Question

A two-product firm faces the following demand and cost functions: \(Q_{1}=40-2 P_{1}-P_{2} \quad Q_{2}=35-P_{1}-P_{2} \quad C=Q_{1}^{2}+2 Q_{2}^{2}+10\) (a) Find the output levels that satisfy the first-order condition for maximum profit. (Use fractions.) (b) Check the second-order sufficient condition. Can you conclude that this problem possesses a unique absolute maximum? (c) What is the maximal profit?

Short answer

Calculate the first-order conditions and check the second-order conditions to confirm local maxima. Substitute to find the maximal profit.

Step-by-step solution

Understand the Functions

The demand functions for the two products are:- For product 1: \(Q_1 = 40 - 2P_1 - P_2\)- For product 2: \(Q_2 = 35 - P_1 - P_2\)The cost function is given as:\[C = Q_1^2 + 2Q_2^2 + 10\]Our goal is to maximize profit \(\pi\), which is revenue minus cost. The revenue for each product is \(P_i \times Q_i\), so total revenue is \((P_1 \times Q_1) + (P_2 \times Q_2)\). The profit function becomes:\[\pi = (P_1 \times Q_1) + (P_2 \times Q_2) - (Q_1^2 + 2Q_2^2 + 10)\]

Substitute Demand Functions

Substitute the demand functions into the profit equation:\[\pi = (P_1 \times (40 - 2P_1 - P_2)) + (P_2 \times (35 - P_1 - P_2)) - (Q_1^2 + 2Q_2^2 + 10)\]Simplify this to express profit in terms of prices \(P_1\) and \(P_2\).

Derive First-order Conditions

Take the partial derivatives of the profit function with respect to \(P_1\) and \(P_2\), and set them to zero:\[\frac{\partial \pi}{\partial P_1} = 0\]\[\frac{\partial \pi}{\partial P_2} = 0\]Solve these two equations simultaneously to find \(P_1\) and \(P_2\). Then, find \(Q_1\) and \(Q_2\) by substituting \(P_1\) and \(P_2\) back into the demand functions.

Check Second-order Conditions

Calculate the second-order derivatives to check if the solution is a maximum. Find:\[\frac{\partial^2 \pi}{\partial P_1^2}\], \[\frac{\partial^2 \pi}{\partial P_2^2}\], and \[\frac{\partial^2 \pi}{\partial P_1 \partial P_2}\].Use the Hessian determinant to check if the condition for a local maximum is satisfied:\[H = \left(\frac{\partial^2 \pi}{\partial P_1^2}\right)\left(\frac{\partial^2 \pi}{\partial P_2^2}\right) - \left(\frac{\partial^2 \pi}{\partial P_1 \partial P_2}\right)^2 > 0\]If satisfied, it indicates the critical points are local maxima.

Calculate Maximal Profit

Substitute the optimal \(P_1\), \(P_2\), \(Q_1\), and \(Q_2\) back into the original profit equation to find the maximal profit:\[\pi^* = (P_1 \times Q_1) + (P_2 \times Q_2) - (Q_1^2 + 2Q_2^2 + 10)\]

Key concepts

Profit Maximization

Profit maximization is a fundamental goal for firms in the realm of mathematical economics. Essentially, it involves finding the best combination of prices and output levels that yield the highest possible profit, which is the difference between total revenue and total cost. In this exercise, the firm sells two products, and the profit function given describes how revenue and costs interact. The formula for profit (") is:

• Total Revenue = Sales Price  Quantity of Product 1 + Sales Price  Quantity of Product 2
• Total Cost = Cost of Producing Quantity 1 plus twice the Cost of Producing Quantity 2, with a fixed cost of 10

To maximize this profit function, we first need to express it in terms of the prices of the products (1 and 2) by substituting the demand functions. This substitution turns the problem into a calculus exercise where first-order and second-order conditions help us find the optimal set points.

Demand Functions

Demand functions reflect how much of each product consumers are willing to buy at various price levels. In this exercise, we have two demand functions for two products:

• For product 1: \(Q_1 = 40 - 2P_1 - P_2\)
• For product 2: \(Q_2 = 35 - P_1 - P_2\)

These equations imply a relationship between the price of each product and the quantity demanded. When the price of one product goes up, the demand for it tends to go down, which is a typical downward-sloping demand relationship. The demand functions are critical in setting up the profit maximization problem because they determine how quantities sold will change as prices change, impacting both total revenue and total profit. By substituting these demand functions into the revenue equations, we can express the profit function solely in terms of prices rather than outputs. This makes it easier to apply calculus methods needed to find the maxima or minima.

Cost Functions

A cost function represents the total cost of production associated with different levels of output. For the firm in this exercise, the cost function is:\[C = Q_1^2 + 2Q_2^2 + 10\]This equation reveals that costs increase with the quantity produced of each product, but the cost of producing the second product increases at double the rate compared to the first product. The constant term "+10" represents fixed costs that don't change regardless of production levels. Understanding this cost function is crucial for profit maximization because it directly affects the profit equation. The aim is to figure out the levels of output (\(Q_1\) and \(Q_2\)) that balance the revenue obtained from selling these products against the costs incurred in producing them. By determining these levels, we achieve the highest possible profit.

First-order Conditions

First-order conditions are the set of criteria used in calculus to find the critical points where a function may reach a high or low value. In the context of profit maximization, they involve taking the derivative of the profit function with respect to each price, \(P_1\) and \(P_2\), and setting them to zero:

• \(\frac{\partial \pi}{\partial P_1} = 0\)
• \(\frac{\partial \pi}{\partial P_2} = 0\)

At these points, any small change in the variables \(P_1\) and \(P_2\) has no immediate effect on profit, revealing they could either be maximum or minimum profit points. Solving these equations gives the values of \(P_1\) and \(P_2\) where profit is extremized. Once found, these points must be tested using second-order conditions to determine if they represent a maximum, ensuring that the chosen price and output levels do indeed maximize profits rather than just balance between profits and losses.

Second-order Conditions

Second-order conditions are used to verify whether the critical points found using the first-order conditions are maxima or minima. In this context, second-order conditions involve calculating the second derivatives of the profit function and using the Hessian determinant:

• Second derivatives \(\frac{\partial^2 \pi}{\partial P_1^2}\) and \(\frac{\partial^2 \pi}{\partial P_2^2}\), along with \(\frac{\partial^2 \pi}{\partial P_1 \partial P_2}\)
• Hessian determinant, \(H = \left(\frac{\partial^2 \pi}{\partial P_1^2}\right)\left(\frac{\partial^2 \pi}{\partial P_2^2}\right) - \left(\frac{\partial^2 \pi}{\partial P_1 \partial P_2}\right)^2 > 0\)

If \(H\) is positive, it indicates the presence of a local maximum. These conditions ensure the solutions derived provide a reliable economic interpretation of being the most efficient prices and corresponding sales volumes maximizing profits, offering a form of certainty about the outcomes derived from the critical points.