Questions: Question number 5. The cost of manufacturing x fishing rods is given by C(x)=700+3x+(1/10000)x^2 Suppose the price of each rod is a function of the number manufactured and is given by p=10-0.0004x If all fishing rods manufactured can be sold, what is the number x that yields maximum profit? 7,000 8,750 11,666 10,000 17,500 None of the above

Solution Steps

To find the number \( x \) that yields maximum profit, we need to first define the profit function \( P(x) \). The profit is given by the revenue minus the cost. The revenue \( R(x) \) is the price per rod \( p \) times the number of rods \( x \). We then take the derivative of the profit function and set it to zero to find the critical points. Finally, we evaluate these points to determine which one gives the maximum profit.

1. Define the cost function \( C(x) \) and the price function \( p(x) \).
2. Define the revenue function \( R(x) = p(x) \cdot x \).
3. Define the profit function \( P(x) = R(x) - C(x) \).
4. Find the derivative of \( P(x) \) and set it to zero to find critical points.
5. Evaluate the critical points to determine the maximum profit.

The cost function for manufacturing \( x \) fishing rods is given by: \[ C(x) = 700 + 3x + \frac{1}{10000}x^2 \] The price per rod as a function of the number manufactured is: \[ p(x) = 10 - 0.0004x \]

The revenue function \( R(x) \) is defined as: \[ R(x) = p(x) \cdot x = x \left(10 - 0.0004x\right) = 10x - 0.0004x^2 \] The profit function \( P(x) \) is then given by: \[ P(x) = R(x) - C(x) = \left(10x - 0.0004x^2\right) - \left(700 + 3x + \frac{1}{10000}x^2\right) \] Simplifying this, we have: \[ P(x) = -0.0001x^2 + 7x - 700 \]

To find the maximum profit, we take the derivative of the profit function: \[ P'(x) = -0.0002x + 7 \] Setting the derivative equal to zero to find critical points: \[ -0.0002x + 7 = 0 \implies x = 7000 \]

Substituting \( x = 7000 \) back into the profit function: \[ P(7000) = -0.0001(7000)^2 + 7(7000) - 700 = 23800 \]

Final Answer