Question

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.

Answer 1

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