The problem in the image states:

Given the cost function: $C(x) = 7250 - 25x + 1.2x^2$ and the demand function: $p(x) = 780 - 8x$

we are asked to find the production level $x$ that will maximize profit.

Solution:

1. Define Profit Function: Profit, $P(x)$, is defined as Revenue $R(x)$ minus Cost $C(x)$. $P(x) = R(x) - C(x)$

2. Determine Revenue: Revenue $R(x)$ is the product of the price $p(x)$ and the quantity $x$. $R(x) = p(x) \cdot x = (780 - 8x)x = 780x - 8x^2$

3. Construct the Profit Function: Substitute $R(x)$ and $C(x)$ into $P(x)$: $P(x) = (780x - 8x^2) - (7250 - 25x + 1.2x^2)$ Simplifying, we get: $P(x) = 780x - 8x^2 - 7250 + 25x - 1.2x^2$ $P(x) = 805x - 9.2x^2 - 7250$

4. Maximize Profit: To maximize profit, take the derivative of $P(x)$ with respect to $x$ and set it equal to zero. $P'(x) = 805 - 18.4x = 0$ Solving for $x$: $x = \frac{805}{18.4} \approx 43.75$

5. Conclusion: The production level $x \approx 43.75$ will maximize profit.

Would you like further details on any steps or have additional questions?

Related Questions

1. How can we verify that this $x$ value indeed maximizes profit?
2. What would be the maximum profit at this production level?
3. How does the marginal cost compare to the marginal revenue at this optimal point?
4. How sensitive is the profit to changes in the cost and demand functions?
5. What happens to the profit if the demand decreases (e.g., due to external factors)?

Tip:

Always check the second derivative to confirm that a critical point is a maximum, as it determines the concavity of the profit function.