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
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
To find the maximum profit and the corresponding price per poster, we need to find the vertex of the parabola. The vertex of a parabola is given by the formula x = -b/2a, where a and b are the coefficients of the quadratic equation in standard form.
In this case, the quadratic equation is p(x) = -10x^2 + 200x - 250, so a = -10 and b = 200. Substituting these values into the formula, we get:
x = -200 / (2 * -10)
x = -200 / -20
x = 10
So, the price per poster for maximum profit is $10.
Now, to find the maximum profit, we substitute x = 10 into the equation p(x) = -10x^2 + 200x - 250:
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 that Max can make in a month is $750.
So, the answer is: $750 at $10 per poster.
To determine the price per poster Max should charge to make the maximum profit and the maximum profit he can make in a month, we need to find the vertex of the parabola described by the function p(x) = -10x^2 + 200x - 250.
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 and b = 200.
x = -200 / (2 * -10)
x = -200 / -20
x = 10
So, the x-coordinate of the vertex is 10.
To find the corresponding y-coordinate (profit), we substitute x = 10 into the equation p(x).
p(10) = -10(10)^2 + 200(10) - 250
p(10) = -10(100) + 200(10) - 250
p(10) = -1000 + 2000 - 250
p(10) = 750
Therefore, the maximum profit Max can make in a month is $750.
To calculate the price per poster, we need to find the value of x in the given range (20 - x) that corresponds to the x-coordinate of the vertex (10).
x = 20 - price per poster
10 = 20 - price per poster
price per poster = 20 - 10
price per poster = 10
So, Max should charge $10 per poster to make the maximum profit, and the maximum profit he can make in a month is $750.
Therefore, the correct answer is:
$750 at $10 per poster