Question:

A manufacturing company produces and sells a certain product. The revenue earned from selling $x$ units of the product is given by the function $R(x) = 5x^2 - 3x + 10$ dollars. The cost of producing $x$ units of the product is given by the function $C(x) = 2x^2 + 4x + 15$ dollars.

1. Find the profit function $P(x)$ in terms of $x$.
2. Find the production level that maximizes the company's profit.

Answer:

1. The profit function $P(x)$ is given by the difference between the revenue function and the cost function:

   $P(x) = R(x) - C(x)$

   Substituting the given functions, we have:

   $P(x) = (5x^2 - 3x + 10) - (2x^2 + 4x + 15)$

   Simplifying, we get:

   $P(x) = 3x^2 - 7x - 5$

2. To find the production level that maximizes the company's profit, we need to find the critical points of the profit function $P(x)$. The critical points occur where the derivative of the function is equal to zero or is undefined.

   First, we find $P'(x)$ by differentiating $P(x)$:

   $P'(x) = \frac{d}{dx}(3x^2 - 7x - 5)$

   $P'(x) = 6x - 7$

   To find the critical points, we set $P'(x) = 0$:

   $6x - 7 = 0$

   Solving for $x$, we get:

   $x = \frac{7}{6}$

   To verify if this value is a maximum, we use the second derivative test. We find $P''(x)$ by differentiating $P'(x)$:

   $P''(x) = \frac{d}{dx}(6x - 7)$

   $P''(x) = 6$

   Since $P''(x) = 6 > 0$, the critical point at $x = \frac{7}{6}$ is a local minimum. Therefore, the production level that maximizes the company's profit is $x = \frac{7}{6}$ units.

Thus, the production level that maximizes the company's profit is $\frac{7}{6}$ units.