Asked by Anonymous
an open too cylindrical container is to have a volume 2744cm^3 what dimensions will minimize the surface area
Answers
Answered by
oobleck
v = πr^2 h, so h = 2744/(πr^2)
The area is
A = 2πr^2 + 2πrh = 2πr^2 + 2πr*2744/(πr^2) = 2πr^2 + 5488/r
set dA/dr = 0 to find that
r = 7∛(4/π)
so h = 2744/(πr^2) = 2744/(π(7∛(4/π))^2) = 14∛(4/π)
The area is
A = 2πr^2 + 2πrh = 2πr^2 + 2πr*2744/(πr^2) = 2πr^2 + 5488/r
set dA/dr = 0 to find that
r = 7∛(4/π)
so h = 2744/(πr^2) = 2744/(π(7∛(4/π))^2) = 14∛(4/π)
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