You have a piece of sheet metal. Your task is to use this material to create a container with a maximum volume?

a) which shape would have the greatest volume: a squared based prism or a cylinder?





b) Justify you answer using a fixed surface area of 3500cm squaresYou have a piece of sheet metal. Your task is to use this material to create a container with a maximum volume?
This one is tricky, its actually easier to do b that a, so i will show you b. Then i will show a without substituting a value





Firstly given a surface area A


Prism:


1 square surface = A/6


one side = 鈭?A/6)





Volume = side^3 = [鈭?A/6)]^3 = (A/6)^(3/2)





If Suface Area = 3500cm^2


Volume = (3500/6)^(3/2) = 14 088.8385cm^3


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Cylinder:


A = 2蟺r^2 + 2蟺rh





Now A is set, but r and h depend on each other, we need an equation for h in terms of r





A = 2蟺r^2 + 2蟺rh





A - 2蟺r^2 = 2蟺rh





h = (A - 2蟺r^2)/2蟺r





now


Volume = 蟺r^2h


eliminate h by substituting the equation for h





Volume = 蟺r^2(A - 2蟺r^2)/2蟺r





simplify/expand





Volume = (Ar - 2蟺r^3)/2





Now we need to find what radius will maximise are Volume


we need to find the derivative V '





V = (Ar - 2蟺r^3)/2





V ' = ( A - 6蟺r^2)/2





Max/Min when V ' = 0





( A - 6蟺r^2)/2 = 0





A = 6蟺r^2





r = 鈭?A/6蟺)





But is it a max or a min, find V ' ' to find out id V ' '(鈭?A/6蟺)) %26lt; 0 its a max





V ' '= -12蟺r/2





V ' '(鈭?A/6蟺)) = -12蟺(鈭?A/6蟺)) /2 %26lt; 0 therefore max volume when r = (鈭?A/6蟺))





So max volume


r = r = (鈭?A/6蟺))


V = (Ar - 2蟺r^3)/2





V = (A鈭?A/6蟺) - 2蟺(鈭?A/6蟺)^3)/2





When A = 3500





V = (3500鈭?3500/6蟺) - 2蟺(鈭?3500/6蟺)^3)/2 = 15897.55186 cm^3





So the cylinder has a greater Volume


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EDIT:





Now part a) Compare without substituting 3500cm


V(Prism) = (A/6)^(3/2) = (A/6)^(1.5)





V(Cylinder) = (A(A/6蟺)^0.5 - 2蟺(A/6蟺)^(1.5) )/2





Comparing these two is too difficult we need to simplify





V(Cylinder) = (A(A/6蟺)^0.5 - 2蟺(A/6蟺)^(1.5) )/2





V(Cylinder) = ( A^1.5 (6蟺)^-0.5 - 2蟺 (6蟺)^-1.5 A^(1.5) )/2





V(Cylinder) = A^1.5 [ (6蟺)^-0.5 - 2蟺 (6蟺)^-1.5 ]/2





V(Cylinder) = A^1.5 x (0.076776)


(correct to 6 decimal places)





V(Prism) = (A/6)^(1.5) = A^1.5 (0.068041)


(correct to 6 decimal places)





And from here you can see V(Prism) %26lt; V(Cylinder)