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Real World Quadratic Functions

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Read the following instructions in order to complete this assignment: 1. Solve problem 56 on pages 666-667 of Elementary and Intermediate Algebra. 2. Write a two to three page paper that is formatted in APA style and according to the Math Writing Guide. Format your math work as shown in the example and be concise in your reasoning. In the body of your essay, please make sure to include:

o An explanation of the basic shape and location of the graph and what it tells us about the Profit function.
o Your solutions (it is a double question) to the above problem, making sure to include all mathematical work for both problems.
o Discuss how and why this is important for managers to know, and explain what could happen if the ideal conditions were not met.
The paper must be at least two pages in length and formatted according to APA style. Cite your resources in text and on the reference page. For information regarding APA samples and tutorials, visit the Ashford Writing Center, within the Learning Resources tab on the left navigation toolbar. Carefully review the Grading Rubric for the criteria that will be used to evaluate your assignment. Real World Quadratic Functions

Solving real world quadratic problems is mandatory for business professionals and managers. Profit functions routinely show up in their work tasks and these professionals must know how to look at and interpret such things. Profit functions can come in many forms and this paper will seek to describe the quadratic type. Question number 56 on page 667 states the problem, “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  = −25�! + 300.

What number of clerks will maximize the profit, and what is the maximum possible profit?” So in this problem the store manager is to find the maximum profit P using the number of clerks x.Since it is a quadratic function then it has the shape of a parabola, and since the value of a is negative, then it is a parabola opening downwards. This implies there must be a maximum as a parabola opening downwards has a highest point. Furthermore, since the graph is located in the first quadrant, then it has a maximum somewhere in this quadrant. A graph of this equation looks like,

The point x at which the function reaches a maximum can be found using the following method:
� = −25�! + 300� The original problem
� = − !
!! This equation will find the value x at which
P is maximum
� = − !””
! !!” Plug in the values a = -25, b = 300
� = !””
!” Simplify
� = 6 Final solution
So the maximum profit will occur when x = 6. This profit can now be found by plugging in the value x = 6,
� = −25 6 ! + 300(6) Plug in x = 6
� = −25 36 + 1800 Simplify
� = $900 Final answer
Hence the maximum profit is $900.This information is vital to the managers since it will maximize profits. Any other choice would only cause a loss of profit as is evident from the graph. It slopes downward on both sides of x = 6. So any other choice would mean an opportunity cost. For instance if the choice was made to use x = 5, then there still may be a profit, but it would not be as large as it could have been if x = 6 was chosen. This is just one example of where a profit function could be a valuable asset to any business. Profit functions allow a person to make the best decision about their everyday choices. It provides an easy way of seeing just how their choices will affect their outcomes.

DugopolsQuadratici,M.(2012).Elementaryandintermediatealgebra(4thed.).NewYork, NY:McGraw-HillPublishing.

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