Real World Quadratic Functions
Real World Quadratic Functions
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 = ? 25×2 + 300x. What number of clerks will maximize the profit, and what is the maximum possible profit? In order to find the point at which profit is maximized, I must find the critical points of the first derivative of the equation. Coefficient of x^2 is negative, so the maximum value of P will be found at the parabola’s vertex.
I also know that it has to be symmetric, so the average of or midpoint between the roots of the function will give me the x value of the vertex. The xcoordinate of the vertex is given by: x = b/2·a, and so the maximum value of P will be found at x = b/2·a My equation is already in the standard form: P=ax^2+bx+c, p = 25×2 + 300x quadratic equation Therefore I can find the max profit by finding the value of x of the axis of symmetry and find the vertex with that: this is a quadratic equation with a negative coefficient of x^2, so I know that the max is on the axis of symmetry.
The formula for the axis of symmetry; x = b/ (2a), in this equation a = 25, b = 300 X = – 300 / 2 · (25) I will Simplify the equation x = 300/ 50 Divide x = 6 The basic shape of the graph in this equation of a parabola that opens downwards (coefficient of x^2 is negative) so the maximum value of P will be found at the parabola’s vertex. The parabola will cross the x axis at 0 and 6.
To maximize profits, the manager should employ 6 clerks. The maximum profit can be found by substituting 6 for x in the original equation for P. P = 25 · (6)^2 + 300 · (6) Plug in 6 P = 25 · (36) + 1800 Multiply and add P = 900 + 1800 Add the equation P = 900 Maximum profit is 900 The graph represents the max occurs when x = 6 clerks which is the axis of symmetry and the vertex is the max profit of 900.
In conclusion the graph shows that there will be a maximum profit 900 in profit made with having 6 clerks. What I learned is that the graph of the profit function is a convex down parabola because of the negative lead coefficient. Hence, understanding that the vertex of the parabola is a maximum. All that was needed to solve the equation was finding the X coordinate of the vertex. The coordinates of the vertex ycoordinate of vertex will give me the maximum value
B

Subject: Derivative, Function,

University/College: University of Chicago

Type of paper: Thesis/Dissertation Chapter

Date: 22 July 2016

Words:

Pages:
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