The daily profit p in dollars of a company making tables is described by the function upper p left parenthesis x right parenthesis equals negative 6 x squared plus 312 x minus 3762p(x)=−6x2+312x−3762, where x is the number of tables that are manufactured in 1 day. the maximum profit of the company occurs at the vertex of the parabola. how many tables should be made per day in order to obtain the maximum profit for the company? what is the maximum profit?

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.

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