Let's analyze the problem.

The profit function given is:

$P(x) = -10x^2 + 1620x - 49000$

To maximize the monthly rental profit, we need to find the value of $x$ that gives the maximum point of this quadratic function. Since the coefficient of $x^2$ is negative ($-10$), the parabola opens downward, and the maximum profit occurs at the vertex of the parabola.

Step 1: Finding the Vertex

The vertex of a quadratic function $ax^2 + bx + c$ occurs at:

$x = -\frac{b}{2a}$

Here, $a = -10$ and $b = 1620$. Substituting these values, we get:

$x = -\frac{1620}{2(-10)} = \frac{1620}{20} = 81$

Therefore, the maximum profit occurs when 81 units are rented out. This is consistent with the answer already provided.

Step 2: Finding the Maximum Profit

We now need to find the maximum profit by substituting $x = 81$ into the profit function $P(x)$:

$P(81) = -10(81)^2 + 1620(81) - 49000$

Calculating each term:

1. $-10 \cdot 81^2 = -10 \cdot 6561 = -65610$
2. $1620 \cdot 81 = 131220$

Thus:

$P(81) = -65610 + 131220 - 49000 = 16610$

The maximum monthly profit is $16,610, not $32,220 as mentioned in the image.

Summary:

• Units to rent out for maximum profit: 81
• Maximum monthly profit: $16,610

Would you like further details or explanations on any step?

Here are 5 related questions to explore this topic further:

1. How can we find the maximum point of a quadratic function if it does not open downward?
2. What happens to the profit function if the cost per unit changes?
3. How does the coefficient of $x^2$ affect the shape of the parabola?
4. Can the profit function have more than one maximum point?
5. How do real-world constraints (like a limit on the number of units) affect the calculation of maximum profit?

Tip: For quadratic functions, always look for the vertex to find maximum or minimum values.