The problem in the image asks to find the level of production that yields the maximum profit given the profit function:

$P(x) = 120x - x^2 - 1100$

Here, $x$ is the level of production, and $P(x)$ is the profit function. To determine the production level that yields the maximum profit, we need to:

1. Find the critical points by taking the derivative of $P(x)$ and setting it equal to zero.
2. Determine the nature of the critical points by using either the second derivative test or analyzing the concavity.
3. Calculate the maximum profit by substituting the production level into the profit function.

Let's start with step 1 by finding the derivative of $P(x)$.

$P'(x) = 120 - 2x$

Now, let's set this equal to zero to find the critical point.

$120 - 2x = 0$ $2x = 120$ $x = 60$

So, the production level that could potentially maximize profit is $x = 60$.

Step 2: To confirm that this is a maximum, we take the second derivative of $P(x)$:

$P''(x) = -2$

Since $P''(x)$ is negative, the profit function is concave down, confirming that $x = 60$ gives a maximum profit.

Step 3: Now, let's calculate the maximum profit by substituting $x = 60$ into the original profit function:

$P(60) = 120(60) - (60)^2 - 1100$ $P(60) = 7200 - 3600 - 1100$ $P(60) = 2500$

Thus, the maximum profit is $2,500 at a production level of 60 units.

The correct answer is:

• Production levels of 60 yield a maximum profit of $2,500.

Would you like further clarification on any part of the solution?

Here are 5 follow-up questions based on this problem:

1. How do we know that the second derivative being negative confirms a maximum?
2. What would happen if the second derivative were positive?
3. Can you apply the same process to find a minimum instead of a maximum?
4. How would changes in the coefficient of $x^2$ affect the shape of the profit curve?
5. Why do we set the first derivative equal to zero to find critical points?

Tip: When solving optimization problems, always check the second derivative to confirm if the critical point is a maximum or minimum.