Website Profit   You operate a gaming website, www.mudbeast.net, where users must pay a small fee to log on. When you charged $3 the demand was 510 log-ons per month. When you lowered the price to $2.50, the demand increased to 765 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)?
$

### Part (a): Constructing the Linear Demand Function and Revenue Function

You are given two points on the demand function, where $ x $ is the price per log-on and $ d(x) $ is the number of log-ons:

- When the price was $3, the demand was 510 log-ons per month, i.e., $ (x_1, d_1) = (3, 510) $.
- When the price was $2.50, the demand was 765 log-ons per month, i.e., $ (x_2, d_2) = (2.50, 765) $.

Since the demand function is linear, it takes the form $ d(x) = mx + b $, where $ m $ is the slope and $ b $ is the intercept.

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

The slope $ m $ of the linear demand function is given by:
\[
m = \frac{d_2 - d_1}{x_2 - x_1} = \frac{765 - 510}{2.50 - 3} = \frac{255}{-0.5} = -510
\]

#### Step 2: Write the equation of the demand function

Using the point-slope form of the line, $ d(x) = m x + b $, we now substitute $ m = -510 $ and one of the points (for example, $ (3, 510) $) to solve for $ b $.

\[
510 = -510(3) + b
\]
\[
510 = -1530 + b
\]
\[
b = 2040
\]

Thus, the demand function is:
\[
d(x) = -510x + 2040
\]

#### Step 3: Construct the revenue function $ R(x) $

Revenue $ R(x) $ is the product of the price per log-on $ x $ and the demand $ d(x) $, i.e.,
\[
R(x) = x \cdot d(x)
\]
Substitute $ d(x) = -510x + 2040 $ into the revenue formula:
\[
R(x) = x(-510x + 2040)
\]
\[
R(x) = -510x^2 + 2040x
\]

### Part (b): Constructing the Profit Function

Your monthly profit $ P(x) $ is the difference between the revenue and the cost. The only cost is the $40 monthly fee charged by your Internet provider. Thus, the profit function is:

\[
P(x) = R(x) - 40
\]
Substitute $ R(x) = -510x^2 + 2040x $ into the profit formula:
\[
P(x) = (-510x^2 + 2040x) - 40
\]
\[
P(x) = -510x^2 + 2040x - 40
\]

### Step 3: Maximizing the Profit

To find the price $ x $ that maximizes the profit, we take the derivative of $ P(x) $ and set it equal to zero.

\[
P'(x) = \frac{d}{dx}(-510x^2 + 2040x - 40) = -1020x + 2040
\]
Set $ P'(x) = 0 $ to find the critical point:
\[
-1020x + 2040 = 0
\]
\[
1020x = 2040
\]
\[
x = \frac{2040}{1020} = 2
\]

Thus, the price that maximizes the profit is **$2**.

### Step 4: Calculate the Maximum Profit

Substitute $ x = 2 $ into the profit function $ P(x) = -510x^2 + 2040x - 40 $ to find the maximum profit:

\[
P(2) = -510(2)^2 + 2040(2) - 40
\]
\[
P(2) = -510(4) + 4080 - 40
\]
\[
P(2) = -2040 + 4080 - 40 = 2000
\]

Thus, the largest possible monthly profit is **$2000**.

### Summary of Answers:

(a) The monthly revenue as a function of the log-on fee is:
\[
R(x) = -510x^2 + 2040x
\]

(b) The monthly profit as a function of the log-on fee is:
\[
P(x) = -510x^2 + 2040x - 40
\]

- The log-on fee to maximize profit is $ x = 2 $ dollars.
- The largest possible monthly profit is **$2000**.

---

Would you like more details on any step or further clarifications?

### Related Questions:
1. What happens to the profit if the log-on fee is set to $2.50 instead of $2?
2. How would the demand function change if new data indicated different log-on demand at higher or lower prices?
3. If the fixed monthly cost increases, how will the profit-maximizing price change?
4. What is the break-even price, i.e., the price at which profit is zero?
5. How would a shift in customer behavior (e.g., increased demand elasticity) affect the demand function?

### Tip:
When optimizing revenue or profit functions, always look for ways to derive the maximum by taking derivatives and solving for critical points.

You operate a gaming website, www.mudbeast.net, where users must pay a small fee to log on. When you charged $3 the demand was 510 log-ons per month. When you lowered the price to $2.50, the demand increased to 765 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 construct a linear demand function and optimize profit for www.mudbeast.net. This problem includes the calculation of revenue and profit functions, and uses derivatives to find the log-on fee that maximizes profit. Suitable for Grades 10-12.

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