The maximum area of a rectangle inscribed in a circle of radius \( 3\sqrt{2} \) cm, is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 36 Sq. units
A rectangle inscribed in a circle has its diagonal along a diameter. Let the sides be x and y. With \( r=3\sqrt{2} \), the diameter is \( 6\sqrt{2} \), so
\( \displaystyle x^2+y^2=(6\sqrt{2})^2=72 \)
\( \displaystyle y=\sqrt{72-x^2} \)
The area is \( A=x\sqrt{72-x^2} \). It is easier to maximise \( S=A^2 \), which is largest at the same x:
\( \displaystyle S=x^2(72-x^2)=72x^2-x^4 \)
\( \displaystyle \frac{dS}{dx}=144x-4x^3=4x(36-x^2) \)
\( \displaystyle \frac{dS}{dx}=0\ \Rightarrow\ x=6 \)
\( \displaystyle \frac{d^2S}{dx^2}=144-12x^2 \)
\( \displaystyle =144-432 \)
\( \displaystyle =-288\lt 0\ \text{at}\ x=6 \)
So S (and hence A) is maximum at x = 6. Then
\( \displaystyle y=\sqrt{72-36}=6 \)
\( \displaystyle A=6\times 6=36 \)
The largest rectangle is a square of side 6, with area 36.
Shortcut: The largest rectangle in a circle of radius r is a square with diagonal 2r, so its area is \( \frac{(2r)^2}{2}=2r^2 \).
\( \displaystyle 2r^2=2(3\sqrt{2})^2=2\times 18=36 \)