SOLUTION: A store sells lecture notes and the monthly revenue, R, of this store can be modelled by the function R(x)=3000+500x-100x², where x is the peso increase over Php 4. What is the max
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-> SOLUTION: A store sells lecture notes and the monthly revenue, R, of this store can be modelled by the function R(x)=3000+500x-100x², where x is the peso increase over Php 4. What is the max
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Question 1105573: A store sells lecture notes and the monthly revenue, R, of this store can be modelled by the function R(x)=3000+500x-100x², where x is the peso increase over Php 4. What is the maximum revenue? Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! I do not know what "x is the peso increase over Php 4." Therefore, I will solve for x itself.
maximum revenue with a quadratic is at the vertex, which is -b/2a.
-100x^2+500x+3000
b=500
a=-100
-b/2a=-500/-200=5/2 or 2.5
R(2.5)=-625+1250+3000=3625 ANSWER.
