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The image describes a problem involving revenue maximization for a housing development. Here’s a breakdown of the problem:

### Problem Summary:
- If 40 houses are built, they will sell at \$400,000 each.
- If 80 houses are built, they will sell at \$320,000 each.
- We need to:
  1. Obtain a linear demand equation for the price, $ p(q) $, where $ p $ is the price of a house and $ q $ is the number of houses built.
  2. Determine how many houses should be built to achieve the largest revenue.
  3. Find the maximum possible revenue.

### Step 1: Finding the Linear Demand Equation

We are given two points that represent the relationship between the number of houses $ q $ and the price $ p $:
- When $ q = 40 $, $ p = 400,000 $.
- When $ q = 80 $, $ p = 320,000 $.

We can treat these as points on a line: $ (q_1, p_1) = (40, 400,000) $ and $ (q_2, p_2) = (80, 320,000) $.

The slope $ m $ of the linear equation is:
\[
m = \frac{p_2 - p_1}{q_2 - q_1} = \frac{320,000 - 400,000}{80 - 40} = \frac{-80,000}{40} = -2,000
\]

Now, using the point-slope form of a linear equation $ p - p_1 = m(q - q_1) $, we can plug in one of the points, say $ (40, 400,000) $, and the slope $ m = -2,000 $:
\[
p - 400,000 = -2,000(q - 40)
\]
Simplifying this:
\[
p = -2,000q + 80,000 + 400,000
\]
\[
p = -2,000q + 480,000
\]
Thus, the linear demand equation is:
\[
p(q) = -2,000q + 480,000
\]

### Step 2: Maximizing Revenue

Revenue $ R $ is the product of the price $ p(q) $ and the quantity $ q $, i.e.:
\[
R(q) = p(q) \times q = (-2,000q + 480,000) \times q
\]
Expanding this:
\[
R(q) = -2,000q^2 + 480,000q
\]
This is a quadratic equation, and since the coefficient of $ q^2 $ is negative, it represents a parabola that opens downward. The maximum revenue occurs at the vertex of this parabola.

The formula for the vertex of a quadratic equation $ ax^2 + bx + c $ is given by:
\[
q = -\frac{b}{2a}
\]
Here, $ a = -2,000 $ and $ b = 480,000 $, so:
\[
q = -\frac{480,000}{2 \times -2,000} = \frac{480,000}{4,000} = 120
\]

Thus, the company should build **120 houses** to maximize revenue.

### Step 3: Finding the Maximum Revenue

Substitute $ q = 120 $ into the revenue equation $ R(q) = -2,000q^2 + 480,000q $:
\[
R(120) = -2,000(120)^2 + 480,000(120)
\]
\[
R(120) = -2,000(14,400) + 480,000(120)
\]
\[
R(120) = -28,800,000 + 57,600,000
\]
\[
R(120) = 28,800,000
\]

Thus, the largest possible revenue is **\$28,800,000**.

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Would you like more details or have any questions on this solution? Here are some related questions:

1. How does changing the slope of the demand curve affect the maximum number of houses?
2. How can the profit equation be derived if the cost per house is also provided?
3. What are the economic implications of building more than 120 houses?
4. Can this problem be solved using a different method, like calculus optimization?
5. How does the quadratic nature of the revenue function determine the maximum?

**Tip:** When maximizing or minimizing quadratic functions, always check the sign of the quadratic term to determine if it opens upward or downward!

Pack-Em-In has another development in the works. If it builds 40 houses in this development, it will be able to sell them at $400,000 each. If it builds 80 houses, it will get only $320,000 each. Obtain a linear demand equation. Let p be the price of a house and q the number of houses. Determine how many houses it should build to get the largest revenue. What is the largest possible revenue (in dollars)?

Learn how to maximize revenue for a housing development by finding the linear demand equation and solving a quadratic equation. This problem covers key algebraic concepts, including linear equations, the quadratic formula, and revenue maximization strategies.

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