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Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

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Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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Revenue In Exercises 39 and 40, find p 1  and  p 2 , the prices per unit (in dollars), so as to maximize the total revenue R   =   x 1 p 1 +   x 2 p 2 , where x 1  and  x 2 are the numbers of units sold, for a retail outlet that sells two substitute products with the given demand functions.

x 1 = 1000 2 p 1 + p 2 ,   x 2 = 1500 + 2 p 1 1.5 p 2

To determine

To calculate: The values of p1 and p2 to maximize the total revenue which is provided by R=x1p1+x2p2, where p1, p2 are the prices per unit (in dollars) and x1, x2 are number of units sold. The provided demand functions are x1=10002p1+p2 and x2=1500+2p11.5p2.

Explanation

Given Information:

The total revenue is provided by R=x1p1+x2p2, where p1, p2 are the prices per unit (in dollars) and x1, x2 are number of units sold. The demand functions are x1=10002p1+p2 and x2=1500+2p11.5p2.

Formula used:

If the function y=f(x) has a maximum at a point then their first partial derivatives fx(x) and fy(x) at that point are zero.

Calculation:

Consider the total revenue, R=x1p1+x2p2 and demand functions, x1=10002p1+p2, x2=1500+2p11.5p2

Substitute x1=10002p1+p2 and x2=1500+2p11.5p2 in R=x1p1+x2p2,

R=(10002p1+p2)p1+(1500+2p11.5p2)p2=(1000p12p12+p1p2)+(1500p2+2p1p21.5p22)=1000p12p12+3p1p2+1500p21.5p22

Now, the first partial derivatives are,

Rp1=p1[1000p12p12+3p1p2+1500p21

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