(4)³+2000 4 = 600.5310 r = 5, A= 2π(5)³+2000 5 = 557.0796 r = 5, A= 2π(5)³+2000 5 = 557.0796 r = 6, A= 2π(6)³+2000 6 = 559.5280 r = 7, A= 2π(7)³+2000 7 = 593.5904 r = 8, A= 2π(8)³+2000 8 = 652.1239 r = 9, A= 2π(9)³+2000 9 = 731.1602 The value of r that falls between 4 and 6 will produce the minimum surface area of the aluminium sheet used. r A 4.5 571.6789 4.6 567.7348 4.7 564.3275 4.8 561.4313 4.9 559.0225 5 557.0796 5.1 555.5825 5.2 554.5127 5.3 553.8532 5.4 553.5881 5.5 553.7027 5.6 554.1835 5.7 555.0179 5.8 556.1939 The lowest surface area is obtained when r=5.4 By substituting r=5.4 into the equation (1) h = 1000 π(5.4)² h = 10.92 h = 10.92 r 5.4 = 2.022,therefore approximately h=2r Surfeca area is minimum,when r=5.4,h=10.92 and h = 2.022 r (b) Alternative method (by substituting the value of h) V= πr²h πr²h = 1000 r² = 1000 πh ………………(1) A = 2πr² + 2πrh = 2π(1000) + 2πh (1000) πh πh = 2000+ 20h#8730; 10πh …………(2) h if h = 1,A = 2000+20 #8730; 10π = 2112.0998 h = 2,A = 2000+40 #8730; 20π 2 = 1158.5331 h = 3,A = 2000+60 #8730; 30π 3 = 860.8293 h = 4,A = 2000+80 #8730; 40π 4 = 724.1996 h = 5,A = 2000+100 #8730; 50π 5 = 650.6628 h = 6,A = 2000+120 #8730; 60π 6 = 607.9207 h = 7,A = 2000+140 #8730; 70π 7 = 582.3025 h = 8,A = 2000+160 #8730; 80π 8 = 567.0662 h = 9,A = 2000+180 #8730; 90π 9 = 558.5217 h = 10,A = 2000+200 #8730; 100π 10 = 554.4908 h = 11, A = 2000+220 #8730; 110π 11 = 553.6112 h = 12, A = 2000+240 #8730; 120π 12 = 554.9918 h = 13, A = 2000+260 #8730; 130π 13 = 558.0278 h = 14, A = 2000+280 #8730; 140π 14 = 562.2963 The value of h that lies between 10 and 12 will produce the least surface area r A 10.1 554.2786 10.2 554.0966 10.3 553.9436 10.4 553.8188 10.5 553.7211 10.6 553.6498 10.7 553.604 10.8 553.5828 10.9 553.5855 11 553.6112 11.1 553.6594 11.2 553.7292 11.3 553.82 11.4 553.9312 Minimum surface area occurs when h= 10.8 By substituting h=10.8 into the equation (1) r² = 1000 ……………(1) πh r = #8730;1000 πh = #8730;1000 10.8π = 5.429 h = 10.8 r 5.429 = 1.989 = 2 Surface area is minimum,when h=10.8, r=5.429 and h=2 r Method 2: Calculus πr² + 2πrh …………… (1)#61472; A = 2 πr²h = 1000 h = 1000 π r² &nbs p; ………………….(2) Substituting equation (2) into (1): A = 2πr² + 2πr (1000) r² #61552; A=2πr² + 2000 r A’(r) = 4πr-2000 r² When A’(r) = 0.4πr = 2000 r² r³ = 500 π r = 500 π r = 5.419 h= 1000 π x 5.419² = 10.84 h =10.84 r 5.419 ~2 A”(r) =4π + 4000 r³ A”(5.419)>0 (because r>0), therefore A is aluminium. The sirface area is minimum, when r = 5.419, h=10.84 and h =2 r Method 3: Graphical Method By plotting the surface area A, against radius, r, for which A=2πr²+200 , the following graph is plotted. r r A 1 2006.283 2 1025.133 3 723.2153 4 600.531 5 557.0796 6 559.528 7 593.5904 8 652.1239 9 731.1602 Graph Area (A) Versus Radius (r) r A 5.1 555.5825 5.2 554.5137 5.3 553.8532 5.4 553.5881 5.5 553.7027 5.6 554.1835 5.7 555.0179 5.8 556.1939 5.9 557.7007 Graph Area (A) Versus Radius (r) r A 5.38 553.6102 5.39 553.5972 5.4 553.5881 5.41 553.5827 5.42 553.5811 5.43 553.5832 5.44 553.5891 5.45 553.5988 5.46 553.6122 5.47 553.6293 5.48 553.6501 From the graph, the value of r = 5.42 h = 1000 πr² = 1000 π(5.42)² = 10.84 h = 10.84 r 5.42 = 2 Surface area is minimum, when r = 5.4, h = 10.84 and h = 2 r To verify the answer , the following method can be used . Method: Construct a Table r A &nbs p; 5 557.0796 &nb sp; The table shows that the minimum surface area occurs 5.05 556.2765 when the value of r falls between 5.4 and 5.5. Therefore 5.1 555.5825 & nbsp; the minimum surface area lines between 553.5881 and 5.15 554.9953 553.7027. 5.2 554.5127 & nbsp; 5.25 554.1327 5.3 553.8532 & nbsp; 5.35 553.6722 5.4 553.5881 & nbsp; 5.45 553.5986 5.5 553.7027 & nbsp; 5.55 553.8982 5.6 554.1835 & nbsp; 5.65 554.5573 r A &nbs p; 5.41 553.5827 To find the more precise value of r and A, the values 5.42 553.5811 of r from 5.41 to 5.54 are used. 5.43 553.5832 5.44 553.5891 From the table, the least value for surface area is 5.45 553.5988 553.5811 and it occurs when r = 5.42. 5.46 553.6122 5.47 553.6293 5.48 553.6501 5.49 553.6746 5.5 553.7027 & nbsp; 5.51 553.7345 5.52 553.77 &n bsp; 5.53 553.8091 5.54 553.8518 r A &nbs p; 5.411 553.5823 ; To find the more precise of value r and A, the values of r 5.412 553.582 From 5.411 to 5.424 are used. 5.413 553.5818 ; 5.414 553.5816 ; From the table, the least value for surface area is 553.581 5.415 553.5814 ; and it occurs when r = 5.419 5.416 553.5812 ; 5.417 553.5811 ; 5.418 553.5811 ; 5.419 553.581 5.42 553.5811 5.421 553.5811 ; 5.422 553.5812 ; 5.423 553.5813 ; 5.424 553.5815 ; Solution for Question 2 Method 1:Calculus Area of aluminium sheet used, A = 2(2r 2r) + 2πrh……….(1) πr2h = 1000………………..(2) A = 8r2 +2πr( ) =8r2 +( ) = 8r2 +2000r -1 16r -2000r -2 When A is minimum, 0. 16r-2000r -2 = 0 16r = r3 = =125 r = 5 h = = =12.73 = =2.546 Minimum area of aluminium sheet used, when r = 5, h = 12.73 and =2.546. Method 2 : Draw a Graph By plotting the surface area A, against radius, r, for which A=8r2 + 2000r -1 , the following graph is produced. r A 1 2008 2 1032 3 738.6667 4 628 5 600 6 621.3333 7 677.7143 8 762 9 870.2222 Graph Area (A) Versus Radius (r) r A 4.7 602.2519 4.8 600.9867 4.9 600.2433 5 600 5.1 600.2369 5.2 600.9354 5.3 602.0785 5.4 603.6504 5.5 605.6364 From the graph, value of r = 5. h = = =12.73 = =2.546 Minimum area of aluminium sheet used, when r = 5, h = 12.73 and =2.546. To verify the answer, the following method can be used. Method: Construct a table r A &nbs p;Table shows that the minimum 1 2008 & nbsp;surface area occurs when the 1.2 1678.187 & nbsp; value of r falls between 4.8 and 1.4 1444.251 & nbsp; 5.2. 