Question

A Norman window has the shape of a semicircle atop a rectangle so that the diameter of the semicircle is equal to the width of the rectangle. What is the area of the largest possible Norman window with a perimeter of 31 feet?

Answer 1

So the first step is always to draw a diagram of your problem.

We're given a constraint of 31 feet for a perimeter, and told that we want the largest, so maximizing problem, Area when it has 31 ft as a perimeter. So let's set a few equations up. Our perimeter will be the circumference of half a circle + the sides of our rectangle.
\[P=y+y+x+\frac{ 1 }{ 2 }\left( 2 \pi \frac{ x }{ 2 } \right)===>P=2y+x+\frac{ \pi x }{ 2 }\]
Now we need an area equation, so we can add the area of the rectangle+ the area of half of a circle (top of the norman window).
\[A=x*y+\frac{ 1 }{ 2 }\pi \left( \frac{ x }{ 2 } \right)^{2}==simplify==> A=x*y+\frac{ \pi x^2 }{ 8 }\]
Since the question is asking for the area, we can use the given perimeter and put it in terms of y, then substitute it into our area equation.
\[31=2y+x+\frac{ \pi x }{ 2 }===>y=\frac{ 31 }{ 2 }-\frac{ x }{ 2 }-\frac{ \pi x }{ 4 }\]
Now we sub. y into our area equation.
\[ A=x*\left( \frac{ 31 }{ 2 }-\frac{ x }{ 2 }-\frac{ \pi x }{ 4 }\right)+\frac{ \pi x^2 }{ 8 } ==> A=\frac{ 31x }{ 2 }-\frac{ x^2 }{ 2 }-\frac{ \pi x^2 }{ 4 }+\frac{ \pi x^2 }{ 8 }\]
We can combine like terms to simplify our area just a little more.
\[A=\frac{ 31x }{ 2 }-\frac{ x^2 }{ 2 }-\frac{ \pi x^2 }{ 8 }\]
This is usually the hard part of the problem. All that's left to do is to take the derivative of our area, find our critical points and check to see whether the points are a max or min with A''(x). I'll let you try to work that part out, or you can come back for more help if you need to.