Question

A closed box, whose lenght is twice its width, is to have a surface area of 192 square inches. Find the dimensions of the box when the volume is maximum

Answer 1

According to the question, the closed box represents a cuboid, which has a top as it is referred to as a closed box.

Let width of the box be $x$ inches. Then the length of the box is 2$x$ inches.

Now, using the formulae for the total surface area($S$) of a cuboid we get:-

$$S=2(lb+bh+lh)\\ \Rightarrow 2(2x.x+x.h+2x.h)=192\\ \Rightarrow 2(2x^2+xh+2xh)=192\\ \Rightarrow 2x^2+3xh=96\\ \Rightarrow 3xh=96-2x^2\\ \Rightarrow h=\frac{96}{3x}-\frac{2x^2}{3x}\\~\\ \Rightarrow h=\frac{32}{x}-\frac{2x}{3}$$

So we get the height of the box, in terms of $x$ .

Now, we have to have the volume($V$) to be maximum.

$$V=lbh\\~\\ \Rightarrow V=2x.x.(\frac{32}{x}-\frac{2x}{3})\\~\\ \Rightarrow V=2x^2(\frac{32}{x}-\frac{2x}{3})\\~\\ \Rightarrow V=64x-\frac{4x^3}{3}\\~\\$$

Now, after differentiating both sides wrt x, we get:-

$$\frac{dV}{dx}=64-4x^2\\~\\$$

Again differentiating both sides wrt x, we get:-

$$\frac{d^2V}{dx^2}=-8x$$

Now, as we clearly see that, $\frac{d^2V}{dx^2}$ is always negative (as x is a dimension and cannot be negative), so $\frac{dV}{dx}=0$ , will give us a value of x which makes V maximum.

Now,

$$\frac{dV}{dx}=0\\ \Rightarrow 64-4x^2=0\\ \Rightarrow 4x^2=64\\ \Rightarrow x^2=16\\ \Rightarrow x=\pm4$$

Here, we neglect the value $x=-4$ as we know, that the width of a box cannot be negative.

So, we have $x=4$ , when the volume of the box is maximum.

Therefore the dimensions of the box are:-

length = $2x=2(4)=8$ inches.

width = $x=4$ inches.

height = $\frac{32}{x}-\frac{2x}{3}=\frac{32}{4}-\frac{2(4)}{3}=8-\frac{8}{3}=\frac{16}{3}$ inches.

and, the maximum volume(V) of the box is:- $64(4)-\frac{4(4)^3}{3}=256-\frac{256}{3}=\frac{512}{3}$ cubic inches.