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Two important and similar ideas to the marginal cost are the marginal revenue an

ID: 2836886 • Letter: T

Question

Two important and similar ideas to the marginal cost are the marginal revenue and profit from producing a certain number of items. Consider the cost function C(x) = 2000 - 4x + 0.1x2 dollars, which gives the cost C to a company of producing x books. In general, the more items one produces, the less revenue one obtains per item from their sale. Suppose that when we produce x books, we can sell them for 56 - 0.1x dollars each. What is the revenue R(x) from selling x books? R(x) = What is the profit P(x) from selling x books? (Consider the revenue and cost functions.) P(x) = What is the actual additional revenue from selling the 101st book? (Round your answer to the nearest cent.) What is the additional profit? (Round your answer to the nearest cent.) What is the rate R'(x) at which the revenue is changing when 100 books are being produced? Include units. How does this compare with part (c)? This is a cents difference from the additional revenue found in part (c). Is the revenue increasing or decreasing with respect to x? What is the rate P(x) at which the profit is changing when 100 books are being produced? Include units. How does this compare with part (c)? This is a cents difference from the additional profit found in part (c). Is the profit increasing or decreasing with respect to x? increasing decreasing For what x would the revenue be a maximum? What is the maximum revenue? For what x would the profit be a maximum? What is the maximum profit?

Explanation / Answer

a)R(x)=(56-0.1x)*x

b)P(x)=R(x)-C(x)

       =(56-0.1x)*x-(2000-4x+0.1x^2)

       =-0.2x^2+60x-2000

c)Additional revenue on selling 101th book = R(101)-R(100) = (56-0.1*101)*101 - (56-0.1*100)*100=35.9$

                                                                                                                                             = 3590 cent

Additional profit = P(101)-P(100) = (-0.2*101^2+60*101-2000)-(-0.2*100^2+60*100-2000) = 19.8$

d)R'(x)=56-0.2x

R'(100)=36$

This is 10 cent difference from part c

Revenue is increasing with x.

e)P'(x)=-0.4x+60

P'(100)=20

This is 20 cent difference from part c

profit is increasing

f)Max revenue is at R'(x)=0

56-0.2x=0

=>x=280

R(280)=(56-0.1*280)*280 =7840$

Max profit is at P'(x)=0

-0.4x+60=0

=>x=150

P(150)=-0.2*150^2+60*150-2000 = 2500$

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