# The light intensity for which P is maximum. ### 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 4.6, Problem 8E
To determine

## To find: The light intensity for which P is maximum.

Expert Solution

The light intensity is maximum at I=2.

### Explanation of Solution

Given:

Given model: P=100II2+I+4.

Calculation:

Differentiate with respect to I,

dPdI=(I2+I+4)×100100I×(2I+1)(I2+I+4)2=100I2+400(I2+I+4)2

For critical points, set dPdI=0

100I2+400(I2+I+4)2=0100I2+400=0I24=0I2=4

Then, I=±2.

Since the light intensity can never be negative, reject I=2.

Thus, a possible light intensity for the maximum P occurrence is at I=2.

Again differentiate dPdI with respect to I,

d2PdI2=(I2+I+4)2×(200I)(100I2+400)×2×(I2+I+4)×(2I+1)(I2+I+4)4=(I2+I+4)×(200I)(100I2+400)(4I+2)(I2+I+4)3=200I32400I800(I2+I+4)3

When I=2, the value of d2pdI2 becomes,

d2pdI2=2002324002800(22+2+4)3=16004800800103=40001000

Therefore, d2pdI2=4<0.

Thus, by second derivative test we have a maximum at I=2.

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