Real World Quadratic Functions



Real World Quadratic Functions

Question 56 (Page # 666 - 667)

In order to find out the particular point for which the profit is maximized, we must find the critical points of the first derivative of the equation.

P = -25x2 + 300x ---------------------- (a)

The graph of P(x) is a parabola opening down and has a maximum point.

The extremes of a function can occur if dP(x)/dx = 0. The first derivative is

dP = -25x + 300

For critical points

0 = -25x + 300

On solving the equation we get,

x = 12

Therefore, 12 labors will be the profit maximizing level for ...