Help with Calculus Optimization Problem! We wish to construct a rectangular auditorium with a stage shaped as a semicircle of radius $r$, as shown in the diagram below (white is the stage and green is the seating area). For safety reasons, light strips must be placed on the perimeter of the seating area. If we have $45\pi + 60$ meters of light strips, what should $r$ be so that the seating area is maximized?

So I first set the width of the seating area to 2r, and the depth to be x. The perimeter would then be  $2x + 2r + \pi r$ = $45\pi + 60$. The problem asks us to maximize the area, though, so it's $2rx - (\pi r^2)/2$. I can solve the equation in terms of x so that it becomes $x = (45\pi + 60 - r(\pi + 2))/2$. Unfortunately, I'm stuck from this point on, so any hint that you could give me would be great. Thanks!

so you are going correctly 
you have $$2x+2r+\pi r=45\pi+60$$
so $$2x=45\pi+60-2r-\pi r$$
now you have $$area=2x*r-\frac{\pi r^2}{2}$$
now replace the value of 2x in the area equation from the 1st equation
to obtain the area in a single variable so that you can calculate the derivative w.r.t.  r.
$$A=(45\pi+60-2r-\pi r)*r-\frac{\pi r^2}{2}$$
the area becomes 
$$A=45\pi r+60r-2r^2-\pi r^2-\frac{\pi r^2}{2}$$
differentiate it w.r.t. r
$$\frac{dA}{dr}=45\pi+60-2\pi r-4r-\pi r=0$$
thus $$r=\frac{45\pi+60}{3\pi+4}=15$$
thus $$r=15$$
thus $$2x=45\pi + 60- 2*15-\pi *15=30+ 30* \pi $$
maximum area is $$A=45*pi*15+60*15-2*15^2-\pi *15^2- \pi * \frac{15^2}{2}$$
also, check the second derivative to confirm that it is indeed maximum.
