Question

The maximum area of the rectangle that can be inscribed in a circle of radius $r$ is

• A. $2r^2\ \text{sq. units}$
• B. $\frac{\pi^2}{4}\ \text{sq. units}$
• C. $\pi r^2\ \text{units}$
• D. $r^3\ \text{sq. units}$

Hint:

A highly useful geometric principle to memorize is that for any given closed bounding curve, the inscribed rectangle with the absolute maximum area is always a perfectly symmetric square. For a circle of radius $r$, the diagonal of this square is the diameter $2r$, making its area equal to $\frac{1}{2} \times d^2 = \frac{1}{2}(2r)^2 = 2r^2$.

Answer 1

The Correct Option is A

Step 1: Set up with the diagonal.
A rectangle inscribed in a circle of radius $r$ has its diagonal equal to the diameter $2r$. Let the sides be $x$ and $y$, so $x^2+y^2=(2r)^2=4r^2$. We want to maximise the area $A=xy$.
Step 2: Work with the square of the area.
Maximising $A$ is the same as maximising $A^2=x^2y^2$, which avoids square roots and trigonometry.
Step 3: Apply the AM-GM inequality.
For the positive numbers $x^2$ and $y^2$, $\dfrac{x^2+y^2}{2}\ge\sqrt{x^2y^2}=xy$.
Step 4: Substitute the constraint.
\[ xy\le\frac{x^2+y^2}{2}=\frac{4r^2}{2}=2r^2. \]
Step 5: Find when equality holds.
AM-GM is an equality only when $x^2=y^2$, i.e. $x=y$. Then $2x^2=4r^2$ gives $x=\sqrt2\,r$, so the optimal rectangle is a square of side $\sqrt2\,r$.
Step 6: Confirm the maximum.
At $x=y=\sqrt2\,r$, $A=(\sqrt2\,r)(\sqrt2\,r)=2r^2$, which is the largest value allowed by Step 4. \[ \boxed{A_{\max}=2r^{2}\ \text{sq. units}} \]