Question

A figure consists of a semi-circle with a rectangle on its diameter. Given the perimeter of the figure, find its dimensions in order that the area may be maximum.

• A. $\frac{2P}{\pi +4}$, $\frac{P}{\pi +4}$
• B. $\frac{P}{\pi +4}$, $\frac{P}{\pi +4}$
• C. $\frac{2P}{\pi -4}$, $\frac{P}{\pi +4}$
• D. $\frac{P}{\pi -4},\frac{2P}{\pi +4}$

Answer 1

Answer: (A)

Let $ABCD$ be a rectangle and let the semi-circle be described on side $AB$ as diameter. Let $AB=2x$ and $AD=2y$. Let $P$ be the perimeter and $A$ be the area of the figure. Then,

$P=2x+4y+\pi x...\left(i\right)$
and, $A=\left(2x\right)\left(2y\right)+\frac{\pi x^2}{2}..\left(ii\right)$
$\Rightarrow A=4xy+\frac{\pi x^2}{2}$
$\Rightarrow A=x\left(P-2x-\pi x\right)+\frac{\pi x^2}{2}$
[Using $\left(i\right)$]
$\Rightarrow A=Px-2x^2-\pi x^2+\frac{\pi x^2}{2}$
$\Rightarrow A=Px-2x^2-\frac{\pi x^2}{2}$
$\Rightarrow \frac{dA}{dx}=P-4x-\pi x$
and $\frac{d^{2}A}{d{x}^2}=-4-\pi$
For maximum or minimum $A$, we must have
$\Rightarrow \frac{dA}{dx}=0$
$\Rightarrow P-4x-\pi x=0$
$\Rightarrow x=\frac{P}{\pi +4}$
Clearly, $\frac{d^{2}A}{d{x}^2}=-4-\pi <0$ for all values of $x$.
Thus, $A$ is maximum when $x=\frac{P}{\pi +4}$ .
Putting $x=\frac{P}{\pi +4}$ in $\left(i\right)$ we get $y=\frac{2P}{2\left(\pi +4\right)}$.
so, dimentions of the figure are $2x=\frac{2P}{\pi +4}$ and $2y=\frac{P}{\pi +4}$.