I will analyze the image content and solve the questions based on the provided data.

Let's begin with the first part:

1. Part a: Price-Demand Function

The price-demand function is given as:

$p(x) = 2000 - 60x$

where $p(x)$ is the price per computer in dollars when $x$ thousand computers are sold.

Revenue Function

The revenue function is given by:

$R(x) = p(x) \cdot x = (2000 - 60x) \cdot x$ Expanding the function:

$R(x) = 2000x - 60x^2$

This is the revenue function in thousands of dollars.

a) Graph the Revenue Function

To plot the revenue function, $R(x) = 2000x - 60x^2$, you would plot values of $R(x)$ for different $x$ (number of computers in thousands).

Let's first compute some key points for the graph:

• For $x = 0$: $R(0) = 2000(0) - 60(0)^2 = 0$
• For $x = 10$: $R(10) = 2000(10) - 60(10)^2 = 20000 - 6000 = 14000$
• For $x = 20$: $R(20) = 2000(20) - 60(20)^2 = 40000 - 24000 = 16000$
• For $x = 30$: $R(30) = 2000(30) - 60(30)^2 = 60000 - 54000 = 6000$

These values will help plot a parabolic curve on the graph, with the maximum revenue between $x = 10$ and $x = 30$. The graph will rise until it hits a maximum, then decrease.

2. Part b: Cost Function

The cost function is given by: $C(x) = 4000 + 50x$ This is a linear function, where the fixed cost is 4000 and the variable cost increases by 50 for each additional $x$ (in thousands of computers).

Let's calculate some values:

• For $x = 0$: $C(0) = 4000$
• For $x = 10$: $C(10) = 4000 + 50(10) = 4500$
• For $x = 20$: $C(20) = 4000 + 50(20) = 5000$
• For $x = 30$: $C(30) = 4000 + 50(30) = 5500$

These values can be used to plot the cost function.

c) Find the Break-Even Points

The break-even points occur when revenue equals cost, i.e., $R(x) = C(x)$.

Set the functions equal to each other: $2000x - 60x^2 = 4000 + 50x$ Rearranging the terms: $-60x^2 + 2000x - 50x - 4000 = 0$ $-60x^2 + 1950x - 4000 = 0$

We can solve this quadratic equation to find the break-even points. I will calculate the roots next.

Let me calculate the break-even points.