Asked by Diandra
When making a cylindrical can with a volume of 500ml , what dimensions will minimize the material to produce each can
Answers
Answered by
oobleck
πr^2 h = 500, so h = 500/(πr^2)
A = 2πr^2 + 2πrh = 2π(r^2 + 500/(πr))
dA/dr = 2π(2r - 500/(πr^2)) = (4πr^3 - 1000)/πr^2
dA/dr = 0 when r = 5∛(2/π), so h = 10∛(2π)
A = 2πr^2 + 2πrh = 2π(r^2 + 500/(πr))
dA/dr = 2π(2r - 500/(πr^2)) = (4πr^3 - 1000)/πr^2
dA/dr = 0 when r = 5∛(2/π), so h = 10∛(2π)
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