Answer: \(592.59 ~in^3\)

Solution:

Volume of the box, \(V = lwh\)

\(V = (20-2x)(20-2x)x\)

\(V = (400 + 4x^2 - 80x)(x)\)

\(V = 400x + 4x^3 - 80x^2\)

Maximize:

\(\frac{dV}{dx} = 400 + 12x^2 - 160x\)

\(0 = 400 + 12x^2 - 160x\)

\(x = \frac{-(-160)- \pm \sqrt{160^2 - 4(12)(400)}}{2(12)}\)

\(x = \frac{160 \pm 180}{24}\)

\(x_1 = 10\)

\(x_2 \approx 3.333\)

10 is not a possible value. Since if \(x = 10\) then \(20-2x = 0\), thus the the box has no width.

Also: \(V'' = 24x-160\)

At \(x = 10\); \(V '' >0\), thus the volume is at minimum.

At \(x \approx 3.33\); \(V '' <0\), thus the volume is at maximum.

\(V = (20-2(3.333))(20-2(3.333))(3.333)\)

\(V = 592.593 ~in^3\)