A piece of cardboard is 15 inches by 30 inches. A square is to be cut from each corner and the sides folded up to make an open-top box. What is the maximum possible volume of the box? Round your answer to the nearest four decimal places.

A rectangular piece of cardboard of side 15 inches by 30 inches is cut in such that a square is cut from each corner. Let x be the side of this square cut. When it was folded to make the box the height of box becomes x, length becomes (30-2x) and the width becomes (15-2x).

Volume is given by

V = [tex]V = Length\times Width\times Height\\V = (30 - 2x)(15-2x)x= 4x^3-90x^2+450x\\So,\\V(x) = 4x^3-90x^2+450x[/tex]

First, we differentiate V(x) with respect to x, to get,

[tex]\frac{d(V(x))}{dx} = \frac{d(4x^3-12x^2+9x)}{dx} = 12x^2 - 180x +450[/tex]

Equating the first derivative to zero, we get,

[tex]\frac{d(V(x))}{dx} = 0\\\\12x^2 - 180x +450 = 0[/tex]

Solving, with the help of quadratic formula, we get,

[tex]x = \displaystyle\frac{5(3+\sqrt{3})}{2}, \frac{5(3-\sqrt{3})}{2}[/tex],

Again differentiation V(x), with respect to x, we get,

[tex]\frac{d^2(V(x))}{dx^2} = 24x - 180[/tex]

At x =

[tex]\displaystyle\frac{5(3-\sqrt{3})}{2}[/tex],

[tex]\frac{d^2(V(x))}{dx^2} < 0[/tex]

Thus, by double derivative test, the maxima occurs at

x = [tex]\displaystyle\frac{5(3-\sqrt{3})}{2}[/tex] for V(x).

Thus, largest volume the box can have occurs when [tex]x = \displaystyle\frac{5(3-\sqrt{3})}{2}}[/tex].

Maximum volume =

[tex]V(\displaystyle\frac{5(3-\sqrt{3})}{2}) = (30 - 2x)(15-2x)x = 649.5191\text{ cubic inches}[/tex]