Math Problem Statement

Profit. The financial department for the company in

Problems 86 and 88 established the following cost function

for producing and selling x thousand notebook computers:

C1x2 = 4,000 + 500x thousand dollars

(A) Write a profit function for producing and selling x thou-

sand notebook computers and indicate its domain.

(B) Complete Table 13, computing profits to the nearest

thousand dollars.

Table 13 Profit

x (thousands) P1x2 (thousand $)

1 -2,560

5

10

15

20

25

(C) Plot the points in part (B) and sketch a graph of the

profit function using these points.

Solution

Let's break down the problem step by step:

Part (A): Write a Profit Function and Indicate Its Domain

1. Cost Function:

Given the cost function for producing and selling xx thousand notebook computers: C(x)=4000+500x (in thousand dollars)C(x) = 4000 + 500x \text{ (in thousand dollars)}

2. Revenue Function:

Assume the selling price per thousand units is pp thousand dollars. The revenue function is: R(x)=p×x (in thousand dollars)R(x) = p \times x \text{ (in thousand dollars)}

3. Profit Function:

The profit function is defined as the difference between revenue and cost: P(x)=R(x)C(x)P(x) = R(x) - C(x) P(x)=p×x(4000+500x)P(x) = p \times x - (4000 + 500x) P(x)=(p500)x4000 (in thousand dollars)P(x) = (p - 500)x - 4000 \text{ (in thousand dollars)}

Domain of the Profit Function:

Since xx represents the number of thousands of notebook computers produced and sold, xx must be a non-negative number. Therefore, the domain of P(x)P(x) is: Domain: x0\text{Domain: } x \geq 0

Part (B): Complete Table 13

To fill out Table 13, we need to compute the profit for the given values of xx. However, the selling price pp is not provided, so we can't compute the exact profit values. If we assume p=1000p = 1000 (as a common price), we can continue.

Using p=1000p = 1000: P(x)=(1000500)x4000P(x) = (1000 - 500)x - 4000 P(x)=500x4000P(x) = 500x - 4000

Now, calculate P(x)P(x) for each value of xx:

P(1)=500(1)4000=5004000=3500 thousand dollarsP(1) = 500(1) - 4000 = 500 - 4000 = -3500 \text{ thousand dollars} P(5)=500(5)4000=25004000=1500 thousand dollarsP(5) = 500(5) - 4000 = 2500 - 4000 = -1500 \text{ thousand dollars} P(10)=500(10)4000=50004000=1000 thousand dollarsP(10) = 500(10) - 4000 = 5000 - 4000 = 1000 \text{ thousand dollars} P(15)=500(15)4000=75004000=3500 thousand dollarsP(15) = 500(15) - 4000 = 7500 - 4000 = 3500 \text{ thousand dollars} P(20)=500(20)4000=100004000=6000 thousand dollarsP(20) = 500(20) - 4000 = 10000 - 4000 = 6000 \text{ thousand dollars} P(25)=500(25)4000=125004000=8500 thousand dollarsP(25) = 500(25) - 4000 = 12500 - 4000 = 8500 \text{ thousand dollars}

Part (C): Plot the Points and Sketch the Graph

Let's summarize the computed values in the table:

xx (thousands)P(x)P(x) (thousand $)
1-3500
5-1500
101000
153500
206000
258500

Finally, the graph of the profit function P(x)=500x4000P(x) = 500x - 4000 is a straight line, with the slope 500 and y-intercept -4000.

Would you like a detailed graph to visualize this, or any additional explanations? Here are 5 related questions to expand on this topic:

  1. What would the profit function look like if the cost per unit increased?
  2. How does changing the selling price affect the profit function?
  3. What is the break-even point for this profit function?
  4. How would the profit change if fixed costs decreased?
  5. Can the profit function be nonlinear, and under what circumstances might this occur?

Tip: Remember, the slope of the profit function corresponds to the difference between the selling price per unit and the variable cost per unit.

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Math Problem Analysis

Mathematical Concepts

Algebra
Profit and Loss
Linear Functions

Formulas

Profit Function: P(x) = Revenue - Cost
Cost Function: C(x) = 4000 + 500x
Revenue Function: R(x) = px
Profit Function: P(x) = (p - 500)x - 4000

Theorems

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Suitable Grade Level

Grades 10-12