Maxima problem? this question is off a textbook, and I've been having a lot of trouble with it:
"A window consists of a rectangle surmounted by a semi-circle having its diameter the width of the rectangle. If the perimeter of the window is $t$ meters, find the greatest possible area of the window."
The final answer is: 
$t^2 / 2(pi + 4)$
Could someone please explain how to do this step by step?
Thanks
 A: We set things up, and leave the differentiation and conclusion to you. 
Let us explore what happens if we let the radius be $x$. Then the curvy part of the semi-circle has perimeter $\pi x$. The base of the window has length $2x$. Since the whole perimeter is $t$, the two sides have combined length $t-\pi x-2x$. It follows that each side of the window has length
$$\frac{t-\pi x-2x}{2}.$$
Next we find a formula for the area $A(x)$ of the window. 
The semi-circle has area $\frac{\pi}{2}x^2$.
The rectangular part of the window has area $(2x)\frac{t-\pi x-2x}{2}=x(t-\pi x -2x)$. It follows that
$$A(x)=\frac{\pi}{2}x^2 +x(t-\pi x -2x).$$
Now do the usual thing: find where $A'(x)=0$.
For completeness, we have to check  whether the maximum is at an end-point, or alternately look at the sign of $A'(x)$ when $A'(x)\ne 0$. 
A: HINT:
If the width $=a$ unit, the height is $=b$ unit,
the perimeter $=\frac{\pi a}2+2b+a=t\implies b=\frac{2t-(\pi+2)a}4$ 
So, the total area $=\frac{\pi\left(\frac a2\right)^2}2+a\cdot b=\frac{\pi\left(\frac a2\right)^2}2+a\cdot\frac{2t-(\pi+2)a}4 =f(a)$ (say)
Use Second derivative test
