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Maximizing the area of a Norman window with fixed perimeter

A Norman window is constructed in the shape of a rectangle topped by a semicircle. The radius of the semicircle equals half the width of the rectangle. The perimeter of the entire window (including the diameter segment where the semicircle meets the rectangle, since that segment is not part of the outer boundary) is fixed at P=20P = 20 ft.

Let xx be the width of the rectangular base (which is also the diameter of the semicircle) and let yy be the height of the rectangular part. The perimeter of the window (two vertical sides, the bottom, and the semicircular arc on top) is fixed at 20 ft, so

2y+x+πx2=20.2y + x + \frac{\pi x}{2} = 20.

The total area of the window is the area of the rectangle plus the area of the semicircle:

A=xy+πx28.A = xy + \frac{\pi x^2}{8}.

Use the perimeter constraint to write AA as a function of xx alone, determine the domain of xx that makes physical sense, and find the value of xx that maximizes the area. What width xx (in feet) maximizes the area of the window?

Hint

Solve the perimeter equation for yy in terms of xx, substitute into the area formula to get A(x)A(x), restrict xx to values that keep y>0y>0 and x>0x>0, then set A′(x)=0A'(x)=0 to find the critical point.

Step-by-step solution

From the constraint 2y+x+πx2=202y + x + \frac{\pi x}{2} = 20, solve for yy: y=20−x−πx22=10−x2−πx4y = \dfrac{20 - x - \frac{\pi x}{2}}{2} = 10 - \dfrac{x}{2} - \dfrac{\pi x}{4}. Since y>0y>0, this requires 0<x<402+π0 < x < \dfrac{40}{2+\pi}, which is the physical domain for xx.

Substitute this yy into A=xy+πx28A = xy + \frac{\pi x^2}{8} to get A(x)=x(10−x2−πx4)+πx28=10x−x22−πx28A(x) = x\left(10 - \frac{x}{2} - \frac{\pi x}{4}\right) + \frac{\pi x^2}{8} = 10x - \frac{x^2}{2} - \frac{\pi x^2}{8}.

Differentiate: A′(x)=10−x−πx4=10−x(1+π4)A'(x) = 10 - x - \frac{\pi x}{4} = 10 - x\left(1+\frac{\pi}{4}\right). Setting A′(x)=0A'(x)=0 gives x=101+π4=404+πx = \dfrac{10}{1+\frac{\pi}{4}} = \dfrac{40}{4+\pi}.

Since A′′(x)=−(1+π4)<0A''(x) = -\left(1+\frac{\pi}{4}\right) < 0, this critical point is a maximum, and it lies within the valid domain 0<x<402+π0 < x < \frac{40}{2+\pi}, confirming it makes physical sense as the optimal width.

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