   Chapter 10, Problem 17RE ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042

#### Solutions

Chapter
Section ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042
Textbook Problem

# Given y =   6400 x   −   18 x 2 − x 3 , find the absolute maximum and minimum for y when (a) 0 ≤ x ≤ 50 and (b) 0   ≤   x   ≤   100 .

(a)

To determine

To calculate: The absolute maximum and minimum for R for the expression R=6400x18x2x33 when 0x50.

Explanation

Given Information:

The provided equation is R=6400x18x2x33 when 0x50.

Formula Used:

To obtain the critical points of a function,

Step 1. Find the first derivative of the function.

Step 2. Set the derivative equal to 0 now obtain the critical points.

For the function f is present, then the value f(a) is the absolute maximum of f

If f(a)f(x) for all x in the domain of f, absolute maximum is the highest point on the graph.

For the function f is present, then the value f(a) is the absolute minimum of f

If f(a)f(x) for all x in the domain of f, absolute minimum is the lowest point on the graph.

Calculation:

Consider the provided equation R=6400x18x2x33,

The critical values are the only values at which the graph can have turning points, the derivative cannot change sign anywhere except at the critical value.

Hence, there will no change in the values of critical values as in the derivative graph.

Take out the first derivative of the equation by the power rule,

R=640036xx2

Put the value of R=0,

R=640036xx2=0=(100+x)(64x)=0

Evaluate the values of x from the equation:

(100+x)(64x)=0x=100,64

Hence, the values of x are x=100,64

(b)

To determine

To calculate: The absolute maximum and minimum for the expression R=6400x18x2x33 when 0x100.

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