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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.

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.

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.

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.