# The function of the area of the window in terms of the width of the window. ### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805 ### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

#### Solutions

Chapter 1.1, Problem 58E
To determine

## To find: The function of the area of the window in terms of the width of the window.

Expert Solution

The formula for the area of the window as a function of the width of the window is A(x)=15xx2(π+4)8_ with the domain 0<x<602+π_ .

### Explanation of Solution

Given:

The window is in the shape of a rectangle surmounted by a semicircle. The Perimeter of the window is 30 ft and the width of the window is denoted as x ft.

Formula used:

Area of the semi-circle or radius r, A=πr22 .

Area of the rectangle of length l and breadth b is A=l×b .

Calculation:

Let the height of the window be h ft.

The diameter of the semi-circle part is x ft as the diameter is same as the width of the window.

Thus, is the radius of the semi-circle part is x2ft .

Area of the rectangular part of width x and height h is A=x×h .

Area of the semi-circle part of radius x2 is as follows:

A=π(x2)22=π(x24)2=πx28

The area of the window is the sum of the area of the rectangular part and the area of the semi-circle part.

Thus, the perimeter of the window is A(x)=xh+πx28 .

Perimeter of the rectangular part of width x and height h is P=2(x+h) .

Perimeter of the semi-circle part of radius x2 is, P=πx2 .

The perimeter of the window is the sum of the perimeter of the rectangle part and the perimeter of the semi-circle part.

Thus, the perimeter of the window is P=2(x+h)+πx2 .

That is, P=2x+2h+πx2

Since the perimeter of the window is 30 ft, the height of the window is computed as follows:

30=x+2h+πx230xπx2=2hh=15x2πx4

Substitute the value of h in A(x)=xh+πx28 and simplify.

A(x)=x(15x2πx4)+πx28=15xx22πx24+πx28=15xx22πx28=15xx2(π+48)

The formula for the area of the window as a function of the width of the window is A(x)=15xx2(π+4)8 .

Domain of the area function:

As width x and height h cannot take negative values, x>0,h>0 .

As h>0,2h>0 which implies, 30xπx2>0 .

Simplify 30xπx2>0 and obtain the required domain.

60>2x+πx60>(2+π)xx<602+π

Combining x>0 and x<602+π, the domain of the area (A) is 0<x<602+π_ .

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