Q.If the sum of the lengths of the hypotenuse and a side of a right-angled triangle is given, show that the area of the triangle is maximum when the angle between them is $\dfrac{\pi}{3}$.

Concept: Lagrange Multipliers (or direct single-variable optimisation).

Let the hypotenuse be $h$ and the given side be $b$, with $h + b = k$ (constant). The third side is $a = \sqrt{h^2 - b^2}$. Area $A = \frac12 b a = \frac12 b \sqrt{h^2 - b^2}$.

Substitute $h = k - b$:

Maximise $A^2$ to simplify: $A^2 = \frac14 b^2 (k^2 - 2kb)$. Differentiate w.r.t. $b$:

$\frac{d(A^2)}{db} = \frac14 \left( 2b(k^2 - 2kb) + b^2(-2k) \right) = \frac14 (2bk^2 - 6kb^2) = 0.$ …

We have a right-angled triangle. Let the hypotenuse be $h$, one leg be $x$, and the other leg be $y$. The given condition is that $x + h$ is fixed — call that constant $k$. The area is $A = \frac12 xy$.

The question asks: when is the area maximum? And specifically, show that the angle between the hypotenuse and the side $x$ (call it $\theta$) is $\pi/3$ at that maximum.

Why use Lagrange multipliers? Because we have a function to maximise (area) subject to a constraint (fixed sum of $x$ and $h$). The method elegantly handles the relation between $x$, $y$, and $h$ via Pythagoras: $h^2 = x^2 + y^2$. So we have two equations linking the three variables — one constraint is explicit ($x + h = k$), the other is implicit (Pythagoras). We can treat this as a constrained optimisation in two variables, or reduce to one variable. Let's do the latter for clarity.

1. Express everything in terms of one variable.

   Let the fixed sum be $S = x + h$. Then $h = S - x$.

   From Pythagoras: $h^2 = x^2 + y^2 \Rightarrow (S - x)^2 = x^2 + y^2$.

   Expand: $S^2 - 2Sx + x^2 = x^2 + y^2 \Rightarrow y^2 = S^2 - 2Sx$.

   So $y = \sqrt{S^2 - 2Sx}$, valid for $0 < x < S/2$ (since $y>0$ requires $S^2 - 2Sx > 0$).

2. Write area as a function of $x$.

   $A(x) = \frac12 x y = \frac12 x \sqrt{S^2 - 2Sx}$.

   We want to maximise $A(x)$ for $x \in (0, S/2)$.

3. Differentiate and set to zero.

   Let $A(x) = \frac12 x (S^2 - 2Sx)^{1/2}$.

   Differentiate using product rule:

   $A'(x) = \frac12 \left[ (S^2 - 2Sx)^{1/2} + x \cdot \frac12 (S^2 - 2Sx)^{-1/2} \cdot (-2S) \right]$

   Simplify:

   $A'(x) = \frac12 \left[ \sqrt{S^2 - 2Sx} - \frac{Sx}{\sqrt{S^2 - 2Sx}} \right]$.

   Set $A'(x) = 0$:

   $\sqrt{S^2 - 2Sx} - \frac{Sx}{\sqrt{S^2 - 2Sx}} = 0$

   $\Rightarrow \sqrt{S^2 - 2Sx} = \frac{Sx}{\sqrt{S^2 - 2Sx}}$

   $\Rightarrow S^2 - 2Sx = Sx$

   $\Rightarrow S^2 = 3Sx$

   $\Rightarrow x = S/3$.

4. Find the corresponding $h$ and $y$.

   $h = S - x = S - S/3 = 2S/3$.

   $y = \sqrt{S^2 - 2Sx} = \sqrt{S^2 - 2S(S/3)} = \sqrt{S^2 - 2S^2/3} = \sqrt{S^2/3} = S/\sqrt{3}$.

5. Find the angle $\theta$ between $x$ and $h$.

   In a right triangle, $\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{x}{h}$.

   So $\cos\theta = \frac{S/3}{2S/3} = \frac12$. …

Method: Optimization with a Geometric (Pythagorean) Constraint, Expressed via an Angle

This is the standard approach whenever a geometry optimization problem gives a sum of two linked lengths as fixed, and asks you to prove the optimum occurs at a particular angle rather than a particular length.

Steps

Step 1: Name the variables and write the constraint

Assign variables to the two quantities whose sum is fixed (here, the hypotenuse and one side of a right triangle), and write the given condition as an equation: (side) + (hypotenuse) = constant.

Step 2: Bring in the geometric relation to introduce the missing side

Use the relevant geometric law — for a right triangle, the Pythagorean theorem — to express the remaining unknown side in terms of the two you already have. This lets you write the quantity to be optimized (here, area) using only the constrained variables.

Step 3: Reduce to a single variable, differentiate, and solve …

Common Mistakes

Mistake 1: Computing $\cos\theta$ using the wrong side of the triangle

A right triangle has two acute angles, and it's easy to compute $\cos\theta$ using the side opposite $\theta$ instead of the side adjacent to the given angle (the one whose sum with the hypotenuse was fixed). Why it's wrong: the problem asks for the angle between the hypotenuse and the given side, so the adjacent-side ratio is the one that matters — using the wrong ratio gives a different angle than the correct $\pi/3$. Correct approach: identify which two sides actually meet at the angle in question before writing $\cos\theta$.

Mistake 2: Differentiating the area directly (with the square root) instead of squaring first …