When is the area of a face of this pyramid a minimum?

When is the area of a face of this pyramid a minimum?

Question

A pyramid of given volume $V$ stands on a horizontal square base of edge $2x$, and its vertex is vertically above the centre of the base. Show that the area $A$ of a sloping triangular face is given by $A^2=\frac{9V^2}{16x^2}+x^4,$ and prove that, as $x$ varies, $A$ is least when the face is equilateral.

Suggestion

What is the formula for the volume of a pyramid?

Sketch a square based pyramid and label its height $h$. Can we now find the area of one of the triangular faces?

Can we create any right-angled triangles where we can use Pythagoras?

…and prove that, as $x$ varies, $A$ is least when the face is equilateral.

Is minimising $A^2$ the same as minimising $A$? How could we go about minimising $A^2(x)$?

Where is the stationary point of $A^2?$

How do we know the stationary point we’ve found is a minimum?

Can we show that for this value of $x$ we have equilateral faces?

Solution

A pyramid of given volume $V$ stands on a horizontal square base of edge $2x$, and its vertex is vertically above the centre of the base. Show that the area $A$ of a sloping triangular face is given by $A^2=\dfrac{9V^2}{16x^2}+x^4,$

The volume $V$ of a square based pyramid is $V=\dfrac{1}{3}bh,$ where $b$ is the area of the base and $h$ is the height.

Where does this formula come from?

To see where this formula comes from (it holds for pyramids with any base, and cones too) consider a unit cube:

Now draw in the four diagonals from each of the pairs of opposite corners like this:

This makes six identical square based pyramids from each of the faces:

Each pyramid has a base of area one and is of height $1/2$. Each one must also have a volume of $1/6$, which is $\dfrac{1}{3}bh$.

We can extend this for cubes of a different side length, and cuboids.

Our pyramid looks like this:

So we have $V=\frac{1}{3}(2x)^2h.$

We need to find the area of a sloping triangular face (which will be the same for all four faces, since the vertex is directly above the centre of the base).

For this, we need the height $y$ of the face, which is related to $h$ and $x$ via $y^2=h^2+x^2,$ which we can see from taking a vertical cross-section of the pyramid.

The area of the face is $A=\frac{1}{2}(2x)y=xy$.

Therefore, $A^2=x^2y^2=x^2(h^2+x^2)=x^2\left(\left(\frac{3V}{4x^2}\right)^2+x^2\right)=x^2\left(\frac{9V^2}{16x^4}+x^2\right)=\frac{9V^2}{16x^2}+x^4,$ as required.

…and prove that, as $x$ varies, $A$ is least when the face is equilateral.

We can say that $A$ has a turning point if and only if $A^2$ does, since we know $A$ is positive and $y=x^2$ is an increasing function for $x > 0$.

If you know the Chain Rule for differentiating, you might think about $\dfrac{d(A^2)}{dx} = 2A\dfrac{dA}{dx}$.

So considering $\frac{d(A^2)}{dx}=4x^3+\frac{9V}{16}\frac{-2}{x^3}=4x^3-\frac{9V^2}{8x^3},$ we find stationary points when $\dfrac{d(A^2)}{dx}=0.$ $\Longrightarrow \qquad 4x^3-\frac{9V^2}{8x^3}=0 \qquad \Longrightarrow \qquad 32x^6=9V^2.$ There will be two real roots of this equation, one negative (which doesn’t interest us) and one positive, so there is one stationary point in the $x>0$ region.

We know $\frac{d(A^2)}{dx}=\frac{32x^6 - 9V^2}{8x^3},$ so if $x$ increases slightly from $\left(\dfrac{9V^2}{32}\right)^\frac{1}{6}$, then $x^6$ increases slightly from $\dfrac{9V^2}{32}$, (since $y = x^6$ is an increasing function for $x > 0$), which tells us that the gradient of $y = A^2$ is positive. Similarly if $x$ decreases slightly from $\left(\dfrac{9V^2}{32}\right)^\frac{1}{6}$, the gradient is negative.

So our single stationary point for $x > 0$ must be a local minimum, and we know that $A^2$ (and hence $A$) is minimized when $x^6=\dfrac{9V^2}{32}$.

Now $h^2 = \dfrac{9V^2}{16x^4}=\dfrac{9V^2}{32x^6}2x^2$, and so if $x^6 = \dfrac{9V^2}{32}$, then $h^2 = 2x^2$, and $y^2 = x^2 + h^2 = 3x^2$.

So $y = \sqrt{3}x$, and each face must be an equilateral triangle.