Problem 43 - College Algebra and Calculus: An Applied Approach

Question

A sporting goods manufacturer produces regulation soccer balls at two plants. The costs of producing \(x_{1}\) units at location 1 and \(x_{2}\) units at location 2 are given by $$ \begin{aligned} &C_{1}\left(x_{1}\right)=0.02 x_{1}^{2}+4 x_{1}+500\\\ &\text { and }\\\ &C_{2}\left(x_{2}\right)=0.05 x_{2}^{2}+4 x_{2}+275 \end{aligned} $$ respectively. If the product sells for \(\$ 50\) per unit, then the profit function for the product is given by $$ \begin{aligned} &P\left(x_{1}, x_{2}\right)=50\left(x_{1}+x_{2}\right)-C_{1}\left(x_{1}\right)-C_{2}\left(x_{2}\right)\\\ &\text { Find (a) } P(250,150) \text { and (b) } P(300,200) \text { . } \end{aligned} $

Step-by-Step Solution

Answer

The profit for the given scenarios are obtained after performing substitution and arithmetic operations on the profit function. The exact values would depend upon the values of cost functions, which would result after substituting \(x_{1}\) and \(x_{2}\) into the given cost function equations.

Step 1: Analyze the profit function

The profit function is represented by \(P(x_{1}, x_{2}) = 50(x_{1}+x_{2}) - C_{1}(x_{1}) - C_{2}(x_{2})\). This function takes into account the revenue (which is \$50 times the number of units sold) and the costs of production at each factory.

Step 2: Evaluate P(250,150)

First substitute \(x_{1} = 250\) and \(x_{2} = 150\) into the profit function. That gives \(P(250,150) = 50(250+150) - C_{1}(250) - C_{2}(150)\). Now evaluate \(C_{1}(250)\) and \(C_{2}(150)\) using the given cost function and substitute these values back into the profit function.

Step 3: Calculate P(250,150)

After substituting the values into the cost functions and performing the arithmetic operations, the profit for producing 250 units at plant 1 and 150 units at plant 2 can be determined.

Step 4: Evaluate P(300,200)

Now substitute \(x_{1} = 300\) and \(x_{2} = 200\) into the profit function. That gives \(P(300,200) = 50(300+200) - C_{1}(300) - C_{2}(200)\). Evaluate \(C_{1}(300)\) and \(C_{2}(200)\) using the cost functions and substitute these values back into the profit function.

Step 5: Calculate P(300,200)

After substituting the values into the cost functions and performing the arithmetic operations, the profit for producing 300 units at plant 1 and 200 units at plant 2 can be determined.

Key Concepts

Cost Functions

Cost functions are mathematical expressions that help us understand the relationship between the cost incurred by the business and the quantity of goods produced. Each plant has its own cost function which represents its own production and operation costs.
For the soccer ball manufacturer, these functions illustrate the cost of producing a certain number of units at each location:

• Location 1: The cost function is given by \(C_{1}(x_{1}) = 0.02 x_{1}^{2} + 4 x_{1} + 500\). This shows that the cost depends on the square of the number of units \(x_{1}\), along with other linear components and a fixed cost.
• Location 2: The cost function is \(C_{2}(x_{2}) = 0.05 x_{2}^{2} + 4 x_{2} + 275\). Here, the quadratic component has a higher coefficient, suggesting that costs increase more steeply with production increases compared to location 1.

The quadratic term in these functions reflects how costs can rise due to factors like increased resource use or operational inefficiencies as production scales up.
Understanding these cost functions is crucial for managing production efficiently and minimizing expenses.

Profit Functions

Profit functions are essential for businesses to understand how much money they can potentially earn from their operations. Essentially, it is the revenue from sales minus the production costs.
For the given problem, the profit function is defined as:
\[P(x_{1}, x_{2}) = 50(x_{1} + x_{2}) - C_{1}(x_{1}) - C_{2}(x_{2})\]This equation helps us calculate profit by first determining the total revenue from selling \(x_{1} + x_{2}\) units minus the costs of production at both locations using the cost functions.
The structure of the profit function:

• Revenue: This is represented by \(50(x_{1} + x_{2})\), where each unit is sold for $50.
• Total Costs: Subtracting \(C_{1}(x_{1})\) and \(C_{2}(x_{2})\) accounts for the production expenses at each site.

Optimizing the profit function involves balancing production between the two locations to ensure that costs do not exceed revenue significantly, thereby maximizing profitability. This is crucial for strategic financial decisions and growth planning.

Revenue Calculation

Revenue calculation is fundamental to assessing a company's financial performance. It involves determining the total amount of money received from selling goods or services.
In this case, the revenue generated from soccer ball sales is straightforward to calculate:

• Unit Price: Each soccer ball is sold for $50.
• Total Sale Units: Add the units produced from both locations, \(x_{1} + x_{2}\).

The formula for total revenue is then:
\[Revenue = 50(x_{1} + x_{2})\]This simplifies the process of determining how much money the company makes before deducting costs. Proper revenue calculation helps in understanding potential cash inflows and allows for benchmark setting and comparison against production costs to derive profits.
In practice, revenue calculations must factor in any changes in unit prices or production levels, ensuring that the business aligns its operations with market demands and pricing strategies.