1.6 1270.48 &n bsp; 1.8 1137.031 & nbsp; 2 1032 & nbsp; 2.1 947.8109 & nbsp; 2.4 879.4133 & nbsp; 2.6 823.3108 & nbsp; 2.8 777.0057 & nbsp; 3 738.6667 &nb sp; 3.2 706.92 &nb sp; 3.4 680.7153 & nbsp; 3.6 659.2356 & nbsp; 3.8 641.8358 & nbsp; 4 628 &n bsp; 4.2 617.3105 & nbsp; 4.4 609.4255 & nbsp; 4.6 604.0626 & nbsp; 4.8 600.9867 & nbsp; 5 600 &n bsp; 5.2 600.9354 & nbsp; 5.4 603.6504 & nbsp; 5.6 608.0229 & nbsp; r A 4.9 600.2433 4.91 600.1968 4.92 600.1553 4.93 600.1187 4.94 600.0871 4.95 600.0604 4.96 600.0386 4.97 600.0271 4.98 600.0096 4.99 600.0024 5 600 5.01 600.0024 5.02 600.0096 5.03 600.0215 To get a more precise value of A and r, a smaller range of r from 4.8 to 5.2 s used. From the table, the minimum area is 600 and this occurs when r = 5. h = = = 12.73 = =2.546 Solution for Question 3 Consider The Following Shapes (1) Equilateral Triangle Method 1:Trigonometry and Calculus tan 60o = x = r Area of PQR = 3 Area of OPQ = 3 2X r = 3 xr = 3 r2 Total area of Aluminium Sheets Used, A = 6 r2 + 2πr( ) = 6 r2 +2000r -1 12 r – 2000r -2 When A is minimum, 0. 12 r = r3 = 96.225 r = 4.582 h = =15.16 =3.309 12 + 4000r -3 (>0, because r>0) Therefore , A is minimum. Minimum area or aluminium sheet used when r = 4.582, h= 15.16 and = 3.309. Method 2: Graphical Method By using the equation A=6#8730;3r²+2000-1, graph A versus r can be plotted to determine the value of r for the minimum value of A. Graph Area (A) Versus Radius (r) r A 1 2010.392 2 1041.568 3 760.1947 4 666.272 5 659.8 6 707.4453 7 794.9223 8 915.088 9 1063.974 From the graph, minimum area occurs when r falls between 4 cm and 5 cm. The following graph is plotted using the values of r between 4.54 and 4.64 Graph Area (A) Versus Radius (r) r A 4.54 654.7244 4.55 654.7008 4.56 654.6836 4.57 654.6726 4.58 654.668 4.59 654.6695 4.6 654.6773 4.61 654.6913 4.62 654.7114 4.63 654.7377 4.64 654.7701 From the graph, area is minimum when r #8776; 4.58 cm. h = 1000 Πr² = 15.17 h = 15.17 r 4.58 = 3.312 Minimun area of aluminium sheet used when r = 4.58, h = 15.17 and h = 3.312 Method 3: Construct a Table A table can be constructed using A = 6 3 r2 + 2000r-1 r h &nbs p;A h/r 1 318.3099 &nb sp; 2010.392 318.3099 2 79.57747 1041.568 &n bsp; 39.78874 3 35.36777 &nb sp; 760.1947 11.78874 4 19.89437 &nb sp; 666.272 11.78926 5 12.7324 &nbs p; 659.8 4.973592 6 8.841941 &nb sp; 707.4453 2.546479 7 6.49612 &nbs p; 794.9223 0.928017 8 4.973592 &nb sp; 915.088 0.621699 By using a more precise range of value r, we get the following tables. r h &nbs p;A h/r 4.1 18.93575 & nbsp; 662.4944 4.618475 4.2 18.04478 & nbsp; 659.5054 4.296376 4.3 17.21525 & nbsp; 657.2644 4.003545 4.4 16.44163 & nbsp; 655.7346 3.736733 4.5 15.71901 & nbsp; 654.8824 3.493113 4.6 15.043 &nb sp; 654.6773 3.270217 4.7 14.40968 & nbsp; 655.0912 3.06589 4.8 13.81553 & nbsp; 656.0983 2.878236 r h &nbs p;A h/r 4.51 15.64938 654.8333 3.469928 4.52 15.58021 654.7906 3.446949 4.53 15.5115 & nbsp; 654.7543 3.424172 4.54 15.44324 654.7244 3.401595 4.55 15.37543 654.7008 3.379216 4.56 15.30807 654.6836 3.357033 4.57 15.24115 654.6726 3.335044 4.58 15.17467 654.668 3.313246 4.59 15.10862 654.6695 3.291638 4.6 15.043 &nb sp; 654.6773 3.270217 4.61 14.97781 654.6913 3.248982 From the above table, minimum area of aluminium sheet used when r = 4.58, h = 15.17 and the ratio of h = 3.312 ( ii ) Regular Hexagon Method 1: Trionometry and Calculus Tan 300 = x r x = r tan 300 = r 3 Area of OPQ & nbsp;= 1 x 2x x r 2 = 1 x 2( r ) x r 2 = r2 3 = 3 r2 3 Area of Hexagon = 6 x 3r2 3 = 2 #8730; 3 r2 Total Area of Aluminium Sheets Used, A = 2(2 #8730; 3 r2) + 2 π r ( 1000 ) = 4 #8730; 3 r2 + 2000r-1 dA = 8 #8730; 3 r – 2000 r-2 dr For minimum surface area, dA = 0 8 #8730;3 r = 2000 r2 r3 = 2000 8 #8730;3 = 250 #8730; 3 r = 5.246 h = 1000 πr² = 1000 π x ( 5.246 ) = 11.57 h = 11.57 r 5.246 = 2.205 Method 2: Graphical Method By using the equation A=4#8730;3r²+2000-1, graph A versus r can be plotted using the value of r from 1 cm to 11 cm. Graph Area (A) Versus Radius (r) r A 1 2006.928 2 1027.712 3 729.0187 4 610.848 5 573.2 6 582.7413 7 625.1863 8 693.392 9 783.3902 10 892.8 11 1020.106 From the graph, the minimum area occurs when the value of r is between 5 cm and 6 cm. By plotting a graph using values of r from 5.18 to 5.3, a more precise value of r can be determined. Graph Area (A) Versus Radius (r) r A 5.18 571.9953 5.19 571.9698 5.2 571.9485 5.21 571.9315 5.22 571.9187 5.23 571.9101 5.24 571.9056 5.25 571.9054 5.26 571.9093 5.27 571.9173 5.28 571.9294 5.29 571.9457 5.3 571.966 From the graph,the area is minimum when r#8776; 5.25. h =1000 πr² =11.57 h =11.57 r 5.25 =2.204 h Minimum area of aluminium sheet used when r =5.25,h=11.57 and r =2.204. Method 3 : Construct a table A table can be constructed by using equation A = 4#8730;3r²+2000r¯¹ and values of r between 1 and 8. r h A h/r 1 318.3 099 2006.928 & nbsp; 318.3099 2 79.57747 &nbs p; 1027.712 39.78874 3 35.36777 &nbs p; 729.0187 11.78926 4 19.89437 &nbs p; 610.848 4.973592 5 12.7324 ; 573.2 2.546479 6 8.841941 &nbs p; 582.7413 1.473657 7 6.49612 ; 625.1863 0.928017 8 4.973592 &nbs p; 693.392 0.621699 Value of r is around 5cm when the area is minimum.Another two tables can be constructed to obtain a more precise