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Find the minimum value of \(x^2 + 2xy + 3y^2 - 6x - 2y,\) over all real numbers \(x\) and \(y.\)

 Apr 16, 2019
 #1
avatar+191 
+1

Try to rewrite the expression as the sum of squares plus a number.

 

Is the answer -11? Inspiration: https://www.quora.com/If-x-and-y-are-real-numbers-then-what-is-the-minimum-value-of-x-2-+4xy+6y-2-4y+4

 

Don't take my word for it, though. wink

 Apr 16, 2019
 #2
avatar+191 
0

I think it's x=-4, y=1. 

 Apr 16, 2019
 #3
avatar+7685 
+1

\(x^2+2xy+3y^2 -6x-2y\\ =(x+y)^2 + 2y^2-2y-6x\\ =(x+y)^2 + 2y(y-1) - 6x\\ \boxed{\text{Minimum occurs when x = -y.}}\\ = 2y^2-2y+6y\\ = 2(y^2+2y)\\ =2((y+1)^2-1)\\ =2(y+1)^2 - 2\\ \text{Minimum value} = -2.\)

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 Apr 18, 2019

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