A profit function is derived from the production cost and revenue function for a given item. The monthly profit function for a certain item is given by P(x)=−0.05x2+500x−100,000, where P is in dollars and x is the number of units sold.

19a. How many units must be sold on a monthly basis to maximize the profit?

19b. Find the maximum profit:

recall how to find the vertex of a parabola. In this case, that will be at

x = -b/2a = 500/0.1

Now use that to find P(x)

To find the number of units that must be sold on a monthly basis to maximize the profit, we need to determine the value of x that corresponds to the maximum point of the profit function P(x).

19a. To find the number of units that maximize the profit, we need to find the vertex of the parabola formed by the profit function P(x). The x-coordinate of the vertex, denoted as x_max, can be found using the formula x_max = -b/2a, where a, b, and c are the coefficients of the quadratic equation Ax^2 + Bx + C = 0. In this case, the equation is P(x) = -0.05x^2 + 500x - 100,000, so a = -0.05 and b = 500.

Plugging these values into the formula, we can find x_max as follows:
x_max = -500 / (2 * (-0.05))
x_max = -500 / (-0.1)
x_max = 5000

Therefore, 5000 units must be sold on a monthly basis to maximize the profit.

19b. To find the maximum profit, we substitute the value of x_max into the profit function P(x) and evaluate it:
P(x_max) = -0.05 * (5000)^2 + 500 * 5000 - 100,000
P(x_max) = -0.05 * 25,000,000 + 2,500,000 - 100,000
P(x_max) = -1,250,000 + 2,500,000 - 100,000
P(x_max) = 1,150,000

Therefore, the maximum profit is $1,150,000.