SOLUTION: Please help, I'm stuck... this is a precalculus problem, but I have problems with the algebra in it, so I thought I'd give this site a shot. The revenue generated by selling X u

Question 68151: Please help, I'm stuck... this is a precalculus problem, but I have problems with the algebra in it, so I thought I'd give this site a shot.
The revenue generated by selling X units of a certain commodity is given by:
R= (-1/5X)^2 + 200X
R is given in dollars. What is the MAXIMUM revenue possible?
I'm supposed to make it quadratic, but I'm confused.
(They want me to use slope formula to find the vertex or something, but I dont get why a vertex would relate to this problem...)
Thanks for your time!

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
R= (-1/5X)^2 + 200X
R is given in dollars. What is the MAXIMUM revenue possible?
--------
The graph of this quadratic is a parabola opening downward with
its vertex at its highest (maximum) point.
--------------
Put the quadratic in vertex form as follows:
R=(-1/5)(x^2-[200/(-1/5)]x)
R+(-1/5)(500^2)=(-1/5)(x^2-1000x+500^2)
R-50000=(-1/5)(x-500)^2
The vertex is at (500,50000)
Conclusion: The maximum Revenue is $50,000
---------------------

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Cheers,
Stan H.
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