Problem 7 - A First Course in Mathematical Modeling

Question

The total cost to manufacture one unit of product \(A\) is \(\$ 3\), and for one unit of product \(B\) it is \(\$ 2\). If \(x\) and \(y\) are the retail prices per unit of \(A\) and \(B\), respectively, then marketing research has established that $$ Q_{A}=2750-700 x+200 y $$ and $$ Q_{B}=2400+150 x-800 y $$ are the quantities of each product that will be sold each day. Find a function \(P(x, y)\) modeling the daily profit and the maximum daily profit.

Step-by-Step Solution

Answer

The maximum profit is calculated when the profit function is maximized using its critical points.

Step 1: Understand the profit function

Profit is calculated as the revenue from selling products minus the cost of manufacturing them. We need to form a profit function \( P(x, y) \) based on this information.

Step 2: Calculate Revenue for Products A and B

Revenue from selling product \( A \) is the retail price \( x \) times the quantity \( Q_A \), i.e., \( x(2750 - 700x + 200y) \). Similarly, for product \( B \), it is \( y(2400 + 150x - 800y) \).

Step 3: Consider the Cost of Manufacturing

The cost for manufacturing product \( A \) is \( 3 \times Q_A \), and for product \( B \) it is \( 2 \times Q_B \).

Step 4: Formulate the Profit Function

The profit function \( P(x, y) \) is given by Total Revenue - Total Cost. So, \( P(x, y) = [x(2750 - 700x + 200y) + y(2400 + 150x - 800y)] - [3(2750 - 700x + 200y) + 2(2400 + 150x - 800y)] \).

Step 5: Simplify the Profit Function

Simplify the profit function by expanding and combining like terms. The full expression becomes:\[ P(x, y) = -700x^2 + 200xy + 150xy - 800y^2 + 2750x + 2400y - 8250 + 4400x + 400y \].Further simplifying, we get:\[ P(x, y) = -700x^2 - 800y^2 + 350xy + 7150x + 2800y - 8250 \].

Step 6: Find Criticial Points

To find critical points, take partial derivatives \( \frac{\partial P}{\partial x} \) and \( \frac{\partial P}{\partial y} \), set them to zero, and solve the system of equations.

Step 7: Check the Nature of Critical Points

Use the second derivative test to determine if the critical point found is a maximum. Calculate the second partial derivatives and check the determinant of the Hessian matrix.

Step 8: Calculate Maximum Profit

Substitute the critical points back into the profit function \( P(x, y) \) to find the maximum profit. Evaluate \( P(x, y) \) at the critical points to obtain the maximum value.

Key Concepts

Revenue Calculation

To understand how much money a company makes from selling its products, we calculate the revenue. It’s simply the total income from selling all units of products.

In the exercise, we have products A and B. The revenue for product A is calculated as the product of its price and quantity sold. Since research indicates that quantity sold of A depends on the prices per unit, i.e., \( Q_{A} = 2750 - 700x + 200y \), the revenue for product A becomes:

• \( x(2750 - 700x + 200y) \)

Where \( x \) is the retail price per unit for product A.

Similarly, the revenue for product B uses the equation \( Q_{B} = 2400 + 150x - 800y \), making its revenue:

• \( y(2400 + 150x - 800y) \)

Where \( y \) is the retail price per unit for product B. By summing up these revenues for A and B, we obtain the total revenue. It offers an essential insight into the potential earnings before deducting any costs.

Cost of Manufacturing

The cost of manufacturing represents the expenses incurred in producing the products.

In our context, it involves evaluating how much it costs to produce each unit of products A and B. For product A, the cost per product unit is \( \\(3 \) multiplied by the number of units \( Q_{A} \). Similarly, for product B, it is \( \\)2 \) times \( Q_{B} \).

• The cost for producing A: \( 3(2750 - 700x + 200y) \)
• The cost for producing B: \( 2(2400 + 150x - 800y) \)

The sum of these equations gives the total manufacturing cost. This calculation helps us realize how much the company spends on producing these items before making any sales. Understanding the total cost is crucial to evaluate profitability correctly.

Critical Points Analysis

Finding critical points involves looking for values of \( x \) and \( y \) where the profit is stationary.

This analysis is crucial as these points could be where the maximum or minimum profit occurs. In mathematical terms, it involves taking partial derivatives of the profit function \( P(x, y) \) with respect to both \( x \) and \( y \), and setting them equal to zero.

• \( \frac{\partial P}{\partial x} = 0 \)
• \( \frac{\partial P}{\partial y} = 0 \)

Solving this system of equations helps in determining the potential critical points. Through critical points analysis, we focus on understanding where the profit doesn't change with small price alterations, possibly hinting at optimal pricing standards.

Second Derivative Test

The second derivative test informs us about the nature of the critical points.

It allows verifying whether these points are maxima, minima, or saddle points. After locating critical points, it's crucial to determine if they offer maximum profit. We do this by calculating the second partial derivatives and constructing the Hessian matrix.

• \( \frac{\partial^2 P}{\partial x^2} \)
• \( \frac{\partial^2 P}{\partial y^2} \)
• \( \frac{\partial^2 P}{\partial x \partial y} \)

Next, we find the determinant of the Hessian matrix. Checking the signs helps decide:

• If the determinant is positive and \( \frac{\partial^2 P}{\partial x^2} \) is negative, it's a local maximum.
• If negative, it's a saddle point.
• If positive and \( \frac{\partial^2 P}{\partial x^2} \) is positive, it indicates a local minimum.

Thus, this test is crucial in ensuring the identified points indeed lead to optimal profits.