BuyFind

Single Variable Calculus: Concepts...

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

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

Answer to Problem 58E

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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