value of r. r h A h/r 4.8 13.81553 576.2878 2.878236 4.9 13.25739& nbsp; 574.5045 ; 2.705589 5 12.7324 ; 573.2 2.546479 5.1 12.23798& nbsp; 572.3541 ; 2.399604 5.2 11.77182& nbsp; 571.9485 ; 2.263811 5.3 11.33179& nbsp; 571.966 2.138073 5.4 10.91598& nbsp; 572.3909 ; 2.021477 5.5 10.52264& nbsp; 573.2084 ; 1.913207 r ; h &nbs p; A &nb sp; h/r 5.2 11.77182 &nbs p; 571.9485 2.263811 5.21 11.72667 571.9315 2.2508 5.22 11.68178 571.9187 2.237889 5.23 11.63715 571.9101 2.225077 5.24 11.59278 571.9056 2.212362 5.25 11.54866 571.9054 2.199744 5.26 11.50479 571.9093 2.187222 5.27 11.46117 571.9173 2.174795 5.28 11.4178 & nbsp; 571.9294 2.162461 Minimum area of aluminium sheet used when r= 5.25,h=11.55 and h =2.2. r Additional Task Considering a regular polygon with n sides . Interior angle,z = 180(n-2) N Tan z = r 2 x x = r tan z ……………………..(1) ² Area of n –sided regular polygon =nxr =nr( r ) tan z ................................... (2) ² Tota; Area of Aluminium Sheets Used, A =2nr( r )+2#12640;rh Tan z ² =2nr² +2πr²(1000) , h=1000 tan z πr² πr² ² = 2nr² + 2000 tan z r ² dA = 4nr - 2000 dr tan z r² ² ; When A is minimum ,dA = 0 Dr 4nr = 2000 tan z r² ² ; z r³ =500 tan 2 n r = 3#8730; 500tan n By varying the values of n ,the corresponding values of z, r, h and h can be found from the following table n Z=((n-2)*180)/n &n bsp; z/n Tanz/2 ; R=((500tan z/2)/n) ^(1/3) h=1000/((22/7)*(r ^2)) h/r 3 60.0 & nbsp;30.0 0.5776 &nb sp; 4.58317552 15.14 754949 3.305034 4 90.0 & nbsp;45.0 1.0006 &nb sp; 5.001053852 12.7 2190936 2.543846 5 108.0 54.0 1.3775 &n bsp; 5.164502417 11. 92939418 2.309883 6 120.0 60.0 1.7337 &n bsp; 5.24727803 11.5 5599132 2.202283 7 128.6 64.3 2.0789 &n bsp; 5.295456734 11. 34667227 2.142718 8 135.0 67.5 2.4175 &n bsp; 5.326142937 11. 21630278 2.105896 9 140.0 70.0 2.7517 &n bsp; 5.34698339 11.1 2903964 2.081368 10 144.0 ; 72.0 3.0830 & nbsp; 5.361841508 11 .06744607 2.044113 11 147.3 ; 73.6 3.4122 & nbsp; 5.372848674 11 .02214555 2.051453 12 150.0 ; 75.0 3.7399 & nbsp; 5.381261993 10 .9877074 2.041846 13 152.3 ; 76.2 4.0665 & nbsp; 5.387863064 10 .96080018 2.03435 14 154.3 ; 77.1 4.3923 & nbsp; 5.393159096 10 .93928395 2.028363 15 156.0 ; 78.0 4.7173 & nbsp; 5.397491134 10 .92173121 2.023483 16 157.5 ; 78.8 5.0419 & nbsp; 5.401095477 10 .90715915 2.019435 17 158.8 ; 79.4 5.3661 & nbsp; 5.404140104 10 .89487269 2.016023 18 160.0 ; 80.0 5.6900 & nbsp; 5.406747113 10 .88436871 2.013109 19 161.1 ; 80.5 6.0136 & nbsp; 5.409006999 10 .87527562 2.010586 20 162.0 ; 81.0 6.3371 & nbsp; 5.410987996 10 .86731406 2.008379 21 162.9 ; 81.4 6.6604 & nbsp; 5.412742349 10 .86027068 2.006427 22 163.6 ; 81.8 6.9836 & nbsp; 5.414310607 10 .85398022 2.004684 23 164.3 ; 82.2 7.3068 & nbsp; 5.41572464 10. 84831306 2.003114 24 165 & nbsp;82.5 7.6299 &nb sp; 5.417009773 10.8 4316636 2.001689 Conclusion The above table shows that as the number of sides of a regular polygon increases,the ratio of h/r approaches 2. 1. Material wastage is reduced if the number of the sides of the aluminium sheet increases (regular polygon shapes) 2. Geometrical shapes that can be fitted into each other without any space in between are a) equilateral triangles b) squares c) regular hexagons 3. Therefore,circles must be cut from the geometrical shapes above to reduce wastage. Among these geometrical shapes,cutting circles from regular hexagons will result in the least wastage of the aluminium sheets.

aluminium can: Solution Method 2: Trial and improvement (a) (by substituting the value of r ) V= πr²h πr²h= 1000 h= 1000 πr² ………………(1) A= 2πr² +2πrh = 2πr²+ 2πr( 1000) πr² = 2πr³+2000 r ……………...(2) If r = 1, A= 2π+2000 1 = 2006.2832 r = 2, A= 2π(2)³+2000 2 = 1025.1327 r = 3, A= 2π(3)³+2000 3 = 723.2153 r = 4, A= 2π(4)³+2000 4 = 600.5310 r = 5, A= 2π(5)³+2000 5 = 557.0796 r = 5, A= 2π(5)³+2000 5 = 557.0796 r = 6, A= 2π(6)³+2000 6 = 559.5280 r = 7, A= 2π(7)³+2000 7 = 593.5904 r = 8, A= 2π(8)³+2000 8 = 652.1239 r = 9, A= 2π(9)³+2000 9 = 731.1602 The value of r that falls between 4 and 6 will produce the minimum surface area of the aluminium sheet used. r A 4.5 571.6789 4.6 567.7348 4.7 564.3275 4.8 561.4313 4.9 559.0225 5 557.0796 5.1 555.5825 5.2 554.5127 5.3 553.8532 5.4 553.5881 5.5 553.7027 5.6 554.1835 5.7 555.0179 5.8 556.1939 The lowest surface area is obtained when r=5.4 By substituting r=5.4 into the equation (1) h = 1000 π(5.4)² h = 10.92 h = 10.92 r 5.4 = 2.022,therefore approximately h=2r Surfeca area is minimum,when r=5.4,h=10.92 and h = 2.022 r (b) Alternative method (by substituting the value of h) V= πr²h πr²h = 1000 r² = 1000 πh ………………(1) A = 2πr² + 2πrh = 2π(1000) + 2πh (1000) πh πh = 2000+ 20h#8730; 10πh …………(2) h if h = 1,A = 2000+20 #8730; 10π = 2112.0998 h = 2,A = 2000+40 #8730; 20π 2 = 1158.5331 h = 3,A = 2000+60 #8730; 30π 3 = 860.8293 h = 4,A = 2000+80 #8730; 40π 4 = 724.1996 h = 5,A = 2000+100 #8730; 50π 5 = 650.6628 h = 6,A = 2000+120 #8730; 60π 6 = 607.9207 h = 7,A = 2000+140 #8730; 70π 7 = 582.3025 h = 8,A = 2000+160 #8730; 80π 8 = 567.0662 h = 9,A = 2000+180 #8730; 90π 9 = 558.5217 h = 10,A = 2000+200 #8730; 100π 10 = 554.4908 h = 11, A = 2000+220 #8730; 110π 11 = 553.6112 h = 12, A = 2000+240 #8730; 120π 12 = 554.9918 h = 13, A = 2000+260 #8730; 130π 13 = 558.0278 h = 14, A = 2000+280 #8730; 140π 14 = 562.2963 The value of h that lies between 10 and 12 will produce the least surface area r A 10.1 554.2786 10.2 554.0966 10.3 553.9436 10.4 553.8188 10.5 553.7211 10.6 553.6498 10.7 553.604 10.8 553.5828 10.9 553.5855 11 553.6112 11.1 553.6594 11.2 553.7292 11.3 553.82 11.4 553.9312 Minimum surface area occurs when h= 10.8 By substituting h=10.8 into the equation (1) r² = 1000 ……………(1) πh r = #8730;1000 πh = #8730;1000 10.8π = 5.429 h = 10.8 r 5.429 = 1.989 = 2 Surface area is minimum,when h=10.8, r=5.429 and h=2 r Method 2: Calculus πr² + 2πrh …………… (1)#61472; A = 2 πr²h = 1000 h = 1000 π r² &nbs p; ………………….