UPSC 2015 Maths Optional Paper 1 Q2b — Step-by-Step Solution

13 marks · Section A

Lagrange's method of multipliers (constrained extrema) · Calculus · asked 8× in 13 yrs · Read the full method →

Question

A conical tent is of given capacity. For the least amount of canvas required for it, find the ratio of its height to the radius of its base.

Constrained optimisation by substitution; minimise $S^2$ instead of $S$ to avoid the square root; verify via second derivative or convexity.

Setup. Let $r=$ base radius, $h=$ height, $\ell=\sqrt{r^2+h^2}=$ slant height.

Volume (given, fixed): $V=\tfrac{1}{3}\pi r^2 h$ .
Canvas (lateral surface only; a tent has no floor canvas and no top): $S=\pi r\ell=\pi r\sqrt{r^2+h^2}$ .

We minimise $S$ subject to $V$ fixed.

From $V=\tfrac{1}{3}\pi r^2 h$ , $h=\dfrac{3V}{\pi r^2}$ .

S^2=\pi^2 r^2(r^2+h^2)=\pi^2 r^4+\pi^2 r^2 h^2=\pi^2 r^4+\pi^2 r^2\!\cdot\!\dfrac{9V^2}{\pi^2 r^4}=\pi^2 r^4+\dfrac{9V^2}{r^2}.

f'(r)=4\pi^2 r^3-\dfrac{18V^2}{r^3}.

Set $f'=0$ :

4\pi^2 r^6=18V^2\;\Longrightarrow\;r^6=\dfrac{9V^2}{2\pi^2}\;\Longrightarrow\;r^2=\left(\dfrac{9V^2}{2\pi^2}\right)^{1/3}.

Use the constraint: $h=\dfrac{3V}{\pi r^2}$ , so

\dfrac{h}{r}=\dfrac{3V}{\pi r^3}.

Square: $\dfrac{h^2}{r^2}=\dfrac{9V^2}{\pi^2 r^6}$ .

From $r^6=9V^2/(2\pi^2)$ : $\dfrac{9V^2}{\pi^2 r^6}=\dfrac{9V^2}{\pi^2}\cdot\dfrac{2\pi^2}{9V^2}=2$ .

So $\dfrac{h^2}{r^2}=2$ , i.e.

\boxed{\;\dfrac{h}{r}=\sqrt{2}.\;}