In Example 4, find the prices that will yield a maximum profit when the total demand is limited to 250 units per year.

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

Given Information:

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

x1=200(p2−p1)x2=500+100p1−180p2

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=R−C

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)(p2−p1)+0.75(500+100p1+180p2)=100p2−100p1+375+75p1−135p2=375−25p1−35p2

And the revenue function is,

R=p1x1+p2x2=p1×200(p2−p1)+p2(500+100p1−180p2)=200p1p2−200p12+500p2+100p1p2−180p22=−200p12−180p22+300p1p2+500p2

Now, the profit function is

   P=R−C     =(−200p12−180p22+300p1p2+500p2)−(375−25p1−35p2)     =−200p12−180p22+300p1p2+500p2−375+25p1+35p2=−200p12−180p22+300p1p2+25p1+535p2−375

Hence the total demand is,

x1+x2=200(p2−p1)+500+100p1−180p2=200p2−200p1+500+100p1−180p2=−100p1+20p2+500

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

Now, the constraint is x1+x=2250,

−100p1+20p2+500−250=0−100p1+20p2+250=0−10(10p1−2p2−25)=0         10p2−2p2−25=0

Let g(p1,p2)=10p1−2p2−25.

Then the new function is

F(p1,p2,λ)=R−λg(p1,p2)=−200p12−180p22+300p1p2+25p1+535p2−375−λ(10p1−2p2−25)

Where, λ is Lagrange multiplier