Max makes and sells posters. The function p(x) = –10x2 + 200x – 250, graphed below, indicates how much profit he makes in a month if he sells the posters for 20 – x dollars each.

The graph of a downward facing parabola is shown in the first quadrant of a coordinate grid. The parabola connects the points left-parenthesis 2 comma 100 right-parenthesis, left-parenthesis 10 comma 750 right-parenthesis, and left-parenthesis 18 comma 100 right-parenthesis.

What should Max charge per poster to make the maximum profit, and what is the maximum profit he can make in a month?

(1 point)
Responses

$800 at $10 per poster
$800 at $10 per poster

$800 at $5 per poster
$800 at $5 per poster

$750 at $10 per poster
$750 at $10 per poster

$750 at $5 per poster

1 answer

To find the price per poster that maximizes Max's profit, we need to find the x-coordinate of the vertex of the parabola represented by the function p(x).

The x-coordinate of the vertex can be found using the formula x = -b/2a, where a, b, and c are the coefficients of the quadratic equation.

In this case, a = -10, b = 200, and c = -250. Substituting these values into the formula, we get:
x = -200 / (2*(-10)) = -200 / (-20) = 10

Therefore, Max should charge $10 per poster to maximize his profit.

To find the maximum profit, we need to substitute the x-coordinate of the vertex (10) into the function p(x):

p(10) = -10(10)^2 + 200(10) - 250
p(10) = -10(100) + 2000 - 250
p(10) = -1000 + 2000 - 250
p(10) = 750

Therefore, the maximum profit Max can make in a month is $750. Therefore, the correct answer is:

$750 at $10 per poster