max makes and sells posters. the function p(x) = -10x^2 + 200x - 250, graphed below how much profit he makes in a month if he sells the poster for 20 - x dollars. What should max charge per poster to make the maximum profit, and what is the maximum profit he can make in a month?
The profit function P(x) is calculated by subtracting the cost function from the revenue function:
P(x) = R(x) - C(x)
Given that the selling price per poster is (20 - x) dollars, the revenue function R(x) is:
R(x) = x * (20 - x) = 20x - x^2
The cost function is given as -10x^2 + 200x - 250. So the profit function P(x) becomes:
P(x) = (20x - x^2) - (-10x^2 + 200x - 250)
= 20x - x^2 + 10x^2 - 200x + 250
= 9x^2 - 180x + 250
To find the price per poster that maximizes the profit, we need to find the x-value of the vertex of the parabola represented by the profit function. The x-coordinate of the vertex can be found using the formula:
x = -b / (2a)
In this case, a = 9 and b = -180. Substituting these values, we get:
x = -(-180) / (2 * 9)
= 180 / 18
= 10
Therefore, Max should charge $10 per poster to make the maximum profit.
To find the maximum profit, we substitute this value of x into the profit function:
P(10) = 9(10)^2 - 180(10) + 250
= 9(100) - 1800 + 250
= 900 - 1800 + 250
= 350
Therefore, the maximum profit Max can make in a month is $350.