The cost of production of an item is given by ﻿C left parenthesis x right parenthesis space equals space 16 comma 000 plus 500 x﻿
The price-demand function of the item is ﻿1700 minus 7 x﻿
a) Find the revenue function R(x). ( 2 marks )
b) Find the profit function P(x).   ( 2 marks )
c) Find P(80), P(90) and P(100).  ( 3 marks )
d) Comparing the three values of P(80), P(90) and P(100) , shows that the profit increases until 90 units and for more than 90 units of production, the profit decreases. Is this statement true or false.  ( 1 mark )
Type your answers clearly in the space provided.

Let's break this down step by step:

Given dataLet's break this down step by step:

Given 
- Cost of production function: \( C(x) = 16,000 + 500x \)
- Price-demand function: \( p(x) = 1700 - 7x \)

### a) Revenue function \( R(x) \)

The revenue function is given by the product of the price-demand function and the number of units produced (i.e., \( x \)):

\[
R(x) = p(x) \cdot x = (1700 - 7x) \cdot x = 1700x - 7x^2
\]

Thus, the revenue function is:

\[
R(x) = 1700x - 7x^2
\]

### b) Profit function \( P(x) \)

The profit function is the difference between revenue and cost:

\[
P(x) = R(x) - C(x)
\]

Substituting the expressions for \( R(x) \) and \( C(x) \):

\[
P(x) = (1700x - 7x^2) - (16,000 + 500x)
\]

Simplifying this:

\[
P(x) = 1700x - 7x^2 - 16,000 - 500x = -7x^2 + 1200x - 16,000
\]

Thus, the profit function is:

\[
P(x) = -7x^2 + 1200x - 16,000
\]

### c) Find \( P(80) \), \( P(90) \), and \( P(100) \)

We substitute \( x = 80 \), \( x = 90 \), and \( x = 100 \) into the profit function \( P(x) \):

1. \( P(80) = -7(80)^2 + 1200(80) - 16,000 \)
2. \( P(90) = -7(90)^2 + 1200(90) - 16,000 \)
3. \( P(100) = -7(100)^2 + 1200(100) - 16,000 \)

Let's calculate these values:

1. \( P(80) = -7(6400) + 1200(80) - 16,000 = -44,800 + 96,000 - 16,000 = 35,200 \)
2. \( P(90) = -7(8100) + 1200(90) - 16,000 = -56,700 + 108,000 - 16,000 = 35,300 \)
3. \( P(100) = -7(10,000) + 1200(100) - 16,000 = -70,000 + 120,000 - 16,000 = 34,000 \)

Thus:
- \( P(80) = 35,200 \)
- \( P(90) = 35,300 \)
- \( P(100) = 34,000 \)

### d) Profit comparison: True or False?

Comparing the values:
- \( P(80) = 35,200 \)
- \( P(90) = 35,300 \)
- \( P(100) = 34,000 \)

The profit increases from \( P(80) \) to \( P(90) \), but it decreases from \( P(90) \) to \( P(100) \). Therefore, the statement that "the profit increases until 90 units and for more than 90 units of production, the profit decreases" is **true**.

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Do you want further details or explanations for any step?

Here are 5 related questions for you to consider:
1. How would the profit function change if the cost per unit increases?
2. What is the maximum profit, and how do you find the optimal production level?
3. How does the price-demand function affect revenue at different levels of production?
4. What is the significance of the coefficient of \( x^2 \) in the profit function?
5. What other real-world factors could affect the accuracy of this model?

**Tip:** To maximize profit, you can find the critical points by setting the derivative of the profit function to zero.

The cost of production of an item is given by C(x) = 16,000 + 500x. The price-demand function of the item is 1700 - 7x. a) Find the revenue function R(x). b) Find the profit function P(x). c) Find P(80), P(90), and P(100). d) Comparing the three values of P(80), P(90), and P(100), shows that the profit increases until 90 units and for more than 90 units of production, the profit decreases. Is this statement true or false?

Solve for the revenue and profit functions using quadratic equations. Learn to find R(x), P(x), and compare P(80), P(90), and P(100) to understand profit trends in production. This problem is ideal for students in Grades 10-12 and covers concepts like revenue, profit, and quadratic equations.

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