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)

$\frac{2P}{\pi + 4}$, $\frac{P}{\pi + 4}$

Answer 2

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, $P = 2x+4y +\pi x \quad...\left(i\right)$ and, $A= \left(2x\right) \left(2y\right) +\frac{\pi x^{2}}{2} \quad..\left(ii\right)$ $\Rightarrow A= 4xy +\frac{\pi x^{2}}{2}$ $\Rightarrow A = x\left( P-2x -\pi x\right) +\frac{\pi x^{2}}{2}$ [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 $\frac{d^{2}A}{dx^{2}} = -4 -\pi$ 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, $\frac{d^{2}A}{dx^{2}} = -4 -\pi < 0$ 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 $2y = \frac{P}{\pi +4 }$.