When is the volume of this cuboid a maximum?

When is the volume of this cuboid a maximum?

Question

A cuboid has a total surface area of $150\,\text{cm}^{2}$ and is such that its base is a square of side $x\,\text{cm}$.

Show that the height, $h\,\text{cm}$, of the cuboid is given by $h=\frac{75-x^2}{2x}$.

Express the volume of the cuboid in terms of $x$. Hence determine, as $x$ varies, its maximum volume and show that this volume is a maximum.

Suggestion

How can we find the surface area of a cuboid from its sides?

Can we write the volume of the cuboid in terms of $x$ and $h$? Can we now substitute for $h$?

How can we find the maximum value?

This applet shows how the shape of the cuboid changes as we vary $x$. The surface area is fixed and the length of the red line is proportional to the volume.

[interactive embed (geogebra-iframe): //www.geogebratube.org/material/iframe/id/qG6ubfmq/width/700/height/400/border/888888/rc/false/ai/false/sdz/false/smb/false/stb/false/stbh/false/ld/false/sri/true/asb/false/at/html5] — not part of the text import; see the source page.

Download GeoGebra file

Solution

A cuboid with side lengths $x$, $x$ and $h$ has a surface area of$S=2x^2+4xh.$

We know that the total surface area should be $150\,\text{cm}^{2}$, so

$$\begin{align*} 150 &= 2x^2+4xh\\ \implies 150 -2x^2 &= 4xh\\ \implies h &= \frac{75-x^2}{2x}. \end{align*}$$

A cuboid with side lengths $x$, $x$ and $h$ has a volume of $V=x^2h$. Now

$$\begin{align*} V &= x^2h\\ &= x^2\bigl(\frac{75-x^2}{2x}\bigr)\\ &= \frac{75}{2}x-\frac{1}{2}x^3. \end{align*}$$

We have $V$ as a function of $x$, so to find the stationary points we differentiate, and set the result equal to zero. We have

$$V' (x) = \frac{75}{2} - \frac{3}{2} x^2.$$

Putting this equal to zero, we see

$$\begin{align*} \frac{75}{2} - \frac{3}{2} x^2 &=0\\ \implies \frac{75}{2} &= \frac{3}{2} x^2 \\ \implies 25 &= x^2 \\ \implies x &= \pm5. \end{align*}$$

Since $x$ is a length, it must be positive, implying $x=5$.

Is our solution a maximum? We know $V = 0.5x(75-x^2)$, and so the equation $V=0$ has three roots, $0$ and $\pm \sqrt{75}$. Given the coefficient of $x^3$ is negative, the curve must look like this:

Thus $x = 5$ gives us a maximum for $V$.

Or we could show $x = 5$ gives a maximum by evaluating the second derivative of $V$, giving

$$\begin{align*} V''(x)&=-3x\\ \implies V''(5) &=-15. \end{align*}$$

The second derivative is negative at $x=5$, so the volume has a maximum here, as required.

Now all that remains is to find the volume of the cuboid when $x=5$, which is given by

$$\begin{align*} V(x) &= \frac{75}{2}x-\frac{1}{2}x^3 \\ \implies V(5) &= \frac{75}{2} \times 5 - \frac{1}{2} \times 5^3 \\ &= \frac{375}{2} - \frac{125}{2} \\ &= \frac{250}{2} \\ &= 125. \end{align*}$$

Therefore the maximum volume of the cuboid is $125\,\text{cm}^{3}$.