(2) Substituting equation (2) into (1): A = 2πr² + 2πr (1000) r² #61552; A=2πr² + 2000 r A’(r) = 4πr-2000 r² When A’(r) = 0.4πr = 2000 r² r³ = 500 π r = 500 π r = 5.419 h= 1000 π x 5.419² = 10.84 h =10.84 r 5.419 ~2 A”(r) =4π + 4000 r³ A”(5.419)>0 (because r>0), therefore A is aluminium. The sirface area is minimum, when r = 5.419, h=10.84 and h =2 r Method 3: Graphical Method By plotting the surface area A, against radius, r, for which A=2πr²+200 , the following graph is plotted. r r A 1 2006.283 2 1025.133 3 723.2153 4 600.531 5 557.0796 6 559.528 7 593.5904 8 652.1239 9 731.1602 Graph Area (A) Versus Radius (r) r A 5.1 555.5825 5.2 554.5137 5.3 553.8532 5.4 553.5881 5.5 553.7027 5.6 554.1835 5.7 555.0179 5.8 556.1939 5.9 557.7007 Graph Area (A) Versus Radius (r) r A 5.38 553.6102 5.39 553.5972 5.4 553.5881 5.41 553.5827 5.42 553.5811 5.43 553.5832 5.44 553.5891 5.45 553.5988 5.46 553.6122 5.47 553.6293 5.48 553.6501 From the graph, the value of r = 5.42 h = 1000 πr² = 1000 π(5.42)² = 10.84 h = 10.84 r 5.42 = 2 Surface area is minimum, when r = 5.4, h = 10.84 and h = 2 r To verify the answer , the following method can be used . Method: Construct a Table r A &nbs p; 5 557.0796 &nb sp; The table shows that the minimum surface area occurs 5.05 556.2765 when the value of r falls between 5.4 and 5.5. Therefore 5.1 555.5825 & nbsp; the minimum surface area lines between 553.5881 and 5.15 554.9953 553.7027. 5.2 554.5127 & nbsp; 5.25 554.1327 5.3 553.8532 & nbsp; 5.35 553.6722 5.4 553.5881 & nbsp; 5.45 553.5986 5.5 553.7027 & nbsp; 5.55 553.8982 5.6 554.1835 & nbsp; 5.65 554.5573 r A &nbs p; 5.41 553.5827 To find the more precise value of r and A, the values 5.42 553.5811 of r from 5.41 to 5.54 are used. 5.43 553.5832 5.44 553.5891 From the table, the least value for surface area is 5.45 553.5988 553.5811 and it occurs when r = 5.42. 5.46 553.6122 5.47 553.6293 5.48 553.6501 5.49 553.6746 5.5 553.7027 & nbsp; 5.51 553.7345 5.52 553.77 &n bsp; 5.53 553.8091 5.54 553.8518 r A &nbs p; 5.411 553.5823 ; To find the more precise of value r and A, the values of r 5.412 553.582 From 5.411 to 5.424 are used. 5.413 553.5818 ; 5.414 553.5816 ; From the table, the least value for surface area is 553.581 5.415 553.5814 ; and it occurs when r = 5.419 5.416 553.5812 ; 5.417 553.5811 ; 5.418 553.5811 ; 5.419 553.581 5.42 553.5811 5.421 553.5811 ; 5.422 553.5812 ; 5.423 553.5813 ; 5.424 553.5815 ; Solution for Question 2 Method 1:Calculus Area of aluminium sheet used, A = 2(2r 2r) + 2πrh……….(1) πr2h = 1000………………..(2) A = 8r2 +2πr( ) =8r2 +( ) = 8r2 +2000r -1 16r -2000r -2 When A is minimum, 0. 16r-2000r -2 = 0 16r = r3 = =125 r = 5 h = = =12.73 = =2.546 Minimum area of aluminium sheet used, when r = 5, h = 12.73 and =2.546. Method 2 : Draw a Graph By plotting the surface area A, against radius, r, for which A=8r2 + 2000r -1 , the following graph is produced. r A 1 2008 2 1032 3 738.6667 4 628 5 600 6 621.3333 7 677.7143 8 762 9 870.2222 Graph Area (A) Versus Radius (r) r A 4.7 602.2519 4.8 600.9867 4.9 600.2433 5 600 5.1 600.2369 5.2 600.9354 5.3 602.0785 5.4 603.6504 5.5 605.6364 From the graph, value of r = 5. h = = =12.73 = =2.546 Minimum area of aluminium sheet used, when r = 5, h = 12.73 and =2.546. To verify the answer, the following method can be used. Method: Construct a table r A &nbs p;Table shows that the minimum 1 2008 & nbsp;surface area occurs when the 1.2 1678.187 & nbsp; value of r falls between 4.8 and 1.4 1444.251 & nbsp; 5.2. 1.6 1270.48 &n bsp; 1.8 1137.031 & nbsp; 2 1032 & nbsp; 2.1 947.8109 & nbsp; 2.4 879.4133 & nbsp; 2.6 823.3108 & nbsp; 2.8 777.0057 & nbsp; 3 738.6667 &nb sp; 3.2 706.92 &nb sp; 3.4 680.7153 & nbsp; 3.6 659.2356 & nbsp; 3.8 641.8358 & nbsp; 4 628 &n bsp; 4.2 617.3105 & nbsp; 4.4 609.4255 & nbsp; 4.6 604.0626 & nbsp; 4.8 600.9867 & nbsp; 5 600 &n bsp; 5.2 600.9354 & nbsp; 5.4 603.6504 & nbsp; 5.6 608.0229 & nbsp; r A 4.9 600.2433 4.91 600.1968 4.92 600.1553 4.93 600.1187 4.94 600.0871 4.95 600.0604 4.96 600.0386 4.97 600.0271 4.98 600.0096 4.99 600.0024 5 600 5.01 600.0024 5.02 600.0096 5.03 600.0215 To get a more precise value of A and r, a smaller range of r from 4.8 to 5.2 s used. From the table, the minimum area is 600 and this occurs when r = 5. h = = = 12.73 = =2.546 Solution for Question 3 Consider The Following Shapes (1) Equilateral Triangle Method 1:Trigonometry and Calculus tan 60o = x = r Area of PQR = 3 Area of OPQ = 3 2X r = 3 xr = 3 r2 Total area of Aluminium Sheets Used, A = 6 r2 + 2πr( ) = 6 r2 +2000r -1 12 r – 2000r -2 When A is minimum, 0. 