To solve this problem, we need to follow a structured approach to determine the number of units that maximize profits for the monopoly, and then find the maximum profit.

1. Determine the Revenue Function:

The demand function is given as $p = 5200 - \frac{1}{2}x^2$, where $p$ is the price per unit and $x$ is the quantity of units sold. The revenue $R$ is the product of price $p$ and the number of units sold $x$:

$R(x) = p \cdot x = \left(5200 - \frac{1}{2}x^2\right) \cdot x$

Simplifying the revenue function:

$R(x) = 5200x - \frac{1}{2}x^3$

2. Determine the Cost Function:

The average cost function is given by $C = 3010 + 2x$. The total cost $TC(x)$ is the product of the average cost and the number of units $x$:

$TC(x) = C \cdot x = (3010 + 2x) \cdot x = 3010x + 2x^2$

3. Determine the Profit Function:

The profit $P(x)$ is the revenue minus the total cost:

$P(x) = R(x) - TC(x)$

Substituting the expressions for $R(x)$ and $TC(x)$:

$P(x) = \left(5200x - \frac{1}{2}x^3\right) - \left(3010x + 2x^2\right)$

Simplifying:

$P(x) = 5200x - \frac{1}{2}x^3 - 3010x - 2x^2$

$P(x) = (5200 - 3010)x - \frac{1}{2}x^3 - 2x^2$

$P(x) = 2190x - \frac{1}{2}x^3 - 2x^2$

4. Maximize the Profit:

To maximize profit, we take the derivative of the profit function $P(x)$ with respect to $x$ and set it equal to zero.

First, find the derivative:

$P'(x) = 2190 - \frac{3}{2}x^2 - 4x$

Set $P'(x) = 0$ to find the critical points:

$2190 - \frac{3}{2}x^2 - 4x = 0$

Multiply through by 2 to eliminate the fraction:

$4380 - 3x^2 - 8x = 0$

Rearrange this quadratic equation:

$3x^2 + 8x - 4380 = 0$

Solve using the quadratic formula:

$x = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 3 \cdot (-4380)}}{2 \cdot 3}$

$x = \frac{-8 \pm \sqrt{64 + 52560}}{6}$

$x = \frac{-8 \pm \sqrt{52624}}{6}$

$x = \frac{-8 \pm 229.34}{6}$

Taking the positive root (since quantity $x$ cannot be negative):

$x = \frac{-8 + 229.34}{6} \approx 37.22$

Thus, $x \approx 37$ units.

5. Find the Maximum Profit:

Now, substitute $x = 37$ into the profit function to find the maximum profit:

$P(37) = 2190(37) - \frac{1}{2}(37)^3 - 2(37)^2$

$P(37) = 81030 - \frac{1}{2}(50653) - 2(1369)$

$P(37) = 81030 - 25326.5 - 2738$

$P(37) \approx 52965.5$

Final Answer:

• The number of units that maximizes profits is approximately 37 units.
• The maximum profit is approximately $52,965.50.

Would you like to see further details, or ask any additional questions? Here are 5 related questions to explore further:

1. How would changes in the demand function impact the profit-maximizing quantity?
2. What happens if production limits are increased beyond 100 units?
3. How do you solve a maximization problem when a constraint like production capacity is present?
4. Can we use numerical methods to solve this instead of the quadratic formula?
5. What is the role of the second derivative test in confirming whether the profit is maximized?

Tip: Always check for potential constraints (like production limits) when solving for optimal quantities, as they can affect the final solution.