SOLUTION: I have a homeowrk question that states: find the maximum revenue for the revenue function R(x)=490x-0.07x^2. How do I work this problem out? I have tried everything.

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Question 86040: I have a homeowrk question that states: find the maximum revenue for the revenue function R(x)=490x-0.07x^2. How do I work this problem out? I have tried everything.
Answer by jim_thompson5910(35256)   (Show Source): You can put this solution on YOUR website!


Make into into a fraction


Rearrange the terms
Now lets convert this to vertex form to find the max revenue
Solved by pluggable solver: Completing the Square to Get a Quadratic into Vertex Form


Start with the given equation



Subtract from both sides



Factor out the leading coefficient



Take half of the x coefficient to get (ie ).


Now square to get (ie )





Now add and subtract this value inside the parenthesis. Doing both the addition and subtraction of does not change the equation




Now factor to get



Distribute



Multiply



Now add to both sides to isolate y



Combine like terms




Now the quadratic is in vertex form where , , and . Remember (h,k) is the vertex and "a" is the stretch/compression factor.




Check:


Notice if we graph the original equation we get:


Graph of . Notice how the vertex is (,).



Notice if we graph the final equation we get:


Graph of . Notice how the vertex is also (,).



So if these two equations were graphed on the same coordinate plane, one would overlap another perfectly. So this visually verifies our answer.






So the maximum revenue that can be generated is $857,500 (the y-coordinate of the vertex) by selling 3,500 units (the x-coordinate of the vertex)

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