12 r = r3 = 96.225 r = 4.582 h = =15.16 =3.309 12 + 4000r -3 (>0, because r>0) Therefore , A is minimum. Minimum area or aluminium sheet used when r = 4.582, h= 15.16 and = 3.309. Method 2: Graphical Method By using the equation A=6#8730;3r²+2000-1, graph A versus r can be plotted to determine the value of r for the minimum value of A. Graph Area (A) Versus Radius (r) r A 1 2010.392 2 1041.568 3 760.1947 4 666.272 5 659.8 6 707.4453 7 794.9223 8 915.088 9 1063.974 From the graph, minimum area occurs when r falls between 4 cm and 5 cm. The following graph is plotted using the values of r between 4.54 and 4.64 Graph Area (A) Versus Radius (r) r A 4.54 654.7244 4.55 654.7008 4.56 654.6836 4.57 654.6726 4.58 654.668 4.59 654.6695 4.6 654.6773 4.61 654.6913 4.62 654.7114 4.63 654.7377 4.64 654.7701 From the graph, area is minimum when r #8776; 4.58 cm. h = 1000 Πr² = 15.17 h = 15.17 r 4.58 = 3.312 Minimun area of aluminium sheet used when r = 4.58, h = 15.17 and h = 3.312 Method 3: Construct a Table A table can be constructed using A = 6 3 r2 + 2000r-1 r h &nbs p;A h/r 1 318.3099 &nb sp; 2010.392 318.3099 2 79.57747 1041.568 &n bsp; 39.78874 3 35.36777 &nb sp; 760.1947 11.78874 4 19.89437 &nb sp; 666.272 11.78926 5 12.7324 &nbs p; 659.8 4.973592 6 8.841941 &nb sp; 707.4453 2.546479 7 6.49612 &nbs p; 794.9223 0.928017 8 4.973592 &nb sp; 915.088 0.621699 By using a more precise range of value r, we get the following tables. r h &nbs p;A h/r 4.1 18.93575 & nbsp; 662.4944 4.618475 4.2 18.04478 & nbsp; 659.5054 4.296376 4.3 17.21525 & nbsp; 657.2644 4.003545 4.4 16.44163 & nbsp; 655.7346 3.736733 4.5 15.71901 & nbsp; 654.8824 3.493113 4.6 15.043 &nb sp; 654.6773 3.270217 4.7 14.40968 & nbsp; 655.0912 3.06589 4.8 13.81553 & nbsp; 656.0983 2.878236 r h &nbs p;A h/r 4.51 15.64938 654.8333 3.469928 4.52 15.58021 654.7906 3.446949 4.53 15.5115 & nbsp; 654.7543 3.424172 4.54 15.44324 654.7244 3.401595 4.55 15.37543 654.7008 3.379216 4.56 15.30807 654.6836 3.357033 4.57 15.24115 654.6726 3.335044 4.58 15.17467 654.668 3.313246 4.59 15.10862 654.6695 3.291638 4.6 15.043 &nb sp; 654.6773 3.270217 4.61 14.97781 654.6913 3.248982 From the above table, minimum area of aluminium sheet used when r = 4.58, h = 15.17 and the ratio of h = 3.312 ( ii ) Regular Hexagon Method 1: Trionometry and Calculus Tan 300 = x r x = r tan 300 = r 3 Area of OPQ & nbsp;= 1 x 2x x r 2 = 1 x 2( r ) x r 2 = r2 3 = 3 r2 3 Area of Hexagon = 6 x 3r2 3 = 2 #8730; 3 r2 Total Area of Aluminium Sheets Used, A = 2(2 #8730; 3 r2) + 2 π r ( 1000 ) = 4 #8730; 3 r2 + 2000r-1 dA = 8 #8730; 3 r – 2000 r-2 dr For minimum surface area, dA = 0 8 #8730;3 r = 2000 r2 r3 = 2000 8 #8730;3 = 250 #8730; 3 r = 5.246 h = 1000 πr² = 1000 π x ( 5.246 ) = 11.57 h = 11.57 r 5.246 = 2.205 Method 2: Graphical Method By using the equation A=4#8730;3r²+2000-1, graph A versus r can be plotted using the value of r from 1 cm to 11 cm. Graph Area (A) Versus Radius (r) r A 1 2006.928 2 1027.712 3 729.0187 4 610.848 5 573.2 6 582.7413 7 625.1863 8 693.392 9 783.3902 10 892.8 11 1020.106 From the graph, the minimum area occurs when the value of r is between 5 cm and 6 cm. By plotting a graph using values of r from 5.18 to 5.3, a more precise value of r can be determined. Graph Area (A) Versus Radius (r) r A 5.18 571.9953 5.19 571.9698 5.2 571.9485 5.21 571.9315 5.22 571.9187 5.23 571.9101 5.24 571.9056 5.25 571.9054 5.26 571.9093 5.27 571.9173 5.28 571.9294 5.29 571.9457 5.3 571.966 From the graph,the area is minimum when r#8776; 5.25. h =1000 πr² =11.57 h =11.57 r 5.25 =2.204 h Minimum area of aluminium sheet used when r =5.25,h=11.57 and r =2.204. Method 3 : Construct a table A table can be constructed by using equation A = 4#8730;3r²+2000r¯¹ and values of r between 1 and 8. r h A h/r 1 318.3 099 2006.928 & nbsp; 318.3099 2 79.57747 &nbs p; 1027.712 39.78874 3 35.36777 &nbs p; 729.0187 11.78926 4 19.89437 &nbs p; 610.848 4.973592 5 12.7324 ; 573.2 2.546479 6 8.841941 &nbs p; 582.7413 1.473657 7 6.49612 ; 625.1863 0.928017 8 4.973592 &nbs p; 693.392 0.621699 Value of r is around 5cm when the area is minimum.Another two tables can be constructed to obtain a more precise value of r. r h A h/r 4.8 13.81553 576.2878 2.878236 4.9 13.25739& nbsp; 574.5045 ; 2.705589 5 12.7324 ; 573.2 2.546479 5.1 12.23798& nbsp; 572.3541 ; 2.399604 5.2 11.77182& nbsp; 571.9485 ; 2.263811 5.3 11.33179& nbsp; 571.966 2.138073 5.4 10.91598& nbsp; 572.3909 ; 2.021477 5.5 10.52264& nbsp; 573.2084 ; 1.913207 r ; h &nbs p; A &nb sp; h/r 5.2 11.77182 &nbs p; 571.9485 2.263811 5.21 11.72667 571.9315 2.2508 5.22 11.68178 571.9187 2.237889 5.23 11.63715 571.9101 2.225077 5.24 11.59278 571.9056 2.212362 5.25 11.54866 571.9054 2.199744 5.26 11.50479 571.9093 2.187222 5.27 11.46117 571.9173 2.174795 5.28 11.4178 & nbsp; 571.9294 2.162461 Minimum area of aluminium sheet used when r= 5.25,h=11.55 and h =2.2. r Additional Task Considering a regular polygon with n sides . Interior angle,z = 180(n-2) N Tan z = r 2 x x = r tan z ……………………..(1) ² Area of n –sided regular polygon =nxr =nr( r ) tan z ................................... (2) ² Tota; Area of Aluminium Sheets Used, A =2nr( r )+2#12640;rh Tan z ² =2nr² +2πr²(1000) , h=1000 tan z πr² πr² ² = 2nr² + 2000 tan z r ² dA = 4nr - 2000 dr tan z r² ² ; When A is minimum ,dA = 0 Dr 4nr = 2000 tan z r² ² ; z r³ =500 tan 2 n r = 3#8730; 500tan n By varying the values of n ,the corresponding values of z, r, h and h can be found from the following table n Z=((n-2)*180)/n &n bsp; z/n Tanz/2 ; R=((500tan z/2)/n) ^(1/3) h=1000/((22/7)*(r ^2)) h/r 3 60.0 & nbsp;30.0 0.5776 &nb sp; 4.58317552 15.14 754949 3.305034 4 90.0 & nbsp;45.0 1.0006 &nb sp; 5.001053852 12.7 2190936 2.543846 5 108.0 54.0 1.3775 &n bsp; 5.164502417 11. 92939418 2.309883 6 120.0 60.0 1.7337 &n bsp; 5.24727803 11.5 5599132 2.202283 7 128.6 64.3 2.0789 &n bsp; 5.295456734 11. 