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Question 59633This question is from textbook Elementry and Intermediate Algebra
: The daily profit of a company P is related to the number of employees (x) working on that day and is represented by the equation P = - 25x2 + 300x.
What number of employees will maximize the profit and what will be the maximum profit?
This question is from textbook Elementry and Intermediate Algebra
Answer by josmiceli(19441)   (Show Source):
You can put this solution on YOUR website! 
The negative sign in front of the  term means the graph will
rise, then peak, then fall.
There should be two points where the profit,P, is zero.
One point is where x=0, which means no employees. That
makes sense, doesn't it?
This is the point (0, 0)
The other zero-profit point is where the graph rises and then falls
back to the x-axis, where P, the vertical axis, is zero.
This is true when x=0, as I just showed, and also when (-25x + 300) = 0
So, with 12 employees, the profit is back to zero again.
This is the point (12, 0)
The vertex, or peak of the graph is mid-way between these two
x-values, x=0 and x=12
That means x = 6 should give us the max profit
This is at the point (6, 900)
We can check that, with 6 employees, the profit is maximum, and it
is 900. Try putting x = 5.9 and x = 6.1 into the equation and see
what you get for P. I did and I got
P(5.9) = 899.75
P(6.1) = 899.75
So the profit drops slightly on either side of x = 6. That is a strong
indication that (6, 900) is a maximum.
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