A manufacturer of tennis rackets finds that the total cost of manufacturing x rackets/day is given by
0.0001x2 + 4x + 500
dollars. Each racket can be sold at a price of p dollars, where
p = −0.0004x + 9.
Find an expression giving the daily profit P for the manufacturer, assuming that all the rackets manufactured can be sold. Hint: The total revenue is given by the total number of rackets sold multiplied by the price of each racket. The profit is given by revenue minus cost. (Simplify your answer completely.)

To find an expression for the daily profit \( P \), we need to follow these steps:

1. **Revenue Function:**
   The revenue \( R \) is given by the number of rackets sold, \( x \), multiplied by the price per racket, \( p \).

   Since the price function is \( p = -0.0004x + 9 \), the total revenue is:
   \[
   R(x) = x \cdot p = x(-0.0004x + 9)
   \]
   Expanding this expression, we get:
   \[
   R(x) = -0.0004x^2 + 9x
   \]

2. **Cost Function:**
   The total cost \( C(x) \) is given as:
   \[
   C(x) = 0.0001x^2 + 4x + 500
   \]

3. **Profit Function:**
   The profit \( P(x) \) is the difference between the revenue and the cost:
   \[
   P(x) = R(x) - C(x)
   \]
   Substituting the expressions for \( R(x) \) and \( C(x) \):
   \[
   P(x) = \left( -0.0004x^2 + 9x \right) - \left( 0.0001x^2 + 4x + 500 \right)
   \]
   Now, simplify the expression:
   \[
   P(x) = -0.0004x^2 + 9x - 0.0001x^2 - 4x - 500
   \]
   Combine like terms:
   \[
   P(x) = (-0.0004x^2 - 0.0001x^2) + (9x - 4x) - 500
   \]
   \[
   P(x) = -0.0005x^2 + 5x - 500
   \]

Thus, the simplified expression for the daily profit \( P(x) \) is:
\[
P(x) = -0.0005x^2 + 5x - 500
\]

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Would you like further details or have any questions about this process?

Here are 5 related questions to extend your understanding:
1. How does the profit change when manufacturing 100 rackets per day?
2. What number of rackets maximizes the profit, and how do you find it?
3. What is the break-even point where the profit is zero?
4. How does an increase in the cost per racket affect the profit function?
5. How would a change in price per racket impact the revenue?

**Tip:** To maximize profit, find the critical points of the profit function by taking its derivative and setting it equal to zero.

A manufacturer of tennis rackets finds that the total cost of manufacturing x rackets/day is given by 0.0001x^2 + 4x + 500 dollars. Each racket can be sold at a price of p dollars, where p = −0.0004x + 9. Find an expression giving the daily profit P for the manufacturer, assuming that all the rackets manufactured can be sold.

Explore how to optimize profit in tennis racket manufacturing by using quadratic functions. Learn how to calculate daily profit using cost and revenue equations, with step-by-step explanations. This problem covers key algebraic concepts like profit functions and quadratic optimization, ideal for students in grades 10-12.

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