34667227 2.142718 8 135.0 67.5 2.4175 &n bsp; 5.326142937 11. 21630278 2.105896 9 140.0 70.0 2.7517 &n bsp; 5.34698339 11.1 2903964 2.081368 10 144.0 ; 72.0 3.0830 & nbsp; 5.361841508 11 .06744607 2.044113 11 147.3 ; 73.6 3.4122 & nbsp; 5.372848674 11 .02214555 2.051453 12 150.0 ; 75.0 3.7399 & nbsp; 5.381261993 10 .9877074 2.041846 13 152.3 ; 76.2 4.0665 & nbsp; 5.387863064 10 .96080018 2.03435 14 154.3 ; 77.1 4.3923 & nbsp; 5.393159096 10 .93928395 2.028363 15 156.0 ; 78.0 4.7173 & nbsp; 5.397491134 10 .92173121 2.023483 16 157.5 ; 78.8 5.0419 & nbsp; 5.401095477 10 .90715915 2.019435 17 158.8 ; 79.4 5.3661 & nbsp; 5.404140104 10 .89487269 2.016023 18 160.0 ; 80.0 5.6900 & nbsp; 5.406747113 10 .88436871 2.013109 19 161.1 ; 80.5 6.0136 & nbsp; 5.409006999 10 .87527562 2.010586 20 162.0 ; 81.0 6.3371 & nbsp; 5.410987996 10 .86731406 2.008379 21 162.9 ; 81.4 6.6604 & nbsp; 5.412742349 10 .86027068 2.006427 22 163.6 ; 81.8 6.9836 & nbsp; 5.414310607 10 .85398022 2.004684 23 164.3 ; 82.2 7.3068 & nbsp; 5.41572464 10. 84831306 2.003114 24 165 & nbsp;82.5 7.6299 &nb sp; 5.417009773 10.8 4316636 2.001689 Conclusion The above table shows that as the number of sides of a regular polygon increases,the ratio of h/r approaches 2. 1. Material wastage is reduced if the number of the sides of the aluminium sheet increases (regular polygon shapes) 2. Geometrical shapes that can be fitted into each other without any space in between are a) equilateral triangles b) squares c) regular hexagons 3. Therefore,circles must be cut from the geometrical shapes above to reduce wastage. Among these geometrical shapes,cutting circles from regular hexagons will result in the least wastage of the aluminium sheets.

aluminium can: Solution Method 2: Trial and improvement (a) (by substituting the value of r ) V= πr²h πr²h= 1000 h= 1000 πr² ………………(1) A= 2πr² +2πrh = 2πr²+ 2πr( 1000) πr² = 2πr³+2000 r ……………...(2) If r = 1, A= 2π+2000 1 = 2006.2832 r = 2, A= 2π(2)³+2000 2 = 1025.1327 r = 3, A= 2π(3)³+2000 3 = 723.2153 r = 4, A= 2π(4)³+2000 4 = 600.5310 r = 5, A= 2π(5)³+2000 5 = 557.0796 r = 5, A= 2π(5)³+2000 5 = 557.0796 r = 6, A= 2π(6)³+2000 6 = 559.5280 r = 7, A= 2π(7)³+2000 7 = 593.5904 r = 8, A= 2π(8)³+2000 8 = 652.1239 r = 9, A= 2π(9)³+2000 9 = 731.1602 The value of r that falls between 4 and 6 will produce the minimum surface area of the aluminium sheet used. r A 4.5 571.6789 4.6 567.7348 4.7 564.3275 4.8 561.4313 4.9 559.0225 5 557.0796 5.1 555.5825 5.2 554.5127 5.3 553.8532 5.4 553.5881 5.5 553.7027 5.6 554.1835 5.7 555.0179 5.8 556.1939 The lowest surface area is obtained when r=5.4 By substituting r=5.4 into the equation (1) h = 1000 π(5.4)² h = 10.92 h = 10.92 r 5.4 = 2.022,therefore approximately h=2r Surfeca area is minimum,when r=5.4,h=10.92 and h = 2.022 r (b) Alternative method (by substituting the value of h) V= πr²h πr²h = 1000 r² = 1000 πh ………………(1) A = 2πr² + 2πrh = 2π(1000) + 2πh (1000) πh πh = 2000+ 20h#8730; 10πh …………(2) h if h = 1,A = 2000+20 #8730; 10π = 2112.0998 h = 2,A = 2000+40 #8730; 20π 2 = 1158.5331 h = 3,A = 2000+60 #8730; 30π 3 = 860.8293 h = 4,A = 2000+80 #8730; 40π 4 = 724.1996 h = 5,A = 2000+100 #8730; 50π 5 = 650.6628 h = 6,A = 2000+120 #8730; 60π 6 = 607.9207 h = 7,A = 2000+140 #8730; 70π 7 = 582.3025 h = 8,A = 2000+160 #8730; 80π 8 = 567.0662 h = 9,A = 2000+180 #8730; 90π 9 = 558.5217 h = 10,A = 2000+200 #8730; 100π 10 = 554.4908 h = 11, A = 2000+220 #8730; 110π 11 = 553.6112 h = 12, A = 2000+240 #8730; 120π 12 = 554.9918 h = 13, A = 2000+260 #8730; 130π 13 = 558.0278 h = 14, A = 2000+280 #8730; 140π 14 = 562.2963 The value of h that lies between 10 and 12 will produce the least surface area r A 10.1 554.2786 10.2 554.0966 10.3 553.9436 10.4 553.8188 10.5 553.7211 10.6 553.6498 10.7 553.604 10.8 553.5828 10.9 553.5855 11 553.6112 11.1 553.6594 11.2 553.7292 11.3 553.82 11.4 553.9312 Minimum surface area occurs when h= 10.8 By substituting h=10.8 into the equation (1) r² = 1000 ……………(1) πh r = #8730;1000 πh = #8730;1000 10.8π = 5.429 h = 10.8 r 5.429 = 1.989 = 2 Surface area is minimum,when h=10.8, r=5.429 and h=2 r Method 2: Calculus πr² + 2πrh …………… (1)#61472; A = 2 πr²h = 1000 h = 1000 π r² &nbs p; ………………….(2) Substituting equation (2) into (1): A = 2πr² + 2πr (1000) r² #61552; A=2πr² + 2000 r A’(r) = 4πr-2000 r² When A’(r) = 0.4πr = 2000 r² r³ = 500 π r = 500 π r = 5.419 h= 1000 π x 5.419² = 10.84 h =10.84 r 5.419 ~2 A”(r) =4π + 4000 r³ A”(5.419)>0 (because r>0), therefore A is aluminium. The sirface area is minimum, when r = 5.419, h=10.84 and h =2 r Method 3: Graphical Method By plotting the surface area A, against radius, r, for which A=2πr²+200 , the following graph is plotted. r r A 1 2006.283 2 1025.133 3 723.2153 4 600.531 5 557.0796 6 559.528 7 593.5904 8 652.1239 9 731.1602 Graph Area (A) Versus Radius (r) r A 5.1 555.5825 5.2 554.5137 5.3 553.8532 5.4 553.5881 5.5 553.7027 5.6 554.1835 5.7 555.0179 5.8 556.1939 5.9 557.7007 Graph Area (A) Versus Radius (r) r A 5.38 553.6102 5.39 553.5972 5.4 553.5881 5.41 553.5827 5.42 553.5811 5.43 553.5832 5.44 553.5891 5.45 553.5988 5.46 553.6122 5.47 553.6293 5.48 553.6501 From the graph, the value of r = 5.42 h = 1000 πr² = 1000 π(5.42)² = 10.84 h = 10.84 r 5.42 = 2 Surface area is minimum, when r = 5.4, h = 10.84 and h = 2 r To verify the answer , the following method can be used . Method: Construct a Table r A &nbs p; 5 557.0796 &nb sp; The table shows that the minimum surface area occurs 5.05 556.2765 when the value of r falls between 5.4 and 5.5. Therefore 5.1 555.5825 & nbsp; the minimum surface area lines between 553.5881 and 5.15 554.9953 553.7027. 