A company's profit on phone sales is modeled by the function P(x)= -x^2+90x+497,975 where x is the price of a phone. To the nearest dollar, what price gives the maximum profit? Show all work.

The answer is $752 per phone, but I don't know how.

To find the price that gives the maximum profit, we first need to understand that the maximum profit occurs at the vertex of the quadratic function.

The profit function is represented by the equation P(x) = -x^2 + 90x + 497,975. This is a quadratic function with a negative coefficient, meaning the graph of the function will be a downward-opening parabola.

The x-coordinate of the vertex represents the price that gives the maximum profit. To find this x-coordinate, we can use the formula x = -b/2a, where a, b, and c are the coefficients in the quadratic function.

In this case, a = -1 (the coefficient of the x^2 term), b = 90 (the coefficient of the x term), and c = 497,975 (the constant term).

Substituting these values into the formula, we have x = -90 / (2 * -1) = -90 / -2 = 45.

So, the x-coordinate of the vertex is 45. However, since this represents the price of a phone, it cannot be negative. Therefore, we disregard the negative value and take the positive value, which is 45.

To find the maximum profit, we substitute this value back into the profit function: P(45) = -(45)^2 + 90(45) + 497,975.

Evaluating this expression gives us P(45) = -2025 + 4050 + 497,975 = 500,000.

Therefore, the maximum profit occurs when the price of a phone is $752 (rounded to the nearest dollar).