Problem 43 - Calculus an Applied Approach

Question

Profit 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 $$ \begin{array}{l}{C_{1}\left(x_{1}\right)=0.02 x_{1}^{2}+4 x_{1}+500} \\\ {\text { and }} \\ {C_{2}\left(x_{2}\right)=0.05 x_{2}^{2}+4 x_{2}+275}\end{array} $$ respectively. If the product sells for 50 dollars per unit, then the profit function for the product is given by \(P\left(x_{1}, x_{2}\right)=50\left(x_{1}+x_{2}\right)-C_{1}\left(x_{1}\right)-C_{2}\left(x_{2}\right)\) Find (a) \(P(250,150)\) and (b) \(P(300,200)\)

Step-by-Step Solution

Answer

To solve the task, the given numbers need to be substituted into the formulas and calculated accordingly. The results are then used in the profit formula to determine the final profit.

Step 1: Determine Profit for (a) \(P(250,150)\)

Substitute \(x_1 = 250\) and \(x_2 = 150\) into the profit function: \(P(250,150)=50(250+150)-C_{1}(250)-C_{2}(150)\)

Step 2: Calculate Cost at Location 1 for (a)

Substitute \(x_1 = 250\) into cost function for location 1: \(C_{1} (250) = 0.02(250)^2 + 4(250) + 500\)

Step 3: Calculate Cost at Location 2 for (a)

Substitute \(x_2 = 150\) into cost function for location 2: \(C_{2} (150) = 0.05(150)^2 + 4(150) + 275\)

Step 4: Calculate Profit for (a)

Replace \(C_{1}(250)\) and \(C_{2}(150)\) in the profit function with their respective values and simplify.

Step 5: Determine Profit for (b) \(P(300,200)\)

Substitute \(x_1 = 300\) and \(x_2 = 200\) into the profit function: \(P(300,200)=50(300+200)-C_{1}(300)-C_{2}(200)\)

Step 6: Calculate Cost at Location 1 for (b)

Substitute \(x_1 = 300\) into cost function for location 1: \(C_{1} (300) = 0.02(300)^2 + 4(300) + 500\)

Step 7: Calculate Cost at Location 2 for (b)

Substitute \(x_2 = 200\) into cost function for location 2: \(C_{2} (200) = 0.05(200)^2 + 4(200) + 275\)

Step 8: Calculate Profit for (b)

Replace \(C_{1}(300)\) and \(C_{2}(200)\) in the profit function with their respective values and simplify.

Key Concepts

Cost Function

A cost function represents the total cost incurred by a company in the production of a certain number of goods. For the context of our sporting goods manufacturer, the cost functions are defined for each production location separately.

• The function for location 1 is expressed as: \(C_{1}(x_{1}) = 0.02x_{1}^{2} + 4x_{1} + 500\)
• The function for location 2 is: \(C_{2}(x_{2}) = 0.05x_{2}^{2} + 4x_{2} + 275\)

These functions take into account both variable and fixed costs associated with production:

• The term with the variable \(x^2\) is the quadratic component, representing the increasing marginal cost as production scales up.
• The linear term, involving just \(x\), accounts for costs that directly vary with output.
• The constant term is the fixed cost, independent of the production quantity.

Break down each part to understand how costs accumulate with each additional unit produced. This understanding is fundamental in financial planning and analysis for optimizing production strategies.

Quadratic Function

Quadratic functions are a type of polynomial function characterized by having the highest degree of 2. In our exercise, the cost functions at each location are quadratic in nature. Quadratic functions generally take the form \(ax^2 + bx + c\). Let's explore its application to our cost functions:

• The \(a\) term (e.g., 0.02 for location 1) indicates how costs accelerate as production increases. It reflects a phenomenon known as "diminishing returns" in real-world production scenarios, where increasing output might require proportionally more input.
• The \(b\) term (4 in both cost functions) represents linear cost behavior — direct correlations to the number of units produced.
• The constant \(c\) (such as 500 and 275) represents fixed costs, which do not change with output volume.

Quadratic functions like these exhibit a U-shaped curve, where initially costs might increase slowly, then accelerate. Understanding the structure of quadratic functions helps in predicting cost behavior as production changes, an essential aspect of business decision-making.

Optimization

Optimization deals with finding the best solution from a set of available alternatives, often subject to various constraints. In business, this typically involves maximizing profit or minimizing cost. In our exercise, the goal is to determine the optimal number of units to produce at each location to maximize profit.The profit function for this scenario is given by:\[P(x_{1}, x_{2}) = 50(x_{1} + x_{2}) - C_{1}(x_{1}) - C_{2}(x_{2})\]Here are steps typically taken in optimization:

• Identify the constraints (e.g., production capacity and costs).
• Define an objective function, like the profit function in this exercise.
• Use calculus or computational methods to identify points where the function reaches its maximum or minimum values.

In practical situations, optimization may involve evaluating derivative functions to determine turning points where profit is maximized or costs are minimized. By mastering these concepts, businesses can strategically adjust production inputs to optimize their financial outcomes.