5.2 554.5127 & nbsp; 5.25 554.1327 5.3 553.8532 & nbsp; 5.35 553.6722 5.4 553.5881 & nbsp; 5.45 553.5986 5.5 553.7027 & nbsp; 5.55 553.8982 5.6 554.1835 & nbsp; 5.65 554.5573 r A &nbs p; 5.41 553.5827 To find the more precise value of r and A, the values 5.42 553.5811 of r from 5.41 to 5.54 are used. 5.43 553.5832 5.44 553.5891 From the table, the least value for surface area is 5.45 553.5988 553.5811 and it occurs when r = 5.42. 5.46 553.6122 5.47 553.6293 5.48 553.6501 5.49 553.6746 5.5 553.7027 & nbsp; 5.51 553.7345 5.52 553.77 &n bsp; 5.53 553.8091 5.54 553.8518 r A &nbs p; 5.411 553.5823 ; To find the more precise of value r and A, the values of r 5.412 553.582 From 5.411 to 5.424 are used. 5.413 553.5818 ; 5.414 553.5816 ; From the table, the least value for surface area is 553.581 5.415 553.5814 ; and it occurs when r = 5.419 5.416 553.5812 ; 5.417 553.5811 ; 5.418 553.5811 ; 5.419 553.581 5.42 553.5811 5.421 553.5811 ; 5.422 553.5812 ; 5.423 553.5813 ; 5.424 553.5815 ; Solution for Question 2 Method 1:Calculus Area of aluminium sheet used, A = 2(2r 2r) + 2πrh……….(1) πr2h = 1000………………..(2) A = 8r2 +2πr( ) =8r2 +( ) = 8r2 +2000r -1 16r -2000r -2 When A is minimum, 0. 16r-2000r -2 = 0 16r = r3 = =125 r = 5 h = = =12.73 = =2.546 Minimum area of aluminium sheet used, when r = 5, h = 12.73 and =2.546. Method 2 : Draw a Graph By plotting the surface area A, against radius, r, for which A=8r2 + 2000r -1 , the following graph is produced. r A 1 2008 2 1032 3 738.6667 4 628 5 600 6 621.3333 7 677.7143 8 762 9 870.2222 Graph Area (A) Versus Radius (r) r A 4.7 602.2519 4.8 600.9867 4.9 600.2433 5 600 5.1 600.2369 5.2 600.9354 5.3 602.0785 5.4 603.6504 5.5 605.6364 From the graph, value of r = 5. h = = =12.73 = =2.546 Minimum area of aluminium sheet used, when r = 5, h = 12.73 and =2.546. To verify the answer, the following method can be used. Method: Construct a table r A &nbs p;Table shows that the minimum 1 2008 & nbsp;surface area occurs when the 1.2 1678.187 & nbsp; value of r falls between 4.8 and 1.4 1444.251 & nbsp; 5.2. 1.6 1270.48 &n bsp; 1.8 1137.031 & nbsp; 2 1032 & nbsp; 2.1 947.8109 & nbsp; 2.4 879.4133 & nbsp; 2.6 823.3108 & nbsp; 2.8 777.0057 & nbsp; 3 738.6667 &nb sp; 3.2 706.92 &nb sp; 3.4 680.7153 & nbsp; 3.6 659.2356 & nbsp; 3.8 641.8358 & nbsp; 4 628 &n bsp; 4.2 617.3105 & nbsp; 4.4 609.4255 & nbsp; 4.6 604.0626 & nbsp; 4.8 600.9867 & nbsp; 5 600 &n bsp; 5.2 600.9354 & nbsp; 5.4 603.6504 & nbsp; 5.6 608.0229 & nbsp; r A 4.9 600.2433 4.91 600.1968 4.92 600.1553 4.93 600.1187 4.94 600.0871 4.95 600.0604 4.96 600.0386 4.97 600.0271 4.98 600.0096 4.99 600.0024 5 600 5.01 600.0024 5.02 600.0096 5.03 600.0215 To get a more precise value of A and r, a smaller range of r from 4.8 to 5.2 s used. From the table, the minimum area is 600 and this occurs when r = 5. h = = = 12.73 = =2.546 Solution for Question 3 Consider The Following Shapes (1) Equilateral Triangle Method 1:Trigonometry and Calculus tan 60o = x = r Area of PQR = 3 Area of OPQ = 3 2X r = 3 xr = 3 r2 Total area of Aluminium Sheets Used, A = 6 r2 + 2πr( ) = 6 r2 +2000r -1 12 r – 2000r -2 When A is minimum, 0. 12 r = r3 = 96.225 r = 4.582 h = =15.16 =3.309 12 + 4000r -3 (>0, because r>0) Therefore , A is minimum. Minimum area or aluminium sheet used when r = 4.582, h= 15.16 and = 3.309. Method 2: Graphical Method By using the equation A=6#8730;3r²+2000-1, graph A versus r can be plotted to determine the value of r for the minimum value of A. Graph Area (A) Versus Radius (r) r A 1 2010.392 2 1041.568 3 760.1947 4 666.272 5 659.8 6 707.4453 7 794.9223 8 915.088 9 1063.974 From the graph, minimum area occurs when r falls between 4 cm and 5 cm. The following graph is plotted using the values of r between 4.54 and 4.64 Graph Area (A) Versus Radius (r) r A 4.54 654.7244 4.55 654.7008 4.56 654.6836 4.57 654.6726 4.58 654.668 4.59 654.6695 4.6 654.6773 4.61 654.6913 4.62 654.7114 4.63 654.7377 4.64 654.7701 From the graph, area is minimum when r #8776; 4.58 cm. h = 1000 Πr² = 15.17 h = 15.17 r 4.58 = 3.312 Minimun area of aluminium sheet used when r = 4.58, h = 15.17 and h = 3.312 Method 3: Construct a Table A table can be constructed using A = 6 3 r2 + 2000r-1 r h &nbs p;A h/r 1 318.3099 &nb sp; 2010.392 318.3099 2 79.57747 1041.568 &n bsp; 39.78874 3 35.36777 &nb sp; 760.1947 11.78874 4 19.89437 &nb sp; 666.272 11.78926 5 12.7324 &nbs p; 659.8 4.973592 6 8.841941 &nb sp; 707.4453 2.546479 7 6.49612 &nbs p; 794.9223 0.928017 8 4.973592 &nb sp; 915.088 0.621699 By using a more precise range of value r, we get the following tables. r h &nbs p;A h/r 4.1 18.93575 & nbsp; 662.4944 4.618475 4.2 18.04478 & nbsp; 659.5054 4.296376 4.3 17.21525 & nbsp; 657.2644 4.003545 4.4 16.44163 & nbsp; 655.7346 3.736733 4.5 15.71901 & nbsp; 654.8824 3.493113 4.6 15.043 &nb sp; 654.6773 3.270217 4.7 14.40968 & nbsp; 655.0912 3.06589 4.8 13.81553 & nbsp; 656.0983 2.878236 r h &nbs p;A h/r 4.51 15.64938 654.8333 3.469928 4.52 15.58021 654.7906 3.446949 4.53 15.5115 & nbsp; 654.7543 3.424172 4.54 15.44324 654.7244 3.401595 4.55 15.37543 654.7008 3.379216 4.56 15.30807 654.6836 3.357033 4.57 15.24115 654.6726 3.335044 4.58 15.17467 654.668 3.313246 4.59 15.10862 654.6695 3.291638 4.6 15.043 &nb sp; 654.6773 3.270217 4.61 14.97781 654.6913 