Question

An open box of maximum volume is to be made from a square piece of cardboard, 24 inches on each side, by cutting equal squares from the corners and turning up the sides to make the box.
(a) Express the volume V of the box as a function of x, where x is edge length of the square cut-outs.
(b) What are the dimensions of the box that enclose the largest possible volume? State your answer in the form length by width by height.
(c) What is the maximum volume?

Answer 1

A.) Let the length of the sides of the bottom of the box be y and z, and let the length of the sides of the square cut-outs be x, then
V = xyz . . . (1)
2x + y = 24 => y = 24 - 2x . . . (2)
2x + z = 24 => z = 24 - 2x . . . (3)

Putting (2) and (3) into (1), gives:
V = x(24 - 2x)(24 - 2x) = x(24 - 2x)^2 = x(576 - 96x + 4x^2)
V = 4x^3 - 96x^2 + 576x

b.) For maximum volume, dV/dx = 0
dV/dx = 12x^2 - 192x + 576 = 0
x^2 - 16x + 48 = 0
(x - 4)(x - 12) = 0
x = 4 or x = 12
but x = 12 is unrearistice
Therefore, x = 4.
y = z = 24 - 2(4) = 24 - 8 = 16

Therefore, the dimensions of the box that enclose the largest possible volume is 16 inches by 16 inches by 4 inches.

c.) Maximum volume = 16 x 16 x 4 = 1024 cubic inches.