Problem 21 - Applied Calculus

Question

A company operates two plants which manufacture the same item and whose total cost functions are $$C_{1}=8.5+0.03 q_{1}^{2} \quad \text { and } \quad C_{2}=5.2+0.04 q_{2}^{2}$$ where \(q_{1}\) and \(q_{2}\) are the quantities produced by each plant. The company is a monopoly. The total quantity demanded, \(q=q_{1}+q_{2},\) is related to the price, \(p,\) by $$p=60-0.04 q$$ How much should each plant produce in order to maximize the company's profit?\(^{11}\)

Step-by-Step Solution

Answer

Plant 1 should produce about 827 units; Plant 2, about 620 units.

Step 1: Define the Total Cost and Revenue Functions

To maximize profit, we need to define the total cost (TC) and total revenue (TR) functions. The total cost is given by summing the cost functions of both plants:\[TC = C_1 + C_2 = (8.5 + 0.03q_1^2) + (5.2 + 0.04q_2^2)\]The total revenue is found using the price equation and total quantity demanded:\[TR = p \times q = (60 - 0.04q) \times q = 60q - 0.04q^2\]where \(q = q_1 + q_2\).

Step 2: Define the Profit Function

Profit is total revenue minus total cost:\[Profit = TR - TC = (60q - 0.04q^2) - (8.5 + 5.2 + 0.03q_1^2 + 0.04q_2^2)\]Simplifying, the profit function becomes:\[Profit = 60q - 0.04q^2 - 13.7 - 0.03q_1^2 - 0.04q_2^2\]

Step 3: Differentiate the Profit Function with Respect to Quantities

To find maximum profit, we need to take the partial derivatives of the profit function with respect to \(q_1\) and \(q_2\) and set them equal to zero. The partial derivatives are:\[\frac{\partial \text{Profit}}{\partial q_1} = 60 - 0.08q_1 - 0.08q_2 - 0.06q_1 \quad (1)\]\[\frac{\partial \text{Profit}}{\partial q_2} = 60 - 0.08q_1 - 0.08q_2 - 0.08q_2 \quad (2)\]Setting these equations equal to zero, we find the critical points.

Step 4: Solve the Equations Simultaneously

Setting each partial derivative equal to zero, we have:Equation (1):\[60 = 0.08(q_1 + q_2) + 0.06q_1\]Equation (2):\[60 = 0.08(q_1 + q_2) + 0.08q_2\]Solve these equations simultaneously to find \(q_1\) and \(q_2\). First, let's ensure consistency by equalling both expressions:\[0.06q_1 = 0.08q_2\]Thus, \(q_1 = \frac{4}{3} q_2\). Substitute \(q_1 = \frac{4}{3} q_2\) into any of the original equations to find \(q_2\), then use it to find \(q_1\).

Step 5: Determine Exact Quantities

Substituting \(q_1 = \frac{4}{3} q_2\) into the rearranged equation from Step 4:\[60 = 0.08\left(\frac{4}{3}q_2 + q_2\right) + 0.08q_2\]Solve for \(q_2\):\[60 = \left(\frac{32}{24}q_2 + 0.08q_2\right) + 0.08q_2\]\[60 = 0.09667q_2\]\[q_2 \approx 620.41\]Now we find \(q_1\) using \(q_1 = \frac{4}{3}q_2\), resulting in:\[q_1 \approx 827.21\]

Step 6: Conclusion

The calculated values suggest plant 1 should produce approximately 827 units, while plant 2 should produce around 620 units to maximize profit.

Key Concepts

Cost Functions

Cost functions describe the total cost involved in producing goods. In our case, each plant has its own cost function, representing the production costs based on quantity. The cost functions are

• \(C_1 = 8.5 + 0.03 q_1^2\)
• \(C_2 = 5.2 + 0.04 q_2^2\)

These equations tell us how the cost increases as more items get produced. The squared term indicates the cost rises faster than the production increases (non-linear), reflective of factors like increased labor or raw material expenses after a certain threshold. Adding both cost functions gives the Total Cost (TC) for the entire operation, which is crucial for evaluating profitability.
Understanding these functions helps managers predict expenses for future production levels, making them critical for operational planning and decision-making.

Total Revenue

Total revenue (TR) is the total income a company receives from selling its products. For our monopoly situation, the total revenue is defined by the product of price and quantity sold. The price function is given by the equation, \(p = 60 - 0.04q\), which shows how pricing changes with demand.
This leads us to calculate the total revenue:

• \(TR = p imes q = (60 - 0.04q) imes q = 60q - 0.04q^2\)

The term \(60q\) shows potential revenue at initial price levels, while \(-0.04q^2\) captures the decrease in price as sales quantity increases. Total revenue is crucial since it helps in determining profit by comparing it with total costs. In economics, especially for monopolies, understanding how quantity affects revenue through pricing strategies is essential for maximizing profit.

Partial Derivatives

Partial derivatives are a part of calculus focused on finding the rate of change of a function concerning one variable while keeping others constant. They play a key role in optimizing functions, such as profit in business analysis.
For our exercise, we find partial derivatives of the profit function concerning quantities \(q_1\) and \(q_2\). This indicates how changes in production at each plant affect overall profit.

• \(\frac{\partial \text{Profit}}{\partial q_1}\) and \(\frac{\partial \text{Profit}}{\partial q_2}\)

By setting these partial derivatives to zero, we find critical points for maximization, allowing managers to decide the best production quantities at each plant.
This concept is essential in areas like finance and operations where optimizing output results in maximum economic benefit.

Monopoly

A monopoly exists when a firm is the sole producer of a good or service, with the power to influence market prices. This differs from competitive markets where many companies produce similar products. In monopolies, profit maximization requires careful planning of production quantities and pricing strategies to optimize revenue.
For this problem, the company is a monopoly, so it sets the price based on the quantity it decides to produce and sell. The downward sloping demand equation \(p = 60 - 0.04q\) is typical in monopolistic markets, where price decreases as quantity sold increases.

• Price setting power lies solely with the monopoly.
• Demand determines how price changes with quantity.

Understanding how monopolies operate helps students grasp broader economic implications, such as market interventions and economy-wide effects.