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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

If club members charge $5 admission to a classic car show, 1000 people will attend, and for each $1 increase in price, 100 fewer people will attend. What price will give the maximum revenue for the show? Find the maximum revenue.

To determine

To calculate: The price that yields maximum revenue and the maximum revenue for $5 admission charge 1000 people attends the show.

Explanation

Given Information:

For $5 admission charge 1000 people attends the show and for each $1 increase in price reduces 100 people from the crowd.

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:

Assume that x is the increase in the price, then the revenue can be written as:

Revenue=(number of persons attending show)(price paid by each person)=(1000100x)(5+x)R(x)=100x2+500x+5000

As, the number of persons attending show cannot be negative so,

1000100x0100x1000x10

Therefore, 0x10.

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)=100x2+500x+5000ddx(R)=ddx(100x2+500x+5000)

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

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