MAT 222 Intermediate Algebra week 4 ass

Topics: Elementary algebra, Polynomial, Mathematics Pages: 4 (395 words) Published: December 29, 2014
MAT 222 Intermediate Algebra
Mariya Ivanova
Melody Grinter

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In the following paragraphs I will refer to the graph I sketched for problem 56.I then will describe the basic shape of the graph and where it is located in the Cartesian plane... Following the graph description I will demonstrate my solution for both the number of clerks that will maximize the profit as well as the total maximum profit possible, making sure to include all mathematical work and an explanation for each step. For the final step of my assignment I will analyze why this information is important for managers to know, and explain what could happen if the ideal conditions were not met. Problem 56 on page 666 of the Dugopolski, M. (2012). Elementary and intermediate algebra (4th zed.). New York, NY: McGraw-Hill Publishing. The line for the graph would go through 0 and go upward on the right side of the graph curve over at y900 go down and meet at x112. The max will occur when x =6 this would be the axis of symmetry and the vertex is the max which is the 900 in the line.. 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? P=-25x^2+300x is a Quadratic Equation

x=-b/ (2a) Axis of symmetry p=-25x^2+300x, a=-25, b=300
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x=
x=+6 clerks to maximize the profit
Next you need to replace the x with the 6 that was in the original equation to get the maximum possible profit. P= -25() + 300(6)
P= -25(36) +1800
P=-900+1800
P=$900 would be the maximum profit
The vertex would be important to know because without this information the managers would not know how many clerks they would need to help reach the maximum profit and this would lead to a poor profit....
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