Q.Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.

Take a rectangle inscribed in a circle of radius $r$; its diagonal is the diameter $2r$.

Step 1 — One variable. Let one side be $x$; the other side is $\sqrt{(2r)^2-x^2}=\sqrt{4r^2-x^2}$. Area

Maximise $A^2=x^2(4r^2-x^2)=4r^2x^2-x^4$ (same maximiser, easier to differentiate).

Step 2 — Differentiate. With $f(x)=4r^2x^2-x^4$,

$f'(x)=8r^2x-4x^3=4x(2r^2-x^2)=0\Rightarrow x^2=2r^2\Rightarrow x=r\sqrt2.$ …

The idea

Every rectangle inscribed in a circle has the circle's diameter as its diagonal. That single relation lets us write the area in one variable and maximise it with the derivative (standard CBSE method).

Set up

Let the circle have radius $r$, so the diameter is $2r$. If one side of the rectangle is $x$, the diagonal condition $x^2+(\text{other side})^2=(2r)^2$ gives the other side $\sqrt{4r^2-x^2}$. The area is

Work the steps

1. Work with $A^2$ to avoid the square root. Since $A>0$, maximising $A$ is the same as maximising

2. Differentiate:

3. Solve $f'(x)=0$: since $x>0$, we need $2r^2-x^2=0$, i.e. $x^2=2r^2$, so $x=r\sqrt2$.
4. Second-derivative test: …

Method: Optimization Proofs Using the "Maximize $A^2$" Trick (Inscribed-Figure Problems)

This method proves a geometric optimization claim (e.g. "the square has maximum area among inscribed rectangles") by expressing the objective in terms of a single geometric parameter and avoiding messy square-root differentiation.

Steps

Step 1: Use the geometric constraint to relate the two dimensions

For a rectangle inscribed in a circle of radius $r$, the diagonal of the rectangle equals the circle's diameter, $2r$. If one side is $x$, the other side is determined by the Pythagorean relation:

Step 2: Write the objective (area) as a function of the single variable

Step 3: Maximize $A^2$ instead of $A$ directly

Since $A(x) > 0$ on the valid domain, maximizing $A(x)$ is equivalent to maximizing $A(x)^2$ — and squaring removes the square root, turning the problem into a polynomial that's much easier to differentiate:

$f(x) = A(x)^2 = x^2(4r^2 - x^2)$ …

Common Mistakes

Mistake 1: Differentiating the square-root expression $x\sqrt{4r^2-x^2}$ directly

Why it's wrong: this requires the chain rule on a square root, which is more error-prone (a common slip is mishandling the $\frac{1}{2\sqrt{\cdot}}$ factor) than the equivalent, cleaner polynomial approach. Correct approach: square the objective first, since $A>0$ means maximizing $A^2$ gives the same maximizing $x$, and differentiate the resulting polynomial instead.

Mistake 2: Stopping after finding $x = r\sqrt{2}$ without confirming the shape is a square

Why it's wrong: the question specifically asks to show the maximizing rectangle is a square — finding the optimal $x$ alone doesn't demonstrate that; a student must also substitute back to find the other side and explicitly verify it equals $x$. Correct approach: always complete the geometric conclusion the problem asks for, not just the calculus. …