Question 267713: 3x+y=16
y=16-3x
x(16-3x)=16x-3x^2
x^2 +(16x-3x^2)= x^2+16x+9x^2
10x^2 +256
I get 10x^2-144x+256........the answer is 10x^2-96x+256
How do they get -96x??
This is a minimization/maximization of quadratic functions problem
Thanks for any/all help!
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! 3x+y=16
y=16-3x
---------- Are these the 2 equations? It's the same one twice.
There's no point in working further.
--> dependent, infinite # of solutions.
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x(16-3x)=16x-3x^2 ?? Where did this come from?
x^2 +(16x-3x^2)= x^2+16x+9x^2
10x^2 +256
I get 10x^2-144x+256........the answer is 10x^2-96x+256
How do they get -96x??
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Restate the problem, if it's different than that above.
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