Let's break this problem into manageable parts. I'll address each sub-question step by step:

Part a: Price-demand equation, $p(x)$

• Given:
  • $p = 324$ when $x = 90$ units rented.
  • For every additional unit rented ($x$), rent decreases by $1.

This forms a linear relationship: $p(x) = 324 - (x - 90)$

Simplify: $p(x) = 414 - x$

Part b: Monthly revenue function, $R(x)$

Revenue $R(x)$ is the product of the number of units rented ($x$) and the rent per unit ($p(x)$): $R(x) = x \cdot p(x)$

Substitute $p(x) = 414 - x$: $R(x) = x(414 - x)$

Simplify: $R(x) = 414x - x^2$

Part c: Maximize the monthly revenue (find $x$)

To maximize revenue, find the vertex of the parabola $R(x) = -x^2 + 414x$. The formula for the vertex (maximum point) of $ax^2 + bx + c$ is: $x = -\frac{b}{2a}$

Here:

• $a = -1$,
• $b = 414$.

Substitute: $x = -\frac{414}{2(-1)} = \frac{414}{2} = 207$

So, 207 units should be rented.

Part d: Maximum monthly revenue

Substitute $x = 207$ into $R(x)$: $R(207) = 414(207) - (207)^2$

Calculate: $R(207) = 85698 - 42849 = 42849$

The maximum revenue is $42,849.

Part e: Rent to maximize revenue

Substitute $x = 207$ into the price-demand equation $p(x) = 414 - x$: $p(207) = 414 - 207 = 207$

The rent should be $207 per unit.

If you'd like, I can provide further clarification or explore related concepts!

Related Questions

1. Can you explain why the revenue function forms a parabola?
2. How would the equations change if the rent decreased by $2 instead of $1 per unit rented?
3. What happens to the maximum revenue if the starting rent is higher than $324?
4. How does the number of units rented affect the profitability of the apartment complex?
5. Can you verify these results with a graph of $R(x)$ vs. $x$?

Tip:

Always double-check your work by substituting results back into the original equations for verification!