A window is to be built in the shape of a semi-circle over a rectangle:

If the outer perimeter is \(10\) feet, maximize the area. Round your answer to the closest tenth of a square foot.
Let’s label the picture.
The outer perimeter, \(P\), is given by

\(P=2r+2x+r\pi \).

Since we know that \(P=10\), we can express \(x\) in terms of \(r\).

\(x=5-r\frac {\pi }{2}-r\).

The area, A , consists of the area of the rectangle + the area of the semicircle:

\(A= 2\cdot r\cdot x+\frac {\pi }{2} r^2\)

We can now express the area \(A\) as a function of \(r\):

\(A(r)=2\cdot r\cdot (5-r\frac {\pi }{2}-r)+\frac {\pi }{2} r^2=10r-\frac {\pi }{2}r^2-2r^2\).

The domain of A is the closed interval \(\left [0,\frac {10}{2+\pi }\right ]\).(Remember: P=10!)

Now we have to find the global maximum of \(A\) on its domain. First, we have to find the critical points of \(A\). Since

\(A'(r)=10-\pi r-4r\),

it follows that the function \(A\) has its only critical point at \(r=\frac {\answer {10}}{4+\pi }\) .

Now, we evaluate \(A\) at the end points and at the critical point and then compare the values.

\(A(0)=0\)

\(A\left (\frac {10}{2+\pi }\right )=\frac {50\pi }{(2+\pi )^2}\)

\(A\left (\frac {\answer {10}}{4+\pi }\right )=\frac {\answer {50}}{4+\pi }\approx 7\)

This seems complicated. It is better to argue that the derivative

\(A'(r)>0\) on \(\left (0,\frac {10}{4+\pi }\right )\) and \(A'(r)<0\) on \(\left (\frac {10}{4+\pi },\frac {10}{2+\pi }\right )\),

since \(A'(r)=10-\pi r-4r=10-r(\pi +4)=-(\pi +4)\left (r-\frac {10}{\pi +4}\right )\).

This implies that the global maximum occurs at the critical point.

\[ A = \answer {7.0} ft^2 \]