UPSC 2024 Maths Optional Paper 1 Q3b — Step-by-Step Solution

20 marks · Section A

Maxima and minima of single-variable functions · Calculus · asked 8× in 14 yrs · Read the full method →

Question

Find the volume of the greatest cylinder which can be inscribed in a cone of height $h$ and semi-vertical angle $\alpha$ .

Technique

Express cylinder dimensions via similar triangles; write $V(r)$ ; use the first-derivative test.

Solution

$Vertical cross-section through the cone's axis. The inscribed coaxial cylinder (radius r, height H) has its top edge touching the slant of the cone at distance x from the apex; similar triangles give r=x\tan\alpha, so x=r\cot\alpha and H=h-x=h-r\cot\alpha. The labelled h, H, 2r, x, and the semi-vertical angle \alpha carry the entire optimisation setup.$

Step 1 — Geometry.

Place the cone with apex at top and axis vertical, base radius $R=h\tan\alpha$ . Inscribe a coaxial cylinder of radius $r$ and height $H$ .

The cylinder’s top edge touches the cone at distance $x$ from the apex, where the cone has radius $x\tan\alpha=r$ , giving $x=r\cot\alpha$ . The cylinder height is

H=h-x=h-r\cot\alpha.

Step 2 — Optimise volume.

V(r)=\pi r^2 H=\pi r^2(h-r\cot\alpha)=\pi h r^2-\pi r^3\cot\alpha.

\frac{dV}{dr}=2\pi h r-3\pi r^2\cot\alpha=\pi r(2h-3r\cot\alpha).

Critical point (other than $r=0$ ): $r^\star=\dfrac{2h\tan\alpha}{3}$ .

$V''(r^\star)=2\pi h-6\pi r^\star\cot\alpha=2\pi h-4\pi h=-2\pi h<0$ : maximum confirmed.

Step 3 — Maximum volume.

At $r^\star=\dfrac{2h\tan\alpha}{3}$ , the cylinder height is $H^\star=h-\dfrac{2h}{3}=\dfrac{h}{3}$ , and

V^\star=\pi(r^\star)^2 H^\star=\pi\cdot\frac{4h^2\tan^2\alpha}{9}\cdot\frac{h}{3}=\frac{4\pi h^3\tan^2\alpha}{27}.

Answer

\boxed{\;V_{\max}=\frac{4\pi h^3\tan^2\alpha}{27},\quad\text{attained at radius }r=\frac{2h\tan\alpha}{3}\text{ and height }H=\frac{h}{3}.\;}