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Let's solve the given problem step by step:

### Part (a): Construct a linear demand function
We are given two data points for the price-demand relationship:
- When the log-on fee $ x = 4 $, the demand is $ 570 $ log-ons per month.
- When the log-on fee $ x = 3.50 $, the demand is $ 855 $ log-ons per month.

To construct the linear demand function, we can use the form of a linear equation:
\[
D(x) = m x + b
\]
where $ D(x) $ represents the demand at a log-on fee $ x $, and $ m $ and $ b $ are constants.

#### Step 1: Find the slope ($ m $)
The slope $ m $ is given by:
\[
m = \frac{\Delta D}{\Delta x} = \frac{855 - 570}{3.50 - 4} = \frac{285}{-0.5} = -570
\]

#### Step 2: Find the intercept ($ b $)
We can use one of the points to find $ b $. Let's use $ (x, D) = (4, 570) $:
\[
570 = -570(4) + b
\]
\[
570 = -2280 + b
\]
\[
b = 2850
\]

Thus, the demand function is:
\[
D(x) = -570x + 2850
\]

The revenue function $ R(x) $ is given by:
\[
R(x) = x \cdot D(x) = x(-570x + 2850) = -570x^2 + 2850x
\]

### Part (b): Monthly Profit Function
The monthly profit is the revenue minus the costs. We are given a fixed monthly cost of $50 to maintain the site.

Thus, the profit function $ P(x) $ is:
\[
P(x) = R(x) - 50 = -570x^2 + 2850x - 50
\]

### Step 3: Find the maximum profit
To maximize the profit, we need to find the critical points of $ P(x) $ by setting its derivative equal to zero:
\[
P'(x) = -1140x + 2850
\]
Set $ P'(x) = 0 $:
\[
-1140x + 2850 = 0
\]
\[
1140x = 2850
\]
\[
x = \frac{2850}{1140} = 2.5
\]

So, the log-on fee that maximizes profit is $2.50.

### Step 4: Calculate the maximum profit
Substitute $ x = 2.5 $ into the profit function:
\[
P(2.5) = -570(2.5)^2 + 2850(2.5) - 50
\]
\[
P(2.5) = -570(6.25) + 7125 - 50
\]
\[
P(2.5) = -3562.5 + 7125 - 50
\]
\[
P(2.5) = 3512.5
\]

The largest possible monthly profit is $3512.50.

---

Would you like more details or have any questions?

### 5 Related Questions:
1. How does changing the fixed monthly cost affect the maximum profit?
2. Can the demand function still be linear if additional data points do not fit the current trend?
3. What happens to the profit if the log-on fee is increased beyond $4?
4. How would the profit function change if there were additional variable costs per log-on?
5. How does calculus help in optimizing business models like this?

### Tip:
When constructing linear functions, always use at least two data points to find the slope and y-intercept accurately.

You operate a gaming website, www.mudbeast.net, where users must pay a small fee to log on. When you charged $4, the demand was 570 log-ons per month. When you lowered the price to $3.50, the demand increased to 855 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.

(b) Your Internet provider charges you a monthly fee of $50 to maintain your site. Express your monthly profit P as a function of the log-on fee x. Determine the log-on fee you should charge to obtain the largest possible monthly profit (in dollars). What is the largest possible monthly profit (in dollars)?

This problem involves constructing a linear demand function and maximizing the monthly profit for a gaming website by setting an optimal log-on fee. Learn how to calculate demand, revenue, and profit functions, and use derivatives to find the maximum profit.

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