# Calculus

A cylindrical drum is made to hold exactly 1m^3 in its interior. Assume that the material for the top and the bottom costs \$20 per m^2, while that for the side costs \$10 per m^2. Determine the radius of the drum that minimizes the cost of the material used.

Any help will be appreciated!

Area= 2PI r^2 + 2PI * r * h
Cost= 20*2PI r^2 + 10*2PI * r * h
but PI r^2 h is = 1m^3

so solve for h in the last equation, put that in the second, then take dCost/dr, set to zero, and solve.

so basically from what im understanding
(1) Area = 2PIr^2 + 2PI*r*h
h = -2PIr^2/(2PI*r)

But Im not understanding the part about where the Area variable goes, i understand subbing the (1) eqn into the second eqn, but I don't see what happens to the Area constant

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