Asked by Thomas
A steel factory wishes to make a cylindrical container, of thin metal, to hold 100cm³, using the least possible area of metal. If the outside surface is S cm² and the radius is r cm, show that S = 2πr² + 200r-1 and hence find the required radius and height for the container. (Leave π in your answer.)
Answers
Answered by
Reiny
let the height be 100 cm^3
πr^2 h = 100
h = 100/(πr^2)
S = 2 circles + the sleeve
= 2πr^2 + 2πrh
= 2πr^2 + 2πr(100/(πr^2))
= 2πr^2 + 200/r
you typed it wrong, it should have been
2πr^2 + 200 r^-1
dS/dr = 4πr - 200/r^2
= 0 for a min of S
4πr = 200/r^2
r^3 = 200/(4π) = 25/π
r = (25/π)^(1/3) , the cuberoot of (25/π)
h = 100/(πr^2)
= I will let you sub in the value of r
πr^2 h = 100
h = 100/(πr^2)
S = 2 circles + the sleeve
= 2πr^2 + 2πrh
= 2πr^2 + 2πr(100/(πr^2))
= 2πr^2 + 200/r
you typed it wrong, it should have been
2πr^2 + 200 r^-1
dS/dr = 4πr - 200/r^2
= 0 for a min of S
4πr = 200/r^2
r^3 = 200/(4π) = 25/π
r = (25/π)^(1/3) , the cuberoot of (25/π)
h = 100/(πr^2)
= I will let you sub in the value of r
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