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Mathematical Applications for the ...

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

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BuyFindarrow_forward

Mathematical Applications for the ...

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

The function R ¯ ( x ) = R ( x ) / x defines the average revenue for selling x units. For

R ( x ) = 2000 x + 20 x 2 x 3

(a) find the maximum average revenue.

(b) show that R ¯ ( x ) attains its maximum at an x- value where R ¯ ( x ) = M R ¯ .

(a)

To determine

To calculate: The maximum average revenue for the function R(x)=2000x+20x2x3 if the average revenue is R¯(x)=R(x)x.

Explanation

Given Information:

The provided revenue function is:

R(x)=2000x+20x2x3

And the average revenue is defined as:

R¯(x)=R(x)x

Formula used:

If f(x) and g(x) are two differentiable functions then by the property of derivative:

ddx(f(x)+g(x))=ddxf(x)+ddxg(x)

And,

ddxxn=nxn1

Where n is a constant and x is the variable.

Calculation:

Average revenue is given as:

R¯(x)=R(x)x=2000x+20x2x3x=2000+20xx2

The absolute maxima and absolute minima will occur only at the critical points. To calculate the critical points of the revenue function, find the first derivative of the function:

R¯(x)=2000+20xx2ddx(R¯)=ddx(2000+20xx2)

Use ddx(f(x)+g(x))=

(b)

To determine

To prove: The revenue function R¯(x) attains its maximum where R¯(x)=(MR) at an x value if revenue function is R(x)=2000x+20x2x3.

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