Real World Quadratic Functions Week 4 A

Real World Quadratic Functions
MAT222: Intermediate Algebra
Argenia L. McCray
Professor: Eric Bienstock
October 27, 2014

Quadratic Functions This week we have been learning the many different quadratic functions. Throughout the world the quadratic functions are being used / or being implicated into their system of employment, business, and in all schools. To say that the quadratic function has limited/ or less of it many possibilities which is available to be used in solving our many problem today issues. I will used one today to help me to solve my prediction and to helping me make some much better decisions. In this lesson we was given this problem to work out, “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 solving this problem it requires all the following steps listed below (Dugopolski, M. 2012).
One way to finding the x-intercept is to do the following
-25x2 + 300x = 0
25x2-300x = 0 we will now divide -1 to both sides to get rid of the negative in front of -25
25x (x-12) = 0 Then we will factored the left side of the problem of our equation
25x = 0 or x-12 = 0 this is called the Zero Factor Property we are using
X = 0 or x = 12 parabola will cross over the x-axis at 0 and 12. I have solved for x. Now we will find the axis symmetry the x = - b/2a. We determine here if the Parabola has a maximum or a minimum value.
At this point we will now solve the equation by breaking down the a = -25, b = 300, c = 0
X =- 300 / (2) (-25) filled in values (removed the parenthesis)
X = -300/-50 problem simplified further (simplify) x = 6 final answer problem has been solved.
In order to maximize the company profit they will needed 6 more clerks.
To find the maximum profits for Charlie to reach, he needs 6 more working clerks:
P = -25(6)2 +300 (6) now which to re-placing 6 for “x” in equation in our problem,
P = -25 (35) + 1800 then getting rid of our exponent by which we will multiplication,
P = -900 + 1800 using this step to multiplied further getting rid of parentheses, maximum value Which is going to be $900.00
P = 900 then our problem is solved.
The $900.00 dollars which will be our profit if 6 more employees will working and service our clients.
The graph if it was available will look as if; the concave is down, and we have increasing the concave is now up, and the decreasing is now on the x-axis. A person which is in a manager position should know this information which will in terms help provided them the quadratic equation when needed. This will limit the hiring for unnecessary staffing and better the amount of profit for their company based on their new clerks which it should help to lessen the company risk. With this equation problem and other real life problems this predictions will possible help us with other solutions by using the quadratic functions when making decisions better and more accurate if we know the formulas which to put in use.

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