Question:
A cardboard box (without a top) is to be constructed from a $30 \text{ cm} \times 30 \text{ cm}$ square sheet of cardboard by cutting 4 square pieces of the same size from the corners of the sheet, followed by folding up the sides. Determine the dimensions of the box with the largest possible volume that can be constructed.
My attempt:
Given that a square of dimension $20 \text{ cm} \times 20 \text{ cm}$ has its four corners cut off squarely such that the dimension of each square corner is $x \text{ cm} \times x \text{ cm}$. Then, the volume of the box formed in the process when folded up is:
$$V = (30 – 2x)^2(x)$$
where $L =$ length $= 30 – 2x$, $B =$ breadth $= 30 – 2x$, and $H =$ height $= x$.
For maximum volume of the box, a condition must hold as shown below through the chain rule:
$$\frac{dv}{dx} = 0$$
Now, differentiating volume, $V(x)$, with respect to $x$:
$$V'(x) = (30 – 2x)^2 – 4x(30 – 2x) = 0$$
$$V'(x) = 30 – 2x – 4x = 0$$
$$30 = 6x$$
$$x = 5 \text{ cm}$$
Also, the breadth and length are each of length $30 – 2(5) \text{ cm} = 30 – 10 \text{ cm} = 20 \text{ cm}$.
Therefore, for greatest volume of the box, the dimension of the box must be $20 \text{ cm} \times 20 \text{ cm} \times 5 \text{ cm}$.
Am I right ?
Best Answer
Your answer is correct, however, there is a mistake in the calculations. Indeed, the volume is given by $V(x)=(30-2x)^2 x$, but the derivative equals $$ \frac{d}{dx}V(x)=(30-2x)^2-4x(30-2x)=12x^2 - 240x + 900=12(5-x)(15-x). $$ Equating $V'(x)=0$ gives you thus two possible solutions. However, since $$ \frac{d^2}{dx^2}V(x)=24(x-10) $$ and you are looking for maximum (so that $V''(x)<0$) the correct solution is $x=5$.