The cost of manufacturing 𝑥
badminton rackets per day is given by the function

𝐶(𝑥)=500+7𝑥+0.0002𝑥^2

Each racket can be sold at a price of 𝑝
dollars, where 𝑝=12−0.0003𝑥
.  Find the number of rackets to be manufactured which maximizes the daily profit for the company

To solve this problem, we need to maximize the **daily profit** for the company, which is the difference between the **revenue** and the **cost** of manufacturing badminton rackets.

### Step 1: Profit Function
The **revenue** \( R(x) \) is the price per racket multiplied by the number of rackets sold, i.e., \( R(x) = p \cdot x \). Given that \( p = 12 - 0.0003x \), the revenue function becomes:

\[
R(x) = (12 - 0.0003x) \cdot x = 12x - 0.0003x^2
\]

The **cost** function \( C(x) \) is given as:

\[
C(x) = 500 + 7x + 0.0002x^2
\]

The **profit function** \( P(x) \) is the difference between the revenue and the cost:

\[
P(x) = R(x) - C(x) = (12x - 0.0003x^2) - (500 + 7x + 0.0002x^2)
\]

Simplify the profit function:

\[
P(x) = 12x - 0.0003x^2 - 500 - 7x - 0.0002x^2
\]

\[
P(x) = (12x - 7x) - (0.0003x^2 + 0.0002x^2) - 500
\]

\[
P(x) = 5x - 0.0005x^2 - 500
\]

### Step 2: Maximizing the Profit
To find the number of rackets that maximizes profit, we need to take the derivative of \( P(x) \) with respect to \( x \), and set it equal to zero.

\[
P'(x) = \frac{d}{dx} \left( 5x - 0.0005x^2 - 500 \right)
\]

\[
P'(x) = 5 - 0.001x
\]

Set the derivative equal to zero to find the critical points:

\[
5 - 0.001x = 0
\]

Solve for \( x \):

\[
0.001x = 5
\]

\[
x = \frac{5}{0.001} = 5000
\]

### Step 3: Verify Maximum Using the Second Derivative
To confirm that this critical point is a maximum, we compute the second derivative of \( P(x) \):

\[
P''(x) = \frac{d}{dx} \left( 5 - 0.001x \right) = -0.001
\]

Since \( P''(x) = -0.001 \) is negative, the function is concave down, and thus, \( x = 5000 \) corresponds to a maximum.

### Conclusion
The number of rackets that should be manufactured to maximize the daily profit is **5000 rackets**.

Would you like further details on any step or have any questions?

Here are 5 related questions:
1. How do you calculate revenue from price and quantity?
2. What is the role of the second derivative in determining maxima or minima?
3. How do you interpret the profit function in terms of revenue and cost?
4. What are the general conditions to maximize a function?
5. What are some real-world factors that could affect the profit function beyond this model?

**Tip**: Always verify whether the critical point is a maximum or minimum by using the second derivative or analyzing the function's behavior around the critical point.

The cost of manufacturing 𝑥 badminton rackets per day is given by the function C(𝑥)=500+7𝑥+0.0002𝑥^2. Each racket can be sold at a price of p dollars, where p=12−0.0003𝑥. Find the number of rackets to be manufactured which maximizes the daily profit for the company.

Explore how to maximize the daily profit of manufacturing badminton rackets by using cost and revenue functions. This step-by-step guide walks through the use of derivatives and quadratic equations to solve an optimization problem. Suitable for Grades 11-12.

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