Real World Quadratic Functions

Real World Quadratic Functions

Quadratic Functions are used every day by, scientists, engineers, and businesses in general. Quadratic functions can help businesses figure out profit and loss, help determine minimum and maximum values. It also shows how math concepts can be combined into a single problem. To solve these problems we use rules for order of operations, solving equations, exponents, and radicals.

The problem given to solve is on page 666 #56. “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 (Dugopolski, 2012)?” This is in the form form ax2 + bx + c = 0. The x-intercepts of the parabola can be found by solving -25x2+300x=0.

-25x2+300x

-25x2+300x divide both sides by –1

-1 -1

25x2-300x factor the left side

25x(x-12) =0 use 0 factor property

25x=0 or x=-12 solve equation

X=0 or x=-12 the parabola will cross the x-axis at 0 and -12

The parabola will be narrow because the quadratic function has a large a value. The a value is negative so the parabola will open up downwards. So there will be a maximum value of the graph at the vertex, this will happen at the x value of –b/(2a). b=300, a=-25.

What number of clerks will maximize the profit?

–b/(2a)

-300/2(-25) the b value is now negative because of the negative in the formula

-300/-50

6 clerks will maximize the profit

What is the maximum possible profit?

P(x)= -25x2+300x original profit function to find P(6)

P(6)= -25(6)2+300(6) substitute the 6 for all x’s

P(6)=-25(36)+1800 the exponent has to be solved first...