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.

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

A cube.

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

A cube with each pair of opposide corners linked by a dotted line.

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

A cube with each pair of opposide corners linked by a dotted line. These lines all meet in the middle and split the inside of the cube into six square based pyramids coming from each face.

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:

Square based pyramid with base side length 2 x and perpendicular height h.

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 pyramid as before but with a shaded isosceles triangle marked out with its vertices being the apex of the pyramid and the midpoints of two opposite sides of the square base. The shaded triangle has height h and base 2x and other sides length y.

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

The same diagram as the previous but with one of the triangular faces of the pyramid shaded blue. This triangular face has height y and base 2x.

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.