 Basement chop bar estimates its daily cost function to be C(x) = xcube -6xsquare + 13x +15 and its revenue function to be R(X) = 28x. At what value ofx is profit maximized and what is the maximum daily profit.

To determine the value of \( x \) at which profit is maximized and to calculate the maximum daily profit, we need to follow these steps:

### 1. Define the Profit Function
The profit function \( P(x) \) is the difference between the revenue function \( R(x) \) and the cost function \( C(x) \).

\[
P(x) = R(x) - C(x)
\]

Given:
- \( C(x) = x^3 - 6x^2 + 13x + 15 \)
- \( R(x) = 28x \)

Thus, the profit function becomes:

\[
P(x) = 28x - (x^3 - 6x^2 + 13x + 15)
\]

Simplifying the profit function:

\[
P(x) = -x^3 + 6x^2 + 15x - 15
\]

### 2. Find the First Derivative of the Profit Function
To find the value of \( x \) that maximizes profit, we need to take the first derivative of the profit function and set it to zero:

\[
P'(x) = \frac{d}{dx}(-x^3 + 6x^2 + 15x - 15)
\]

\[
P'(x) = -3x^2 + 12x + 15
\]

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

\[
-3x^2 + 12x + 15 = 0
\]

Divide the equation by -3 to simplify:

\[
x^2 - 4x - 5 = 0
\]

This is a quadratic equation, and we can solve for \( x \) using the quadratic formula:

\[
x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-5)}}{2(1)}
\]

\[
x = \frac{4 \pm \sqrt{16 + 20}}{2}
\]

\[
x = \frac{4 \pm \sqrt{36}}{2}
\]

\[
x = \frac{4 \pm 6}{2}
\]

This gives us two solutions:

\[
x_1 = \frac{10}{2} = 5 \quad \text{and} \quad x_2 = \frac{-2}{2} = -1
\]

Since \( x \) represents the quantity sold, it must be non-negative, so we discard \( x = -1 \).

Thus, \( x = 5 \).

### 3. Find the Second Derivative to Confirm Maximum
To confirm that this critical point is a maximum, we take the second derivative of \( P(x) \):

\[
P''(x) = \frac{d}{dx}(-3x^2 + 12x + 15)
\]

\[
P''(x) = -6x + 12
\]

Substituting \( x = 5 \) into the second derivative:

\[
P''(5) = -6(5) + 12 = -30 + 12 = -18
\]

Since \( P''(5) < 0 \), the profit function has a maximum at \( x = 5 \).

### 4. Calculate the Maximum Daily Profit
Finally, substitute \( x = 5 \) into the profit function \( P(x) \) to find the maximum daily profit:

\[
P(5) = -(5)^3 + 6(5)^2 + 15(5) - 15
\]

\[
P(5) = -125 + 150 + 75 - 15 = 85
\]

### Answer:
The profit is maximized when \( x = 5 \), and the maximum daily profit is **Ghc 85**.

Would you like more details or have any questions?  

Here are 8 related questions to expand your understanding:

1. How does changing the revenue function to \( R(x) = 30x \) affect the profit maximization point?
2. What happens to the maximum profit if the cost function changes to \( C(x) = x^3 - 4x^2 + 12x + 10 \)?
3. What is the break-even point, where profit is zero?
4. How would increasing the cost function's constant term (e.g., \( +20 \) instead of \( +15 \)) impact the maximum profit?
5. Can there be more than one value of \( x \) that maximizes profit?
6. What is the significance of the second derivative in determining the maximum point?
7. How does the shape of the profit function graph help in understanding profit maximization?
8. How would a change in fixed costs (e.g., \( 15 \) in \( C(x) \)) impact the overall profit?

**Tip**: Always check the second derivative to confirm whether a critical point is a maximum or minimum.

Learn how to maximize daily profit using profit functions. Solve for the optimal quantity to maximize profit given revenue and cost functions. Suitable for advanced high school students and anyone interested in economics and calculus.

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