A close rectangular

Question

A closed rectangular container with a square base is to have a volume of 2250 in3. The material for the top and bottom of the container will cost per $4 in2, and the material for the sides will cost per $2 in2. Find the dimensions of the container of least cost.

Answer 1

As the base is square, let the length and width = b; thus, as the height is h,

V = b²h, and C = 2b²(4) + 4bh(2) = 8b² + 8bh

We need to minimize 8b² + 8bh subject to b²h = 2250, or h = 2250/b²

Minimize C = 8b² + 8bh = 8b² + 8b(2250/b²) = 8b² + 18000/b

dC/db = 16b - 18000/b²

(Note: d²C/db² = 16 + 36000/b³ > 0 on (0, ∞), so if the derivative equals 0 on (0, ∞), this will be a minimum.

16b - 18000/b² = 0

16b³ = 18000

b³ = 18000/16 = 1125

b = 5 * 3^2/3, so

l = 5 * 3^2/3

w = 5 * 3^2/3

h = 2250/b² = 2250/( 5 * 3^2/3)² = 10 * 3^2/3

As a check, b²h = (5 * 3^2/3)²*(10 * 3^2/3) = 25 * 3^4/3 * 10 * 3^2/3 = 250 * 3^6/3 = 250 * 3² = 250 *9 = 2250