`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 equation P= -25x^2+300x. What number of clerks will maximize the profit, and what is the maximum possible profit?
The max. point is the vertex:
Xv = -B/2A = -300 / -50 = 6 Clerks.
P = Yv = -25*6^2 + 300*6 = $900 = Max.
profit.
To find the number of clerks that will maximize the profit and the maximum possible profit, we can use calculus. Specifically, we need to find the derivative of the profit equation with respect to the number of clerks, x, and set it to zero to find the critical points.
Given the profit equation: P = -25x^2 + 300x
Step 1: Take the derivative of the profit equation with respect to x.
P' = dP/dx = -50x + 300
Step 2: Set the derivative equal to zero and solve for x.
-50x + 300 = 0
-50x = -300
x = -300 / -50
x = 6
Step 3: Determine if the critical point is a maximum or minimum.
To determine whether the critical point is a maximum or minimum, we can examine the second derivative of the profit equation. If the second derivative is negative at the critical point, then it is a maximum.
Step 3a: Take the second derivative of the profit equation.
P'' = d^2P/dx^2 = -50
Since the second derivative is a constant (-50) and negative, we can conclude that the critical point is indeed a maximum.
Therefore, the number of clerks that will maximize the profit is x = 6, and the maximum possible profit can be found by substituting this value back into the profit equation:
P = -25x^2 + 300x
P = -25(6)^2 + 300(6)
P = -25(36) + 1800
P = -900 + 1800
P = 900
So, the maximum possible profit is $900.