# The height in order to maximize the intensity of illumination.

### 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 4, Problem 65RE

(a)

To determine

## To find: The height in order to maximize the intensity of illumination.

Expert Solution

The light pole should be 28.284feet in order to maximize the intensity of illumination.

### Explanation of Solution

Given:

The equation is

l=kcosθd2

Concept used:

If the quantities x and y are related by an equation

yα1x

For some constant k0

That is y varies inversely proportional as x or y is directly proportional to x

The constant k is called constant proportionality .

Calculation:

The original equation given in order utilize variable that can be easily solved the given diagram

a2+b2=c2h2+402=d2d=h2+1600cosθ=hdcosθ=hh2+1600

Now after substituting d and cosθ

l=kh(h2+1600)32l=kh(h2+1600)32

Differentiating with respect to x

dldx=3kh2(h2+1600)52+k(h2+1600)32

Using the zero product property of the equation

0=3kh2(h2+1600)52+k(h2+1600)32h=±800h=28.284

The light pole should be 28.284feet in order to maximize the intensity of illumination.

(b)

To determine

### To find: The rate of change of the intensity of the light equation.

Expert Solution

The value for the rate of the intensity of the light at the point on the woman’s back is 2.139e6k

### Explanation of Solution

Given:

The equation is

l=kcosθd2

Concept used:

If the quantities x and y are related by an equation

yα1x

For some constant k0

That is y varies inversely proportional as x or y is directly proportional to x

The constant k is called constant proportionality .

Calculation:

The original equation given in order utilize variable that can be easily solved the given diagram

a2+b2=c2h2+402=d2d=h2+1600cosθ=hdcosθ=hh2+1600

Now after substituting d and cosθ

l=kh(h2+1600)32l=kh(h2+1600)32

Using the zero product property of the equation

0=3kh2(h2+1600)52+k(h2+1600)32h=±800h=28.284

The light pole should be 28.284feet in order to maximize the intensity of illumination.

The equation is

j=h4j=24.284

l=kj(j2+r2)32

Differentiating with respect to t

dldt=32kj(j2+1600)522.40.4dldt=2.139e6k

The value for the rate of the intensity of the light at the point on the woman’s back is 2.139e6k .

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