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.