Calculus, 2ADV 2026 HSC 30

A pyramid \(MABCD\) has its vertex \(M\) vertically above \(O\), the centre of its square base \(ABCD\), as shown. \(AM=\ell\) cm and makes an angle of \(\theta\) with the base.

Let \(V\) be the volume of the pyramid.
Show that \(V=\dfrac{2\ell^3}{3}\sin\theta\,\cos^2\theta\). (3 marks)
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Hence, find the value of \(\theta\), in radians, for which the volume of the pyramid is a maximum. Give your answer correct to 2 decimal places. (4 marks)
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a. \(\text{See worked solutions}\)

b. \(\theta=0.62 \)

Show Worked Solution

a. \(\text{In}\ \triangle MOA:\)

\(\sin \theta = \dfrac{MO}{\ell}\ \ \Rightarrow \ \ MO=\ell\sin\theta\ \ \text{(height)}\)

\(AO=\ell\cos\theta\ \ \Rightarrow\ \ AC=2\ell\cos\theta\ \ \text{(diagonal of base)}\)

\(\text{Using Pythagoras in}\ \triangle ABC\ \ (AB=BC):\)

\(AB^2+BC^2\)	\(=AC^2\)
\(2AB^2\)	\(=4\ell^2\cos^2\theta\)
\(AB^2\)	\(=2\ell^2\cos^2\theta\ \ \text{(area of base)}\)

\(V\)	\(=\dfrac{1}{3}Ah\)
	\(=\dfrac{1}{3} \times 2\ell^2\cos^2\theta \times \ell\sin\theta\)
	\(=\dfrac{2\ell^3}{3}\sin\theta\cos^2\theta\)

b. \(\text{Using the product rule}\ \ \left(0<\theta<\dfrac{\pi}{2}\right):\)

\(\dfrac{dV}{d\theta}\)	\(=\dfrac{2\ell^3}{3}\big(\cos\theta \cdot \cos^2\theta+\sin\theta \cdot 2\cos\theta(-\sin\theta)\big)\)
	\(=\dfrac{2\ell^3}{3}\cos\theta\left(\cos^2\theta-2\sin^2\theta\right)\)

\(\text{Stationary point when}\ \ \dfrac{dV}{d\theta}=0:\)

\(\cos\theta \ne 0\ \ \text{for}\ \ 0<\theta<\dfrac{\pi}{2}\)

\(\cos^2\theta\)	\(=2\sin^2\theta\)
\(\tan^2\theta\)	\(=\dfrac{1}{2}\)
\(\tan\theta\)	\(=\dfrac{1}{\sqrt{2}}\ \ \ (\tan\theta>0)\)
\(\theta\)	\(=0.6154\ldots\)

\(\text{Check nature}\ \ (\cos\theta>0\ \text{so sign depends on}\ \cos^2\theta-2\sin^2\theta):\)

\(\theta=0.5:\ \ \cos^2(0.5)-2\sin^2(0.5)=0.31\ldots>0\)

\(\theta=0.7:\ \ \cos^2(0.7)-2\sin^2(0.7)=-0.24\ldots<0\)

\(\Rightarrow\ \text{MAX at}\ \ \theta=0.6154\ldots\)

\(\therefore\ V\ \text{is a maximum when}\ \ \theta=0.62\ \text{radians (2 d.p.)}\)