Review question

# When is the volume of this pyramid a maximum? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9905

## Solution

A right square pyramidal frame is formed from a rod of length $4a$ by dividing the rod into eight pieces, four being of length $x$ (to make the edges of the square base) and the other four of length $a - x$ (to make the sloping edges). The rods are jointed together without wastage. If the volume of the pyramid is $V$, and $x$ may vary, show that $\dfrac{dV}{dx} = 0$ when $x = 0$ and when $3x^2 - 10ax + 4a^2 = 0.$

The volume $V$ of a right square pyramid is given by the equation $V = \frac{1}{3} \times \text{area of the base} \times \text{height}.$ As the square base comprises four rods, each of length $x$, the area of the base is $x^2$. For the height, consider the following image.

We wish to calculate $h$. By Pythagoras’s theorem, we have \begin{align*} h^2 + \left( \frac{x}{\sqrt{2}} \right)^2 = (a - x)^2 &\implies h^2 = a^2 - 2ax + x^2 - \frac{x^2}{2} \\ &\implies h = \sqrt{a^2 - 2ax + \frac{x^2}{2}}. \end{align*} Thus, we have that \begin{align*} V &= \frac{1}{3} \times x^2 \times \sqrt{a^2 - 2ax + \frac{x^2}{2}} \\ &= \frac{1}{3} \sqrt{ a^2x^4 - 2ax^5 + \frac{x^6}{2} }. \end{align*} By the chain rule, \begin{align*} \frac{dV}{dx} &= \frac{1}{6} \left( a^2x^4 - 2ax^5 + \frac{x^6}{2} \right)^{-1/2} \times \left( 4a^2 x^3 - 10ax^4 + 3x^5 \right) \\ &= \frac{x^3(3x^2 - 10ax + 4a^2)}{6\sqrt{a^2x^4 - 2ax^5 + \frac{x^6}{2}}} \\ &= \frac{x(3x^2 - 10ax + 4a^2)}{6\sqrt{a^2 - 2ax + \frac{x^2}{2}}}. \end{align*}

So we can see that, $\dfrac{dV}{dx} = 0$ when $x = 0$ or $3x^2 - 10ax + 4a^2 = 0$.

Strictly, we should be concerned about the cases when the denominator is close to zero because then our expression is ill-defined. But note that $\sqrt{a^2 - 2ax + \frac{x^2}{2}}$ is the height of the pyramid, and as this approaches zero the volume will be changing rapidly so we can be sure that $\dfrac{dV}{dx} \neq 0$ at this point.

Show that, when $x$ is equal to the smaller of the two roots of this equation,

1. a real pyramid can be formed,

From the quadratic formula, $3x^2 - 10ax + 4a^2 = 0 \iff x = \frac{10a \pm \sqrt{100a^2 - 48a^2}}{6} = \frac{10 \pm \sqrt{52}}{6} a = \frac{5 \pm \sqrt{13}}{3} a.$

The smaller of these roots is $x=\dfrac{5 - \sqrt{13}}{3} a$.

In order to form a real pyramid, $x$ must be between $0$ and $a$. Also, the sloping diagonal, $a-x$, must be longer than the semi-diagonal of the base, $\dfrac{x}{\sqrt{2}}$.

Since $5>\sqrt{13}$, we know that $x$ is positive. We also require \begin{align*} a-x &> \frac{x}{\sqrt{2}} \\ \iff\quad x\left(1+\frac{1}{\sqrt{2}}\right) &< a \\ \iff\quad x &< \left(2-\sqrt{2}\right)a \end{align*}

Comparing these surd expressions directly is tricky, so let’s use a calculator to check. $\dfrac{5 - \sqrt{13}}{3} \approx 0.465 \quad\text{and}\quad 2-\sqrt{2} \approx 0.586$ so the diagonal condition is satisfied and we do have a real pyramid.

1. $V$ has a maximum value.

As $x \to 0$, the square base goes to a point and the volume of the pyramid shrinks to zero.

As $x \to \left(2-\sqrt{2}\right)a$, the sloping diagonal becomes the same length as the semi-diagonal of the base. The height and therefore the volume of the pyramid shrinks to zero.

In between these values, a real pyramid exists and $V$ is positive.

We know that in this range, we have a single stationary point, which must therefore be a maximum.

Another way to check that the volume is maximised at this value is to compute the second derivative, and to verify that it is negative at this value, but this would be demanding…

Note that the question doesn’t ask us to calculate the maximum value of $V$ in terms of $a$, but it’s not hard to do with a calculator – we find $V_{max} \approx 0.0304a^3$.