SOLUTION: find the maximum revenue for the revnue function R(x)= 140x - 0.02 x to the second power.

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: find the maximum revenue for the revnue function R(x)= 140x - 0.02 x to the second power.      Log On


   

Question 2302: find the maximum revenue for the revnue function R(x)= 140x - 0.02 x to the second power.
Answer by matthew_sessoms(39) About Me  (Show Source):
You can put this solution on YOUR website!
You can graph this or use the "complete the square" (CTS) method. I'll do both. I won't go into detail how to "complete the square" but rather show my steps. If you're not familiar with CTS, write back and I'll tell you.


R(x)=140x-.02x^2
y=-.02x^2+140x


Factor out the -.02
y+=+-.02%28x%5E2-7000x%29
y=-.02%28x%5E2-7000x%2B12250000%29%2B245000
y=-.02%28x-3500%29%5E2%2B245000


This is now in the form of: y=a%28x-h%29%5E2%2Bk, where (h,k) is the vertex. "k" is the maximum if "a" is negative. "k" is the minimum if "a" is positive.


So, (3500, 245000) is the vertex.
245000 is the maximum
Here is the graph to prove my statements.

MS