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 22 feet?

Answer 1

1. Check the picture. Let the radius of the semicircle be R, so the width of the rectangle is 2R, and let the height of the rectangle be H.

2. The perimeter of the semicircle is $\frac{1}{2}*circumference = \frac{1}{2}*2 \pi R=\pi R$

the total perimeter is $\pi R+2H+2R=2H+( \pi +2)R$

We can write H in terms of R, in order to decrease the number of variables:

$2H+( \pi +2)R=22$

$2H=22-( \pi +2)R$

$H= \frac{22-( \pi +2)R}{2} =11- \frac{( \pi +2)R}{2}$

3. The total area as a function of R is the area of the semicircle + the area of the rectangle:

$A(R)= \pi R^2+2R*H=\pi R^2+2R*(11- \frac{( \pi +2)R}{2})=$

$\pi R^2+22R-( \pi +2)R^2=\pi R^2+22R- \pi R^2 -2R^2=22R-2R^2$

so $A(R)=22R-2R^2$ is clearly a quadratic function, with its graph a parabola which opens downward (since the coefficient of R^2 is negative)

The maximal value of A(R) is the maximal value the parabola can reach, that is its vertex.

4. $A(R)=22R-2R^2=2R(11-R)$ means the roots are at R=0 and R=11, so the axis of symmetry passes through R=11/2

thus the vertex point is (11/2, A(11/2))=(11, 2*(11/2)(11-11/2))=(11, 11*11/2)=(11,121/2)
=(11, 60.5)

5. Answer: Largest possible area is 60.5 feet squared