Exercise 1.1.72

Question

A Norman window has the shape of a rectangle surmounted by a semicircle. If the perimeter of the window is 30 ft, express the area A of the window as a function of the width $x$ of the window.

Khagan Gulusoy · Accepted Answer

We can divide the window we see in the picture into two parts. The first part has the shape of a semicircle and the second part - of a rectangle. Thus, we can find the area $A$ of the window by adding the areas of the semicircle and the rectangle. But to do that, we must find the height length of the rectangle first.
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The perimeter $P$ of the window is equal to the sum of the perimeters of the rectangle and semicircle. Assume that $y$ is the length of the rectangle. Denote the height of the triangle by $y$. Then, we have:
$$P=x+2y+\pi R,$$
where $R$ is the radius of the semicircle. From the scheme of the window, we can conclude that $R=\frac{x}{2}.$ Thus, we have:
$$P=x+2y+\pi \cdot \frac{x}{2}.$$
From the exercise text, we have $P=30.$ Thus,
$$x+2y+\pi \cdot \frac{x}{2}=30$$
Our goal is to solve this equation for $y$:
$$2y=30-x-\frac{\pi x}{2}$$
$$y=15-\frac{x}{2}-\frac{\pi x}{4}.$$
Now that we have found the length and width of the rectangle and the radius of the semicircle, we can easily find the area $A$ of the window:
$$A=A_{	ext{rectangle}}+A_{	ext{semicircle}}$$
$$A=xy+\frac{\pi R^2}{2}$$
$$A=xy+\frac{\pi \left(\frac{x}{2}ight)^2}{2}$$
$$A=x\left(15-\frac{x}{2}-\frac{\pi x}{4}ight)+\frac{\pi x^2}{8}$$
$$A=15x-\frac{x^2}{2}-\frac{\pi x^2}{4}+\frac{\pi x^2}{8}$$
$$A=15x-\frac{x^2}{2}-\frac{\pi x^2}{8}.$$
Thus, we can define the function $A$ as follows:
$$\forall x\in (0,+\infty):\quad A(x)=15x-\frac{x^2}{2}-\frac{\pi x^2}{8}.$$

Exercise 1.1.72

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Exercise 1.1.72

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