you charged $4 the demand was 590 log-ons per month. When you lowered the price to $3.50, the demand increased to 885 log-ons per month.

(a)

Construct a linear demand function for your website and hence obtain the monthly revenue *R* as a function of the log-on fee *x*.

*R*(*x*) =

(b)

Your Internet provider charges you a monthly fee of $40 to maintain your site. Express your monthly profit *P* as a function of the log-on fee *x*.

*P*(*x*) =

Determine the log-on fee you should charge to obtain the largest possible monthly profit (in dollars).

*x* = $

What is the largest possible monthly profit (in dollars)?

$

To solve this problem, let's tackle it step-by-step.

### (a) Constructing a Linear Demand Function and Monthly Revenue $ R $

We are given two price-demand points:
1. At $ x = 4 $, the demand is 590 log-ons per month.
2. At $ x = 3.50 $, the demand is 885 log-ons per month.

Assuming a linear demand function of the form:
\[
d(x) = mx + b
\]
where $ d(x) $ is the demand as a function of the log-on fee $ x $.

#### Step 1: Find the slope $ m $

Using the two points $(4, 590)$ and $(3.5, 885)$:
\[
m = \frac{885 - 590}{3.5 - 4} = \frac{295}{-0.5} = -590
\]

#### Step 2: Find the $ y $-intercept $ b $

Substitute one of the points, say $(4, 590)$, into the equation $ d(x) = -590x + b $:
\[
590 = -590(4) + b
\]
\[
590 = -2360 + b
\]
\[
b = 2950
\]

Thus, the demand function is:
\[
d(x) = -590x + 2950
\]

#### Step 3: Find the Revenue Function $ R(x) $

Revenue is the product of the price $ x $ and demand $ d(x) $:
\[
R(x) = x \cdot d(x) = x \cdot (-590x + 2950)
\]
\[
R(x) = -590x^2 + 2950x
\]

So, the monthly revenue function is:
\[
R(x) = -590x^2 + 2950x
\]

### (b) Constructing the Profit Function $ P(x) $

Your monthly costs are $40. Profit is revenue minus costs, so:
\[
P(x) = R(x) - 40
\]
\[
P(x) = (-590x^2 + 2950x) - 40
\]
\[
P(x) = -590x^2 + 2950x - 40
\]

Thus, the profit function is:
\[
P(x) = -590x^2 + 2950x - 40
\]

### (c) Maximizing Profit

To find the log-on fee $ x $ that maximizes profit, we need to find the vertex of the parabola represented by $ P(x) = -590x^2 + 2950x - 40 $. The maximum value of a quadratic function $ ax^2 + bx + c $ occurs at $ x = -\frac{b}{2a} $.

#### Step 1: Apply the vertex formula

Here, $ a = -590 $ and $ b = 2950 $:
\[
x = -\frac{2950}{2 \cdot -590} = \frac{2950}{1180} = 2.5
\]

So, the log-on fee $ x $ that maximizes profit is:
\[
x = 2.5
\]

#### Step 2: Calculate the Maximum Profit

Substitute $ x = 2.5 $ into $ P(x) $:
\[
P(2.5) = -590(2.5)^2 + 2950(2.5) - 40
\]
\[
P(2.5) = -590(6.25) + 2950(2.5) - 40
\]
\[
P(2.5) = -3687.5 + 7375 - 40
\]
\[
P(2.5) = 3647.5
\]

### Final Answers

1. The revenue function $ R(x) $ is:
   \[
   R(x) = -590x^2 + 2950x
   \]

2. The profit function $ P(x) $ is:
   \[
   P(x) = -590x^2 + 2950x - 40
   \]

3. The log-on fee that maximizes profit is:
   \[
   x = 2.5
   \]

4. The largest possible monthly profit is:
   \[
   P(2.5) = 3647.5
   \]

Would you like further details on any step, or help with another topic?

Here are some related questions:
1. How does changing the monthly fee affect profit maximization?
2. What if the demand function was not linear? How would that impact revenue and profit functions?
3. How would an increase in the Internet provider's fee change the optimal log-on fee?
4. Can you determine the price elasticity of demand based on the demand function?
5. How would a decrease in log-on price below $2.5 affect the business?

**Tip:** Always double-check calculations for profit maximization, as small errors can affect the optimal solution.

You charged $4; the demand was 590 log-ons per month. When you lowered the price to $3.50, the demand increased to 885 log-ons per month. 

(a) Construct a linear demand function for your website and hence obtain the monthly revenue R as a function of the log-on fee x.

R(x) =

(b) Your Internet provider charges you a monthly fee of $40 to maintain your site. Express your monthly profit P as a function of the log-on fee x.

P(x) =

Determine the log-on fee you should charge to obtain the largest possible monthly profit (in dollars).

x = $

What is the largest possible monthly profit (in dollars)?

$

Learn how to calculate a linear demand function, revenue function, and profit function for a website with log-on fees. Given price-demand points, determine the optimal log-on fee for maximizing profit, including setup for revenue and profit equations. This problem covers key algebra concepts suitable for Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation