Chapter 7.6, Problem 4CP

### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

Chapter
Section

### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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# In Example 4, find the prices that will yield a maximum profit when the total demand is limited to 250 units per year.

To determine

To calculate: The price that will yield a maximum profit when the total demand is limited to 250 units per year.

Explanation

Given Information:

The demand function of the company which makes two substitute products is,

x1=200(p2p1)x2=500+100p1180p2

Where p1  and  p2 are prices per unit in dollar and x1,x2 are the numbers of units sold.

The costs of producing the two products are $0.50 and$0.75 per unit respectively. And the total demand is limited to 250 units per year.

Formula used:

Step 1: Find the cost function and then find the revenue function by using the formula,

R=p1x1+p2x2

Step 2: Then find the profit function by using the formula,

P=RC

Step 3: Find the demand by the expression x1+x2.

Step 4: From then calculate the equation of constant.

Step 5: Find the new function using Lagrange multiplier.

Step 6: Find the maximum value of that function.

Calculation:

Consider the given cost function is,

C=0.5x1+0.75x2=0.5(200)(p2p1)+0.75(500+100p1+180p2)=100p2100p1+375+75p1135p2=37525p135p2

And the revenue function is,

R=p1x1+p2x2=p1×200(p2p1)+p2(500+100p1180p2)=200p1p2200p12+500p2+100p1p2180p22=200p12180p22+300p1p2+500p2

Now, the profit function is

P=RC     =(200p12180p22+300p1p2+500p2)(37525p135p2)     =200p12180p22+300p1p2+500p2375+25p1+35p2=200p12180p22+300p1p2+25p1+535p2375

Hence the total demand is,

x1+x2=200(p2p1)+500+100p1180p2=200p2200p1+500+100p1180p2=100p1+20p2+500

Since the total demand is limited to 250 units per year.

Now, the constraint is x1+x=2250,

100p1+20p2+500250=0100p1+20p2+250=010(10p12p225)=0         10p22p225=0

Let g(p1,p2)=10p12p225.

Then the new function is

F(p1,p2,λ)=Rλg(p1,p2)=200p12180p22+300p1p2+25p1+535p2375λ(10p12p225)

Where, λ is Lagrange multiplier

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