x=26 We have the equation p(x) = -6x^2 + 312x - 3762 for the profit and we want to know at what value of x, we can maximize the profit. Since all quadratic equations make parabolas and since parabola's are symmetrical, let's first use the quadratic formula to get the roots of the equation and then select a point midway between them. The values for the quadratic formula will be A=-6, B=312, and C=-3762, giving the roots of 19 and 33. Let's calculate the average. (19 + 33)/2 = 52/2 = 26 So the maximum profit will happen at x = 26. Let's verify that x = 26 is the best value by replacing x with (26+e) where e is some small error. So let's plug in that value for x in the original equation and see what we get: -6x^2 + 312x - 3762 = -6(26+e)^2 + 312(26+e) - 3762 = -6(676 + 52e + e^2) + 312(26+e) - 3762 = -4056 - 312e - 6e^2 + 8112 + 312e - 3762 = -4056 - 6e^2 + 8112 - 3762 = 294 - 6e^2 And if you look, you'll see that the only e term is -6e^2. And if e is any non-zero value, the value of the term will be negative. And the total value of the equation will be lower. So we've verified that x=26 is the best value to maximize the profit for the company.