Question

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.

Answer 1

Answer:

Maximum volume = 649.519 cubic inches

Step-by-step explanation:

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

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

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

Equating the first derivative to zero, we get,

$\frac{d(V(x))}{dx} = 0\\\\12x^2 - 180x +450 = 0$

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

$x = \displaystyle\frac{5(3+\sqrt{3})}{2}, \frac{5(3-\sqrt{3})}{2}$,

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

$\frac{d^2(V(x))}{dx^2} = 24x - 180$

At x =

$\displaystyle\frac{5(3-\sqrt{3})}{2}$,

$\frac{d^2(V(x))}{dx^2} < 0$

Thus, by double derivative test, the maxima occurs at

x = $\displaystyle\frac{5(3-\sqrt{3})}{2}$ for V(x).

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

Maximum volume =

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