Need help with my writing homework on Real World Quadratic Functions. Write a 500 word paper answering;

Need help with my writing homework on Real World Quadratic Functions. Write a 500 word paper answering; Real World Quadratic Functions Introduction Quadratic functions have wide application in not only science and engineering but also in business. They find wide application in forecasting profits and losses in business and assisting in the determination of maximum and minimum values (Harshberger & Reynolds, 2013). A quadratic function is one that takes the form of f(x) = ax2 + bx + c. In this assignment, the daily profit recorded in the chain store would be determined by the number of clerks employed. The maximum profit would be determined from the profit function, P(x) = -25×2 + 300x, where:

P – daily profit,

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x – number of clerks

P = -25×2 + 300x

The x-intercepts would be deduced from solving -25×2 + 300x = 0. Multiplying through by -1.

25×2 – 300x = 0Factoring the left hand side.

25x(x – 12) = 0Using the Zero Factor Property.

25x = 0 or x – 12 = 0Solving both equations.

x = 0 or x = 12An indication of a parabola intersecting with the x-axis at 0 and 12.

The maximum value for the graph would occur where b = 300 and a = -25. c = 0. Borrowing from Harshberger and Reynolds (2013), with a &gt. 0, the parabola would open upwards as opposed to when a &lt. 0 where it would open downwards. Since in this case a = -25 &lt. 0, the parabola opens downwards with the maximum value of P being found at the vertex of the parabola.

To determine the number of clerks that would maximize profits, the x-coordinate of the vertex would be employed given by:

x = -b/2a = -300/2(-25) = 6

Therefore, with 6 working clerks would be needed for the profits would be maximized.

To determine the maximum possible profits.

P(6) = -25(6)2 + 300(6)

P(6) = -25(36) + 1800

P(6) = -900 + 1800

Thus, a possible maximum profit of 900 would be expected with 6 clerks working.

It would be deduced from this graph that with no clerks working, there would no profits made just as there would be no profits made with the engagement of 12 clerks. Maximum profit would be achieved with 6 clerks working. This function gives a graph whose relevance only holds in the first quadrant.

Importance to managers

Quadratic functions play a critical role in business cases with managers using these functions to determine the amount of workforce or units needed to obtain the greatest possible profit returns (Harshberger & Reynolds, 2013). Similarly, they help in forecasting the expected profits or losses hence assist in planning. These functions vary from one company to another thus the importance of each manager to determine the function that truly reflects the operations of the managed firm.

Conclusion

Quadratic functions would be applied in various real world situations to find meaningful solutions. In this assignment, its usefulness in determining the maximum profit and how to maximize profits in a business entity has been illustrated. From this, it would be noted that solving a quadratic function problem requires the determination of vertex or description of the parabola’s section.

Reference

Harshbarger, R. J. & Reynolds, J. J. (2013). Mathematical applications for the management, life, and social sciences (10th ed.). Boston, MA: Brooks/Cole.

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