SOLUTION: Maximum profit using the quadratic equations, functions, inequalities and their graphs. A chain store manager has been told by the main office that daily profit, P, is related t

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: Maximum profit using the quadratic equations, functions, inequalities and their graphs. A chain store manager has been told by the main office that daily profit, P, is related t      Log On


   

Question 23275: Maximum profit using the quadratic equations, functions, inequalities and their graphs.
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 equations P = -25x^2 + 300x. What number of clerks will maximize the profit, and what is the maximum possible profit?
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Maximise: P+=+-25x%5E2+%2B+300x
This is the equation of a parabola that opens downwards (coefficient of x^2 is negative) so the maximum value of P (the dependent variable) will be found at the parabola's vertex. The x-coordinate of the vertex is given by:
x+=+-b%2F2a and so the maximum value of P will be found at x+=+-b%2F2a
Your equation is already in the standard form: P+=+ax%5E2+%2B+bx+%2B+c (a = -25, b = 300, c = 0) so we can find the x-coordinate at which P will be the maximum.
x+=+-%28300%29%2F2%28-25%29 Simplify.
x+=+-300%2F%28-50%29
x+=+6
To maximise 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%286%29%5E2+%2B+300%286%29
P+=+-25%2836%29+%2B+1800
P+=+-900+%2B+1800
P+=+900
Maximum profit is 900 (dollars ?)