For the solid described below, find the dimensions giving the minimum surface area, given that the volume is 64 cm3. a closed cylinder with radius r cm and height h cm.

The volume of the cylinder is:
 V = pi * r ^ 2 * h = 64
 The surface area is:
 A = 2 * pi * r ^ 2 + 2 * pi * r * h
 We write the area as a function of r:
 A (r) = 2 * pi * r ^ 2 + 2 * pi * r * (64 / (pi * r ^ 2))
 Rewriting:
 A (r) = 2 * pi * r ^ 2 + 2 * (64 / r)
 A (r) = 2 * pi * r ^ 2 + 128 / r
 Deriving:
 A '(r) = 4 * pi * r - 128 / r ^ 2
 We equal zero and clear r:
 0 = 4 * pi * r - 128 / r ^ 2
 128 / r ^ 2 = 4 * pi * r
 r ^ 3 = 128 / (4 * pi)
 r = (128 / (4 * pi)) ^ (1/3)
 r = 2.17 cm
 The height is:
 h = 64 / (pi * r ^ 2)
 h = 64 / (pi * (2.17) ^ 2)
 h = 4.33 cm
 Answer:
 The dimensions giving the minimum surface area are:
 r = 2.17 cm
 h = 4.33 cm