Question

The profit, P, in dollars, a business can be expressed as P = -n^2 + 200n, Where n represents the number of employees.

What is the maximum profit? How many employees are required for this value?

Answer 1

From an algebra perspective, I believe I remember an equation that tells you where the min/max is or a quadratic, I think it was:
\[-b/2a\]

Where\[a^2+b^2+c^2=0\] So in this case, a=-1 and b=200

or in calculus you may find the first derivative
\[P'=-2n+200\]
and the solution(s) of the first derivative will yield the extrem points of the parent function.

Both methods will give you the same result.

Answer 2

The answer is 10,000 but I don't know how to get it though.

Answer 3

Using either method I provided (there are probably more too) will give you the n value (number of employees) that occurs at the maximum.

However the problem is asking for the profit, so you must plug the n value into the original equation to the the max profit

Answer 4

to get the*

Answer 5

I don't know if this is right so you may correct me.We transform the expression in P=n(200-n) and we observe the following phenomenon, while n decreases, 200-n increases and while n increases 200-n decreases.We can easily see that 0<n<200,you can use AM-GM and find that (200+n-n)/2>=sqrt(n*(200-n)) so 100*100=>P P<=10000 and the largest value the P can get is 10000, now you can solve the quadratic equation to find that n=100..Hope i helped (further help: the AM-GM inequallity shows us that the geometrical mean of two numbers is always smaller from their arithmetic mean),

Answer 6

*smaller or equal. (equal only when the two numbers are equal)