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: <span>p(x)= -10x^2 +200x -250 p(x) = -10 (10)</span>² +200 (10) - 250 <span>p(x) = -1000 + 2000 - 250 p(x) = 750
Since the function is being subtracted by 1, this means it will shift 1 unit downwards. When the number being multiplied to the function is less than one, it widens, which in this case the number 0.4 is less than 1.
