Maximum profit. A chain store manager has been told by the main office that daily profit, P, is related to the number of clerks working that day, x, according to the function P = −25x2 + 300x. What number of clerks will maximize the profit, and what is the maximum possible profit?
In order to find the maximum profit and the number of clerks I will first have to find the value of x which will give me the axis and vertex. I will use the formula x=-b/(2a) to solve my quadratic equation.
First the problem already gives me an equation p=-25x^2+300x I will plug the given numbers of this equation into my formula x=-b/(2a) letting a=-25 and b=300.
P=-25x^2+300x original equation
X=-b/(2a) the formula I will use plugging in a=-25 and b=300
X=-300/(2)(-25) first I will foil out the right side
X=-300/-50 then divide
X=6 final six is the number of clerks that will maximize profit
I will then plug 6 into my original equation of p=-25x+300x
P=-25x^2+300x the original equation I will plug into this the 6 clerks
P=-25(6)^2+300(6) I will multiply out to simplest form
P=-25(36)+300(6)
P=-900+1800 then add out
P=900 final 900 is the max profit in dollars
So basically the max occurs at six which is the axis and the 900 would be the vertex. A graph of it would look something like this. Granted this one is not perfect, but it helps to get the general idea. This information is critical for managers to know. It determines the company’s profit and at what amount of employees that can maximize that profit. Companies all use this to determine how they operate what days, what time and how many.

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Reference sheet
Dugopolski, M. (2012). Elementary and intermediate algebra (4th ed.). New York, NY: McGraw-Hill Publishing.