<PRE><B>Question 68151</B>
R= (-1/5X)^2 + 200X 
R is given in dollars. What is the MAXIMUM revenue possible? 
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The graph of this quadratic is a parabola opening downward with
its vertex at its highest (maximum) point.
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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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{{{graph(300,200,-30,700,-100,60000,(-1/5)x^2+200x)}}}
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Cheers,
Stan H.</PRE>