Problem 42 - College Algebra and Calculus: An Applied Approach

Question

A corporation manufactures candles at two locations. The cost of producing \(x_{1}\) units at location 1 is \(C_{1}=0.02 x_{1}^{2}+4 x_{1}+500\) and the cost of producing \(x_{2}\) units at location 2 is \(C_{2}=0.05 x_{2}^{2}+4 x_{2}+275 .\) The candles sell for \(\$ 15\) per unit. Find the quantity that should be produced at each location to maximize the profit \(P=15\left(x_{1}+x_{2}\right)-C_{1}-C_{2}\)

Step-by-Step Solution

Answer

To maximize profit, 275 units should be produced at location 1 and 110 units at location 2.

Step 1: Derive the given functions

Let's first derive the cost function for both locations with respect to \(x_{1}\) and \(x_{2}\) respectively:\n\(C_{1}' = 0.04x_{1} + 4\),\n\(C_{2}' = 0.1x_{2} + 4\).\nWe also formulate the profit function: \n\(P = 15(x_{1} + x_{2}) - (0.02 x_{1}^{2}+4 x_{1}+500) - (0.05 x_{2}^{2}+4 x_{2}+275).\)

Step 2: Sort out the Profit Function and Apply Derivatives

Simplified, the profit function becomes: \(P = 15x_{1} + 15x_{2} - 0.02 x_{1}^2 - 4x_{1} - 500 - 0.05 x_{2}^2 - 4x_{2} - 275.\) \nTo find the maximum profit, we need to differentiate it with respect to both \(x_{1}\) and \(x_{2}\), equate each to zero and then solve the equations concurrently.\nDifferentiate the profit function with respect to \(x_{1}\) and then equate to zero: \(P'_{x_{1}} = 15 - 0.04x_{1} - 4 = 0\).\nDifferentiate the profit function with respect to \(x_{2}\) and then equate to zero: \( P'_{x_{2}} = 15 - 0.1x_{2} - 4 = 0\).

Step 3: Solving the Equations

Solve \( P'_{x_{1}} = 0 \) and \( P'_{x_{2}} = 0 \) concurrently\nThis gives a system of equations:\n\(0.04x_{1} = 11\) for the first equation,\ntherefore \(x_{1} = \frac{11}{0.04} = 275\)\nNext, \(0.1x_{2} = 11\) for the second equation,\ntherefore \(x_{2} = \frac{11}{0.1} = 110\).

Key Concepts

Cost Function

A cost function represents the total cost of producing a certain number of goods. In this exercise, the cost function for each location provides the cost based on the number of units produced. At location 1, the cost function is given by:\[ C_{1}=0.02 x_{1}^{2}+4 x_{1}+500 \]This equation shows that the cost involves not just a fixed cost (500), but also variable costs that depend on how many units \(x_{1}\) are produced. Here, \(0.02 x_{1}^{2}\) represents the increasing cost as more units are produced, which is typical in real-world scenarios due to factors like overtime or resource limitations.
For location 2, the cost is:\[ C_{2}=0.05 x_{2}^{2}+4 x_{2}+275 \]Again, a fixed cost (275) is present, with the variable portion \(0.05 x_{2}^{2}\) representing the scaling factor of production at this site.
Understanding cost functions is crucial for businesses to evaluate how costs rise with increased production, allowing for better budgeting and financial planning.

Derivatives

Derivatives in mathematics are used to determine the rate of change of a function. They are particularly useful for finding the maximum or minimum values of a function, which in business contexts often relate to costs, revenue, or profit.
In this problem, we took the derivative of each cost function to analyze the incremental cost associated with producing an additional unit. For instance, the derivative of the cost function at location 1:\[ C_{1}' = 0.04x_{1} + 4 \]This derivative gives the marginal cost, or the cost of increasing production by one more unit from the current level of production at location 1.
Applying derivatives to the overall profit function allows us to determine where its rate of change reaches zero, which indicates a potential maximum or minimum profit point. In this context, setting the derivative equal to zero helps in figuring out the production level that maximizes profit.

System of Equations

A system of equations consists of multiple equations that are solved together. They are fundamental in solving problems where multiple conditions must be satisfied simultaneously. In this exercise, we use a system of equations to determine the production levels \(x_{1}\) and \(x_{2}\) that maximize profit.
The problem derived two equations by setting the profit function's derivatives to zero:

• \( P'_{x_{1}} = 15 - 0.04x_{1} - 4 = 0 \)
• \( P'_{x_{2}} = 15 - 0.1x_{2} - 4 = 0 \)

Solving these equations concurrently gives:

• \(0.04x_{1} = 11\), leading to \(x_{1} = 275\)
• \(0.1x_{2} = 11\), resulting in \(x_{2} = 110\)

This numerical solution tells us the exact number of units that should be produced at each location for maximum profitability.

Corporate Production

Corporate production involves planning and decision-making activities that manage how goods are produced to meet sales and profit objectives. In this problem, a corporation needs to determine how many candles to produce at each location to maximize profit. It involves analyzing costs, revenues, and the overall market price per unit to strike the best balance between what is produced and its associated costs.
Combining knowledge of cost functions and profit maximization helps businesses optimize their expenditure and resources. Understanding the relationship between production levels and profitability allows companies to adjust operations across multiple locations to achieve the best financial outcomes.
Strategic production decisions incorporate mathematical tools like derivatives and systems of equations, which ensure businesses not only cover costs but also enhance profitability through calculated, evidence-based production plans.