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