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

let the radius of the circle be r and the height of the rectangular part formed on the diameter of circle be h.
let the area of the figure be 
$$\begin{gathered} A=area of semicircle + area of rec\tan gle \\ A=\frac{1}{2}.\pi r^2+\left(2r\right)h .........\left(1\right) \end{gathered}$$
since the perimeter of the figure is fixed , let it be 2k.
$$\begin{gathered} 2k=\pi r+2r+2h \\ k=\frac{\pi r}{2}+r+h \\ h=k-\frac{\pi r}{2}-r .......\left(2\right) \end{gathered}$$
substituting in eq(1):
$$\begin{gathered} A=\frac{1}{2}.\pi r^2+2r*[k-\frac{\pi r}{2}-r] \\ A=\frac{1}{2}.\pi r^2+2rk-\pi r^2-2r^2 \\ A=-\frac{\pi r^2}{2}+2rk-2r^2 .......\left(3\right) \end{gathered}$$
now to find the critical points let us differentiate eq(3) wrt r:
$$\begin{gathered} \frac{dA}{dr}=-\pi r+2k-4r ........\left(4\right) \\ put \frac{dA}{dr}=0 \\ -\pi r+2k-4r=0 \\ (\pi +4)r=2k \\ r=\frac{2k}{\pi +4} \end{gathered}$$
now to check  the maxima or minima we will again differentiate the equation (4):
$$\begin{gathered} \frac{d^{2}A}{d{r}^2}=\frac{d}{dr}\left(\frac{dA}{dr}\right) \\ =\frac{d}{dr}(-\pi r+2k-4r) \\ =-\pi -4 \\ \Rightarrow \frac{d^{2}A}{d{r}^2}<0 \end{gathered}$$
thus $r=\frac{2k}{\pi +4}$ must be a relative maximum.
since there is only one critical point , a relative maximum is also an absolute max.
thus $r=\frac{2k}{\pi +4}$ gives the maximum area in the problem.
the corresponding value of the height is
$$\begin{gathered} h=k-\frac{\pi r}{2}-r \\ =k-\frac{\pi }{2}.\frac{2k}{\pi +4}-\frac{2k}{\pi +4} \\ =k-\frac{\pi k}{\pi +4}-\frac{2k}{\pi +4} \\ =\frac{k\pi +4k-\pi k-2k}{\pi +4} \\ =\frac{2k}{\pi +4} \end{gathered}$$

hope this helps you