UPSC 2026 Maths Optional Paper 1 Q2b — Step-by-Step Solution

15 marks · Section A

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

Question

If the sum of the lengths of the hypotenuse and another side of a right-angled triangle be given, then find the angle between these sides so that the area of the triangle is maximum.

Technique

Reduce to one variable using the given constraint $b+c=k$ ; maximise $A^2$ rather than $A$ (kills the radical, same maximiser since $A>0$ ); first-derivative test for the stationary point and second-derivative test for the maximum; then read off $\cos\theta = b/c$ in the right triangle. A trigonometric parametrisation $b=c\cos\theta$ gives an independent second route.

Solution

Step 1 — Set up notation and identify the angle asked for.

Let $ABC$ be right-angled at $B$ , with

$c = AC$ = hypotenuse,
$b = AB$ = the “another side” whose length is added to the hypotenuse,
$a = BC$ = the remaining leg,
$\theta = \angle BAC$ = the angle between $b$ and $c$ (they meet at $A$ ).

Figure to draw (small, labelled): right triangle, right angle marked at $B$ ; label the horizontal leg $AB=b$ , the vertical leg $BC=a$ , the hypotenuse $AC=c$ ; mark $\theta$ at $A$ between $b$ and $c$ . This one figure is what fixes which angle is being asked for — without it, it is easy to solve for $\angle ACB$ and lose the marks.

By Pythagoras, $a^2=c^2-b^2$ , and the constraint given is

b+c=k\qquad(k>0\ \text{a given constant}).

The area is

A=\tfrac12\,ab=\tfrac12\,b\sqrt{c^2-b^2}.

Step 2 — Reduce to a single variable.

Put $c=k-b$ . Then

c^2-b^2=(k-b)^2-b^2=k^2-2kb,

A(b)=\tfrac12\,b\sqrt{k^2-2kb}.

The admissible range is $b>0$ together with $c>b$ (the hypotenuse is the longest side), i.e. $k-b>b$ , giving

0<b<\tfrac{k}{2}.

On this open interval $A(b)>0$ , while $A\to0$ at both endpoints ( $b\to0^+$ and $b\to k/2^-$ ). Hence a maximum exists in the interior and is attained at a stationary point.

Step 3 — Maximise $A^2$ (removes the radical).

Since $A>0$ , maximising $A$ is equivalent to maximising

F(b):=4A^2=b^2\,(k^2-2kb)=k^2b^2-2kb^3.

F'(b)=2k^2b-6kb^2=2kb\,(k-3b).

Setting $F'(b)=0$ with $b>0$ and $k>0$ :

k-3b=0\quad\Longrightarrow\quad \boxed{b=\tfrac{k}{3}}\ \in\left(0,\tfrac k2\right).\ \checkmark

Step 4 — Second-derivative test.

F''(b)=2k^2-12kb,\qquad F''\!\left(\tfrac{k}{3}\right)=2k^2-12k\cdot\tfrac{k}{3}=2k^2-4k^2=-2k^2<0.

So $F$ , and therefore $A$ , is maximum at $b=k/3$ .

Step 5 — The sides and the required angle.

b=\frac{k}{3},\qquad c=k-b=\frac{2k}{3},\qquad a=\sqrt{c^2-b^2}=\sqrt{\frac{4k^2}{9}-\frac{k^2}{9}}=\frac{k}{\sqrt3}.

In the right triangle, the angle $\theta$ at $A$ between $b$ and $c$ satisfies

\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{b}{c}=\frac{k/3}{2k/3}=\frac12.

\theta=\frac{\pi}{3}=60^\circ.

Note that this is independent of $k$ — the optimal triangle is always a $30^\circ$ – $60^\circ$ – $90^\circ$ triangle, with sides in the ratio $b:a:c=\tfrac{k}{3}:\tfrac{k}{\sqrt3}:\tfrac{2k}{3}=1:\sqrt3:2$ .

The maximum area is

A_{\max}=\tfrac12\cdot\frac{k}{3}\cdot\frac{k}{\sqrt3}=\frac{k^2}{6\sqrt3}=\frac{\sqrt3\,k^2}{18}.

Step 6 — Independent check by a trigonometric parametrisation.

Work directly in $\theta$ . Since $b=c\cos\theta$ and $a=c\sin\theta$ , the constraint $b+c=k$ gives

c(1+\cos\theta)=k\quad\Longrightarrow\quad c=\frac{k}{1+\cos\theta},

and

A(\theta)=\tfrac12\,ab=\tfrac12\,c^2\sin\theta\cos\theta=\frac{k^2}{2}\cdot\frac{\sin\theta\cos\theta}{(1+\cos\theta)^2},\qquad 0<\theta<\tfrac\pi2 .

At $\theta=\pi/3$ : $\sin\theta\cos\theta=\tfrac{\sqrt3}{2}\cdot\tfrac12=\tfrac{\sqrt3}{4}$ and $(1+\cos\theta)^2=(3/2)^2=9/4$ , so

A=\frac{k^2}{2}\cdot\frac{\sqrt3/4}{9/4}=\frac{k^2}{2}\cdot\frac{\sqrt3}{9}=\frac{\sqrt3\,k^2}{18},

matching Step 5 exactly. The two routes agree.

Answer

\boxed{\;\theta=\frac{\pi}{3}=60^\circ\quad\text{(with } b=\tfrac k3,\; c=\tfrac{2k}3,\; A_{\max}=\tfrac{\sqrt3\,k^2}{18}\text{)}\;}