SOLUTION: Given the following revenue and cost functions, find the x-value that makes revenue a maximum. {{{R(x) = 68x - 2x^2;}}}{{{ C(x) = 21x + 97}}} The answer is supposedly 17, but n

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Given the following revenue and cost functions, find the x-value that makes revenue a maximum. {{{R(x) = 68x - 2x^2;}}}{{{ C(x) = 21x + 97}}} The answer is supposedly 17, but n      Log On


   

Question 1041311: Given the following revenue and cost functions, find the x-value that makes revenue a maximum.
R%28x%29+=+68x+-+2x%5E2%3B+C%28x%29+=+21x+%2B+97
The answer is supposedly 17, but no matter what I do I can't figure out how to get it?
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
The cost function is not involved in finding max x for the revenue function. You have two separate functions, R for revenue and C for cost.

Where is R a maximum value?
Between the zeros in the exact middle.
68x-2x%5E2=0
34x-x%5E2=0
x%2834-x%29=0
Zeros are at 0 and at 34. The maximum occurs exactly between these two values!

Answer by MathTherapy(10551) About Me  (Show Source):
You can put this solution on YOUR website!

Given the following revenue and cost functions, find the x-value that makes revenue a maximum.
R%28x%29+=+68x+-+2x%5E2%3B+C%28x%29+=+21x+%2B+97
The answer is supposedly 17, but no matter what I do I can't figure out how to get it?
The revenue function is: R%28x%29+=+68x+-+2x%5E2, and the maximum revenue occurs at 

That's all!