Math Problem Statement

Suppose a company has fixed costs of $49,400 and variable cost per unit of 1/3x + 333 dollars, where x is the total number of units produced. Suppose further that the selling price of its product is 2259-2/3x dollars per unit.

Solution

In this problem, we are given the fixed and variable costs, as well as the selling price per unit. We can break down the question and set up the necessary equations for total cost, total revenue, and eventually profit, depending on what you're looking to solve.

Key components of the problem:

  1. Fixed Costs: These are constant costs that do not depend on the number of units produced, i.e., $49,400.

  2. Variable Costs per Unit: This is given as 13x+333\frac{1}{3}x + 333 dollars, where xx is the number of units produced.

  3. Selling Price per Unit: The selling price per unit is 225923x2259 - \frac{2}{3}x.

Formulas

1. Total Cost:

The total cost C(x)C(x) includes both fixed and variable costs. The variable cost depends on the number of units xx produced.

C(x)=Fixed Cost+Variable CostxC(x) = \text{Fixed Cost} + \text{Variable Cost} \cdot x

C(x)=49,400+(13x+333)xC(x) = 49,400 + \left( \frac{1}{3}x + 333 \right) \cdot x

Expanding the variable cost part:

C(x)=49,400+(13x2+333x)C(x) = 49,400 + \left( \frac{1}{3}x^2 + 333x \right)

Thus, the total cost function is:

C(x)=49,400+13x2+333xC(x) = 49,400 + \frac{1}{3}x^2 + 333x

2. Total Revenue:

The total revenue R(x)R(x) is the selling price per unit times the number of units produced xx:

R(x)=x(225923x)R(x) = x \cdot \left( 2259 - \frac{2}{3}x \right)

Expanding this:

R(x)=2259x23x2R(x) = 2259x - \frac{2}{3}x^2

3. Profit Function:

The profit function P(x)P(x) is the total revenue minus the total cost:

P(x)=R(x)C(x)P(x) = R(x) - C(x)

Substitute R(x)R(x) and C(x)C(x) into this equation:

P(x)=(2259x23x2)(49,400+13x2+333x)P(x) = \left( 2259x - \frac{2}{3}x^2 \right) - \left( 49,400 + \frac{1}{3}x^2 + 333x \right)

Simplify the profit function:

P(x)=2259x23x249,40013x2333xP(x) = 2259x - \frac{2}{3}x^2 - 49,400 - \frac{1}{3}x^2 - 333x

Combine like terms:

P(x)=(2259x333x)(23x2+13x2)49,400P(x) = (2259x - 333x) - \left( \frac{2}{3}x^2 + \frac{1}{3}x^2 \right) - 49,400

P(x)=1926xx249,400P(x) = 1926x - x^2 - 49,400

Summary of the equations:

  • Total Cost: C(x)=49,400+13x2+333xC(x) = 49,400 + \frac{1}{3}x^2 + 333x
  • Total Revenue: R(x)=2259x23x2R(x) = 2259x - \frac{2}{3}x^2
  • Profit: P(x)=1926xx249,400P(x) = 1926x - x^2 - 49,400

Let me know if you'd like to explore further or find specific values like the break-even point or the maximum profit.

Questions:

  1. What is the break-even point for this company?
  2. How many units should be produced to maximize profit?
  3. What is the maximum profit the company can achieve?
  4. At what production level does the company start incurring a loss?
  5. How does an increase in fixed costs affect the profit function?

Tip: In these types of problems, analyzing the profit function for its maximum or minimum often involves taking the derivative and setting it to zero, which helps in identifying the optimal production level.

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Math Problem Analysis

Mathematical Concepts

Cost and Revenue Analysis
Quadratic Equations
Profit Maximization

Formulas

Total Cost: C(x) = 49,400 + (1/3)x^2 + 333x
Total Revenue: R(x) = 2259x - (2/3)x^2
Profit: P(x) = R(x) - C(x)

Theorems

Profit maximization through differentiation
Quadratic optimization

Suitable Grade Level

Grades 10-12