   Chapter 9, Problem 103RE Mathematical Applications for the ...

12th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781337625340

Solutions

Chapter
Section Mathematical Applications for the ...

12th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781337625340
Textbook Problem

In Problems 100-107, cost, revenue, and profit are in dollars and x is the number of units.Profit If the total revenue function for a product is given by R ( x )   =   60 x and the total cost function is given by C   =   200   +   10 x +   0.1 x 2 , what is the marginal profit at x   = 10 ? What does the marginal profit at x =   10 predict ?

To determine

To calculate: The marginal profit when number of units is 10 units and interprets the marginal profit for 10 units where the cost function is C=200+10x+0.1x2 and the revenue function is R(x)=60x.

Explanation

Given Information:

The total revenue for a product is represented by the function, R(x)=60x. The total cost function is represented by the function, C=200+10x+0.1x2.

Formula used:

The simple power rule for the derivative,

ddx(xn)=nxn1

The rule of derivative for constant multiplication,

ddx[kf(x)]=kddx[f(x)]

The sum or difference rule of the derivative,

ddx[f(x)+g(x)]=ddx[f(x)]+ddx[g(x)]

Calculation:

Consider the provided cost function, C(x)=200+10x+0.1x2 and revenue function, R(x)=60x.

The profit function is the difference of revenue function and cost functions,

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

Substitute 60x for R(x) and 200+10x+0.1x2 for C(x).

P(x)=R(x)C(x)=60x(200+10x+0.1x2)=60x20010x0.1x2=50x2000.1x2

To find the maximum profit, differentiate the profit function with respect to x, use the sum or difference of the derivative.

P(x)=ddx(50x2000

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