3.248982 From the above table, minimum area of aluminium sheet used when r = 4.58, h = 15.17 and the ratio of h = 3.312 ( ii ) Regular Hexagon Method 1: Trionometry and Calculus Tan 300 = x r x = r tan 300 = r 3 Area of OPQ & nbsp;= 1 x 2x x r 2 = 1 x 2( r ) x r 2 = r2 3 = 3 r2 3 Area of Hexagon = 6 x 3r2 3 = 2 #8730; 3 r2 Total Area of Aluminium Sheets Used, A = 2(2 #8730; 3 r2) + 2 π r ( 1000 ) = 4 #8730; 3 r2 + 2000r-1 dA = 8 #8730; 3 r – 2000 r-2 dr For minimum surface area, dA = 0 8 #8730;3 r = 2000 r2 r3 = 2000 8 #8730;3 = 250 #8730; 3 r = 5.246 h = 1000 πr² = 1000 π x ( 5.246 ) = 11.57 h = 11.57 r 5.246 = 2.205 Method 2: Graphical Method By using the equation A=4#8730;3r²+2000-1, graph A versus r can be plotted using the value of r from 1 cm to 11 cm. Graph Area (A) Versus Radius (r) r A 1 2006.928 2 1027.712 3 729.0187 4 610.848 5 573.2 6 582.7413 7 625.1863 8 693.392 9 783.3902 10 892.8 11 1020.106 From the graph, the minimum area occurs when the value of r is between 5 cm and 6 cm. By plotting a graph using values of r from 5.18 to 5.3, a more precise value of r can be determined. Graph Area (A) Versus Radius (r) r A 5.18 571.9953 5.19 571.9698 5.2 571.9485 5.21 571.9315 5.22 571.9187 5.23 571.9101 5.24 571.9056 5.25 571.9054 5.26 571.9093 5.27 571.9173 5.28 571.9294 5.29 571.9457 5.3 571.966 From the graph,the area is minimum when r#8776; 5.25. h =1000 πr² =11.57 h =11.57 r 5.25 =2.204 h Minimum area of aluminium sheet used when r =5.25,h=11.57 and r =2.204. Method 3 : Construct a table A table can be constructed by using equation A = 4#8730;3r²+2000r¯¹ and values of r between 1 and 8. r h A h/r 1 318.3 099 2006.928 & nbsp; 318.3099 2 79.57747 &nbs p; 1027.712 39.78874 3 35.36777 &nbs p; 729.0187 11.78926 4 19.89437 &nbs p; 610.848 4.973592 5 12.7324 ; 573.2 2.546479 6 8.841941 &nbs p; 582.7413 1.473657 7 6.49612 ; 625.1863 0.928017 8 4.973592 &nbs p; 693.392 0.621699 Value of r is around 5cm when the area is minimum.Another two tables can be constructed to obtain a more precise value of r. r h A h/r 4.8 13.81553 576.2878 2.878236 4.9 13.25739& nbsp; 574.5045 ; 2.705589 5 12.7324 ; 573.2 2.546479 5.1 12.23798& nbsp; 572.3541 ; 2.399604 5.2 11.77182& nbsp; 571.9485 ; 2.263811 5.3 11.33179& nbsp; 571.966 2.138073 5.4 10.91598& nbsp; 572.3909 ; 2.021477 5.5 10.52264& nbsp; 573.2084 ; 1.913207 r ; h &nbs p; A &nb sp; h/r 5.2 11.77182 &nbs p; 571.9485 2.263811 5.21 11.72667 571.9315 2.2508 5.22 11.68178 571.9187 2.237889 5.23 11.63715 571.9101 2.225077 5.24 11.59278 571.9056 2.212362 5.25 11.54866 571.9054 2.199744 5.26 11.50479 571.9093 2.187222 5.27 11.46117 571.9173 2.174795 5.28 11.4178 & nbsp; 571.9294 2.162461 Minimum area of aluminium sheet used when r= 5.25,h=11.55 and h =2.2. r Additional Task Considering a regular polygon with n sides . Interior angle,z = 180(n-2) N Tan z = r 2 x x = r tan z ……………………..(1) ² Area of n –sided regular polygon =nxr =nr( r ) tan z ................................... (2) ² Tota; Area of Aluminium Sheets Used, A =2nr( r )+2#12640;rh Tan z ² =2nr² +2πr²(1000) , h=1000 tan z πr² πr² ² = 2nr² + 2000 tan z r ² dA = 4nr - 2000 dr tan z r² ² ; When A is minimum ,dA = 0 Dr 4nr = 2000 tan z r² ² ; z r³ =500 tan 2 n r = 3#8730; 500tan n By varying the values of n ,the corresponding values of z, r, h and h can be found from the following table n Z=((n-2)*180)/n &n bsp; z/n Tanz/2 ; R=((500tan z/2)/n) ^(1/3) h=1000/((22/7)*(r ^2)) h/r 3 60.0 & nbsp;30.0 0.5776 &nb sp; 4.58317552 15.14 754949 3.305034 4 90.0 & nbsp;45.0 1.0006 &nb sp; 5.001053852 12.7 2190936 2.543846 5 108.0 54.0 1.3775 &n bsp; 5.164502417 11. 92939418 2.309883 6 120.0 60.0 1.7337 &n bsp; 5.24727803 11.5 5599132 2.202283 7 128.6 64.3 2.0789 &n bsp; 5.295456734 11. 34667227 2.142718 8 135.0 67.5 2.4175 &n bsp; 5.326142937 11. 21630278 2.105896 9 140.0 70.0 2.7517 &n bsp; 5.34698339 11.1 2903964 2.081368 10 144.0 ; 72.0 3.0830 & nbsp; 5.361841508 11 .06744607 2.044113 11 147.3 ; 73.6 3.4122 & nbsp; 5.372848674 11 .02214555 2.051453 12 150.0 ; 75.0 3.7399 & nbsp; 5.381261993 10 .9877074 2.041846 13 152.3 ; 76.2 4.0665 & nbsp; 5.387863064 10 .96080018 2.03435 14 154.3 ; 77.1 4.3923 & nbsp; 5.393159096 10 .93928395 2.028363 15 156.0 ; 78.0 4.7173 & nbsp; 5.397491134 10 .92173121 2.023483 16 157.5 ; 78.8 5.0419 & nbsp; 5.401095477 10 .90715915 2.019435 17 158.8 ; 79.4 5.3661 & nbsp; 5.404140104 10 .89487269 2.016023 18 160.0 ; 80.0 5.6900 & nbsp; 5.406747113 10 .88436871 2.013109 19 161.1 ; 80.5 6.0136 & nbsp; 5.409006999 10 .87527562 2.010586 20 162.0 ; 81.0 6.3371 & nbsp; 5.410987996 10 .86731406 2.008379 21 162.9 ; 81.4 6.6604 & nbsp; 5.412742349 10 .86027068 2.006427 22 163.6 ; 81.8 6.9836 & nbsp; 5.414310607 10 .85398022 2.004684 23 164.3 ; 82.2 7.3068 & nbsp; 5.41572464 10. 84831306 2.003114 24 165 & nbsp;82.5 7.6299 &nb sp; 5.417009773 10.8 4316636 2.001689 Conclusion The above table shows that as the number of sides of a regular polygon increases,the ratio of h/r approaches 2. 1. Material wastage is reduced if the number of the sides of the aluminium sheet increases (regular polygon shapes) 2. Geometrical shapes that can be fitted into each other without any space in between are a) equilateral triangles b) squares c) regular hexagons 3. Therefore,circles must be cut from the geometrical shapes above to reduce wastage. Among these geometrical shapes,cutting circles from regular hexagons will result in the least wastage of the aluminium sheets.