Unit 2 Assignment 4: Real World
Assignment 4: Real World Quadratic Functions
MAT222: Intermediate Algebra
Professor Andrea Grych
Assignment 4: Real World Quadratic Functions
Managers and business people use quadratic equations on a daily basis in order to find out how much of a profit can be made. The following problem is an example of that. On page number 666 of the textbook, problem number 56 (Dugopolski, 2012) states that in order to get maximum profits, 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 solve the first question of what number of clerks will maximize the profit, the function -25x + 300x = 0 must be solved in order to get the x-intercepts of the parabola. …show more content…
-25x2 + 300x = 0 Divide both sides by -1 to get rid of the negative.
25x2 – 300x = 0 Factor the left side of the problem.
25x (x – 12) = 0 Now use the Zero Factor Property.
25x = 0 or x – 12 = 0 Solve each problem. x = 0 or x = 12 This means that the parabola will cross the x-axis at 0 and 12.
Now that we know where the parabola will cross the x –axis at, the next question can be asked. What number of clerks will maximize the profit?
-b / (2a) This is the formula that will be used.
-300 / 2(-25) Solve
-300 / -50
6 6 clerks will maximize the
profit.
Question number two is what is the maximum possible profit when this many clerks are working? We will use the profit function for this and plug in 6 as the x variable.
P(6) = -25(6)2 + 300(6) Solve the exponent first.
P(6) = -25(36) + 1800 Solve the problem.
P(6) = -900 + 1800
P(6) = 900 $900 is the profit expected when 6 clerks are working.
The parabola will be narrow, and open downward. At will have a vertex of 6, and go through the x-axis at 0 and 12.
This information is very important for a manger to know because they will then know how many clerks to have working at the same time in order to maximize profits. If a manager did not know this information, they would possibly have too many clerks working at once which would decrease profits, and eventually this would cause the manager to lose their job. According to the information gathered, there would be no profits if 0 or 12 clerks were working at once. 6 clerks would maximize profits, and anywhere in between would result in some profits made.
References
Dugopolski, M. (2012). Elementary and intermediate algebra (4th ed.). New York, NY: McGraw-Hill Publishing.
MAT222: Intermediate Algebra
Professor Andrea Grych
Assignment 4: Real World Quadratic Functions
Managers and business people use quadratic equations on a daily basis in order to find out how much of a profit can be made. The following problem is an example of that. On page number 666 of the textbook, problem number 56 (Dugopolski, 2012) states that in order to get maximum profits, 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 solve the first question of what number of clerks will maximize the profit, the function -25x + 300x = 0 must be solved in order to get the x-intercepts of the parabola. …show more content…
-25x2 + 300x = 0 Divide both sides by -1 to get rid of the negative.
25x2 – 300x = 0 Factor the left side of the problem.
25x (x – 12) = 0 Now use the Zero Factor Property.
25x = 0 or x – 12 = 0 Solve each problem. x = 0 or x = 12 This means that the parabola will cross the x-axis at 0 and 12.
Now that we know where the parabola will cross the x –axis at, the next question can be asked. What number of clerks will maximize the profit?
-b / (2a) This is the formula that will be used.
-300 / 2(-25) Solve
-300 / -50
6 6 clerks will maximize the
profit.
Question number two is what is the maximum possible profit when this many clerks are working? We will use the profit function for this and plug in 6 as the x variable.
P(6) = -25(6)2 + 300(6) Solve the exponent first.
P(6) = -25(36) + 1800 Solve the problem.
P(6) = -900 + 1800
P(6) = 900 $900 is the profit expected when 6 clerks are working.
The parabola will be narrow, and open downward. At will have a vertex of 6, and go through the x-axis at 0 and 12.
This information is very important for a manger to know because they will then know how many clerks to have working at the same time in order to maximize profits. If a manager did not know this information, they would possibly have too many clerks working at once which would decrease profits, and eventually this would cause the manager to lose their job. According to the information gathered, there would be no profits if 0 or 12 clerks were working at once. 6 clerks would maximize profits, and anywhere in between would result in some profits made.
References
Dugopolski, M. (2012). Elementary and intermediate algebra (4th ed.). New York, NY: McGraw-Hill Publishing.