Chapter 7.5, 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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# Find the prices that will yield a maximum profit for the products in Example 4 when the costs of producing the two products are $0.75 and$0.50 per unit, respectively.

To determine

To calculate: The unit prices that will yield a maximum profit for the products, which are substitute, with the respective demand functions given by,

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

Where p1 and p2 are the prices per unit (in dollar) and x1, x2 are the numbers of unit 5 sold, when the costs of producing the two products are dollar 0.75 and dollar 0.50 per unit, respectively.

Explanation

Given Information:

The demand function for the product are given by,

x1=200(p2p1)and x2=500+100p1180p2

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

The costs of producing the two products are dollar 0.75 and dollar 0.50 per unit respectively.

Formula used:

The demand functions for any pairs of substance products produced by a company is,

x1=f1(p1,p2)x2=f2(p1,p2);

The cost function of the pair of substitute products for the company is,

c=c1x1+c2x2

The reverse function of the pair of substitute products for the company is,

R=p1x1+p2x2

and the profit function for the company is,

P=RC

Where x1 and x2 are the total of the substitute products demanded and sold in the market, p1 and p2 are the per unit prices of the substitute products respectively, c1 and c2 are the per unit cost of producing the substitute products respectively, C and R and P are respectively the total cost, total reverse and total profit for the company.

The following procedure is used to optimise the profit function:

Step 1: Write the primary equation for the quantities that are to be optimised, there the primary equations is the profit function.

Step 2: Express the primary equation, in terms of the quantities that are to the optimised, with help of secondary tertiary and other functions.

Step 3: Find the first partial derivatives of the primary equation with respect to each quantity that are to be optimised, separate by.

Step 4: Set the first partial derivatives to zero find the critical values of the required quantities.

Step 5: Find second partial derivatives to check if the critical values, obtained at step 4, helps to satisfy the required condition.

Step 6: If the required condition is satisfied, then use the critical values to find further results. Else, state that the optimisation cannot be reached.

Calculation:

The primary equation is the profit function,

P=RC

Where P is the total profit, R is the total revenue and c is the total cost.

The secondary equation in the given demand function as,

x1=200(p2p1) and x2=500+100p1180p2

Where x1 and x2 are the total units of the substitute products sold at the market with per unit price p1 and p2 respectively, in dollars.

The tertiary equation is the cost function here, with the given respective per unit costs as shorn below.

c=0.75x1+0.50x2

Where x1 and x2 are the total units of the substitute products produced with total cost C.

The quaternary equation is the revenge function here, with R as total revenue as shown below :

R=p1x1+p2x2

Where x1 and x2 are the total units of the substitute products demand and sold at the market with respective demanded and sold at the market with respective per unit prices p1 and p2, in dollar’s.

x1=200(p1p1) and x2=500+100p1180p2.

Put form the demand function in the tertiary equation to get,

c=0.75(200)(p2p1)+0.50(500+100p1180p2)=2580100p1+60p2

Put x1=200(p2p1) and x2=500+150p1180p2 in the quaternary equation to get,

R=p1(200(p2p1)+p2(500+100p1180p2)=300p1p2200p12180p22+500p2

Put the expressions obtained for C and R in the primary equation to get5 the expressed primary equations

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