A manufacturer wants to make a closed cylindrical tube out of 6pi ft^2 of cardbo

Question

A manufacturer wants to make a closed cylindrical tube out of 6pi ft^2 of cardboard. The tube has length L feet and the radius of the end cap is r feet. Find the values of L and r such that volume enclosed is at a maximum. Please use calculus.

Explanation / Answer

The surface area of the tube must satisfy: 6pi f^2 = SA => 2*pi*r^2 + 2*pi*r*L = 6pi Solving for L: 2*pi[r^2 + rL] = 6pi => r^2 + rL = 3 => rL = 3 - r^2 => L = (3/r) - r So we must maximize volume which is: V = pi*L*r^2 => V = pi*[(3/r) - r]*r^2 => V = pi*[3r - r^3] => V = 3pi*r - pi*r^3 => V' = 3pi - 3pi*r^2 => 0 = 3pi - 3pi*r^2 => 3pi*r^2 = 3pi => r^2 = 1 => r = 1 So L = (3/1) - 1 = 3 - 1 = 2 Thus the values of L and r such that the volume enclosed is at a maximum are: L = 2 r = 1

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