   Chapter 10.3, Problem 40E ### 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

# The monthly demand function for x units of a product sold by a monopoly is p   = 5900   — 1 2 x dollars, and its average cost is C ¯ = 3020 + 2 x dollars. If production is limited to 100 units, find the number of units that maximizes profit. Will the maximum profit result in a profit or loss?

To determine

To calculate: The number of units that maximizes profit and will the maximum profit result in a profit or loss if the monthly demand function for x units of a product sold by a monopoly is p=590012x2 dollars, and its average cost is C¯=3020+2x dollars and the production is limited to 100 units.

Explanation

Given information:

The provided information is:

The monthly demand function for x units of a product sold by a monopoly is p=590012x2 dollars, and its average cost is C¯=3020+2x dollars and the production is limited to 100 units.

Formula used:

The total profit function is:

P(x)=R(x)C(x)

Where R(x)andC(x) are total revenue function and total cost function.

Calculation:

Consider the provided information,

The monthly demand function for x units of a product sold by a monopoly is p=590012x2 dollars, and its average cost is C¯=3020+2x dollars and the production is limited to 100 units.

The monthly demand function for x units of a product sold by a monopoly is:

p=590012x2 dollars

The average cost is given by:

C¯=3020+2x

Total cost function can be written as:

C(x)=C¯x=3020x+2x2

Total revenue function can be written as:

R(x)=p.x=590012x3

P(x) can be defined as:

P(x)=R(x)C(x)

Now, substitute the values of R(x)andC(x) in the profit formula as, P(x)=R(x)C(x).

P(x)=R(x)C(x)=590012x3(3020x2x3)=12x32x2+2880x

Hence, the value of P(x) is P(x)=12x32x2+2880x

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