Calculus problem

Question

What is the maximum volume in cubic inches of an open box to be made from a 16-inch by 30-inch piece of cardboard by cutting out squares of equal sides from the four corners and bending up the sides? The work must include a statement of the function and its derivative.

Answer 1

Let x = length of a side of each square

V(x) = volume = x(16 - 2x)(30 - 2x) = x(4x² - 92x + 480) = 4x³ - 92x² + 480x, 0 < x < 8

V'(x) = 12x² - 184x + 480 = 4(3x² - 46x + 120) = 4(3x - 10)(x - 12)

V'(x) = 0 when x = 12 or 10/3. Since x can't be 12, x = 10/3.

When 0 < x < 10/3, V'(x) > 0. So, V(x) is increasing

When 10/3 < x < 8, V'(x) < 0. So, V(x) is decreasing.

V(x) has a relative (and absolute) maximum when x = 10/3

Maximum volume = V(10/3) = (10/3)(28/3)(70/3) = 19,600 / 27 in² ≈ 725.9 in²