A figure consists of a semicircle with a rectangle on its diameter. Given that the perimeter of the figure is $20 \; m$, find its dimensions in order that its area may be maximum.
Let $r$ be the radius of the semicircle.
Let the dimensions of the rectangle be $b$ and $2r$.
Perimeter of the figure $= \pi r + 2b + 2r$
Given: $\;$ Perimeter $= 20 \; m$
i.e. $\;$ $\pi r + 2b + 2r = 20$
i.e. $\;$ $r \left(2 + \pi\right) + 2b = 20$
i.e. $\;$ $b = 10 - \dfrac{r \left(2 + \pi\right)}{2}$ $\;\;\; \cdots \; (1)$
Area of the figure $= A = \dfrac{\pi r^2}{2} + 2 r b$
i.e. $\;$ $A = \dfrac{\pi r^2}{2} + 2 r \left[10 - \dfrac{r \left(2 + \pi\right)}{2}\right]$ $\;\;\;$ [in view of equation $(1)$]
i.e. $\;$ $A = \dfrac{\pi r^2}{2} + 20 r - r^2 \left(2 + \pi\right)$ $\;\;\; \cdots \; (2)$
For maximum area, $\;$ $\dfrac{dA}{dr} = 0$ $\;\;\; \cdots \; (3)$
$\therefore \;$ Differentiating equation $(2)$ w.r.t $x$ we have,
$\dfrac{dA}{dr} = \dfrac{2 \pi r}{2} + 20 - 2r \left(2 + \pi\right)$
i.e. $\;$ $\dfrac{dA}{dr} = \pi r + 20 - r \left(4 + 2 \pi\right)$
i.e. $\;$ $\dfrac{dA}{dr} = 20 - r \left(\pi + 4\right)$ $\;\;\; \cdots \; (4)$
$\therefore \;$ $\dfrac{dA}{dr} = 0$ $\implies$ $20 - r \left(\pi + 4\right) = 0$
i.e. $\;$ $r = \dfrac{20}{\pi + 4}$
From equation $(4)$, $\;$ $\dfrac{d^2 A}{dr^2} \Big |_{r = \frac{20}{\pi + 4}} = - \left(\pi + 4\right) < 0$
$\implies$ $r = \dfrac{20}{\pi + 4}$ $\;$ gives maximum area.
Substituting the value of $r$ in equation $(1)$ gives
$b = 10 - \dfrac{20 \left(\pi + 2\right)}{2 \left(\pi + 4\right)}$
i.e. $\;$ $b = 10 \left(1 - \dfrac{\pi + 2}{\pi + 4}\right) = \dfrac{20}{\pi + 4}$
$\therefore \;$ For maximum area of the figure, $\;$ radius of semicircle $= \dfrac{20}{\pi + 4} \; m$ $\;$ and the sides of the rectangle are $\;$ $\dfrac{20}{\pi + 4} \; m$, $\;$ $\dfrac{40}{\pi + 4} \; m$