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?
if you know Calculus ....
dP/dx = -50x + 300 = 0 for a max of P
x = 6
P= -25(36) + 300(6) = 900
otherwise complete the square ....
P = -25(x^2 - 12x + 36-36)
= -25( (x-6)^2 - 36)
= -25(x-6)^2 + 900
same as above
To find the number of clerks that will maximize the profit and the maximum possible profit, we need to determine the maximum point on the graph of the profit function.
The profit function is given as P = -25x^2 + 300x, where P represents the daily profit and x represents the number of clerks working that day.
To find the maximum point, we need to find the vertex of the quadratic equation. The x-value of the vertex can be found using the formula x = -b / (2a), where a and b are the coefficients of the quadratic equation ax^2 + bx + c.
In this case, the equation is P = -25x^2 + 300x, so a = -25 and b = 300. Using the formula, we can calculate the x-coordinate of the vertex:
x = -300 / (2*(-25))
x = -300 / (-50)
x = 6
Now that we have the x-coordinate of the vertex, we can substitute this value back into the profit function to find the maximum profit:
P = -25(6)^2 + 300(6)
P = -25(36) + 1800
P = -900 + 1800
P = 900
Therefore, the number of clerks that will maximize the profit is 6, and the maximum possible profit is $900.