Max makes and sells posters. the function p(x)= -10x^2 +200x -250, graphed below, indicates how much profit he makes in a month if he sells the posters for 20-x dollars each. (picture in comments) what should max charge per poster to make the maximum profit, and what is the maximum profit he can make in a month? a: $800 at $10 per poster b: $800 at $5 per poster c: $750 at $10 per poster d: $750 at $5 per poster

Here is our profit as a function of # of posters
p(x) =-10x² + 200x - 250
Here is our price per poster, as a function of the # of posters:
pr(x) = 20 - x
Since we want to find the optimum price and # of posters, let's plug our price function into our profit function, to find the optimum x, and then use that to find the optimum price:
p(x) = -10 (20-x)² + 200 (20 - x) - 250
p(x) = -10 (400 -40x + x²) + 4000 - 200x - 250
Take a look at our profit function. It is a normal trinomial square, with a negative sign on the squared term. This means the curve is a downward facing parabola, so our profit maximum will be the top of the curve.
By taking the derivative, we can find where p'(x) = 0 (where the slope of p(x) equals 0), to see where the top of profit function is.
p(x) = -4000 +400x -10x² + 4000 -200x -250
p'(x) = 400 - 20x -200
0 = 200 - 20x
20x = 200
x = 10                         
p'(x) = 0 at x=10. This is the peak of our profit function. To find the price per poster, plug x=10 into our price function:
price = 20 - x
price = 10
Now plug x=10 into our original profit function in order to find our maximum profit:
p(x)= -10x^2 +200x -250
p(x) = -10 (10)² +200 (10) - 250
p(x) = -1000 + 2000 - 250
p(x) = 750

Correct answer is C)