Question

A company has found that the relationship between the price p and the demand x for a particular product is given approximately by p = 1155 - 0.16x^2. The company also knows that the cost of producing the product is given by C(x) = 810 + 380x. Find P(x), the profit function. Now use the profit function to do the following: (A) Find the average of the x values of all local maxima of P.(B) Find the average of the x values of all local minima of P. (C) Use interval notation to indicate where P(x) is concave up. (D) Use interval notation to indicate where P(x) is concave down.

Answer 1

To find the profit function, subtract the cost function from the revenue function. The average of the x values of all local maxima of P is not possible to find since the revenue function is a quadratic. The profit function is concave down for all values of x.

Step-by-step explanation:

To find the profit function, we need to subtract the cost function from the revenue function. The revenue function is given by p(x), and the cost function is given by C(x). Let's substitute the given functions into the profit function formula:

P(x) = p(x) - C(x)

Since we are given p(x) = 1155 - 0.16x^2 and C(x) = 810 + 380x, we can substitute these values into the profit function formula:

P(x) = 1155 - 0.16x^2 - (810 + 380x)

Simplifying the expression:

P(x) = 345 - 0.16x^2 - 380x

The profit function is P(x) = 345 - 0.16x^2 - 380x.

(A) To find the average of the x values of all local maxima of P, we need to find the critical points of P(x) and determine which ones are local maxima. To find the critical points, we take the derivative of the profit function and set it equal to zero:

P'(x) = -0.32x - 380

Setting P'(x) = 0 and solving for x:

-0.32x - 380 = 0

-0.32x = 380

x = -380 / -0.32 = 1187.5

However, since the revenue function is a quadratic function, it does not have any local maxima or local minima.

(C) To determine where P(x) is concave up, we need to find the intervals where the second derivative of P(x) is positive. The second derivative of P(x) is:

P''(x) = -0.32

Since the second derivative is a constant value of -0.32, P(x) is concave down for all values of x.

(D) Since the second derivative of P(x) is negative for all values of x, P(x) is concave down for all values of x. Therefore, the interval notation for P(x) being concave down is (-∞, ∞).