Solving Real-World Problems Using Algebra.
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?
The Quadratic equation P= -25x2+300x can be modified to the form y= -25x2+300x, where P= y.
Modifying the equation in this form enables us to plot the function y= -25x2+300x
For various values of x where x is a positive integer.
The general equation of a quadratic is given by y= ax2+bx+c where a, b and c are the coefficients of the first term, second term and third term respectively. A positive coefficient for the first term indicates that the vertex is a minimum while a negative coefficient implies a maximum. The equation y= -25x2+300x has a negative coefficient for the first term,-25x2, implying that the vertex to the graph will be a maximum.
Using the formula, x=-b2a we can find the axis of symmetry.
Substituting for a and b yields
The axis of symmetry is at six, 6. The corresponding value for y is obtained by substituting x with 6 in the equation y= -25x2+300x.
The result for y is,
The resulting graph shows a graph that cuts the x-axis at two points, 0 and 12. The axis of the graph is at the point (6,900). The coordinate corresponds to the calculated figures implying that the maximum value for y is when x is 6.
Therefore, this means that the number of employees that result in maximum profits is six.
Using workers less than six results in an understaffed office, which despite optimal performance will not generate maximum profit for the enterprise. Alternatively, employing more than six workers results in higher overheads hence to low profits. The graph depicts a scenario where increasing the number of workers to more than six leads to progressively lower profits that subsequently turn to losses when more than twelve workers are employed.
The use of quadratic equations formulation and solving is important for managers to know because it equips them with a skill that will enable them solve complex relationships in the work place. These relationships affect the realization of maximum profits in the firm and thereby need close understanding. Since most problems can be accommodated by the use of more complex mathematical models, managers need to fully grasp concepts of solving problems based on algebra (Scarlett, 2012).
Scarlett, B. (2012). Back to Basics – the use of algebra in management accounting. Velocity, June 2012. Retrieved from http://www.cimaglobal.com/Thought- leadership/Newsletters/Velocity-e-magazine/Velocity-2012/Velocity-June- 2012/Back-to-basics---the-use-of-algebra-in-management-accounting/