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Find the minmum value of x^2+2xy+2y^2+4y+7 over all real numbers x and y.

 Dec 29, 2020
 #1
avatar+2197 
-1

The minimum is when x=2 and y=-2.

So, the answer is:

2^2+2(2)(-2)+2(-2)^2+4(-2)+7

=3

 Dec 29, 2020
 #2
avatar+117 
+1

 

Given \(x^2+2y^2+2xy+4y+7\)

 

\(x^2+2y^2+2xy+4y+7\)

\(x^2+y^2+2xy+y^2+4y+7\)

\(x^2+y^2+2xy+y^2+4y+4+3\)

\((x+y)^2+(y+2)^2+3\)

 

 

Minimum of square of any number is \(0\). So, minima will be at \(x=2,y=-2\). Therefore, \((2+(-2))^2+(-2+2)^2+3=3\). Therefore, the least value of the expression \(x^2+2y^2+2xy+4y+7\) is \(\boxed3\).

 

Good Job, CalTheGreat (flec)!

 Dec 30, 2020
 #3
avatar+2197 
-1

Thanks, crypto.

 

Source for crypto's answer: https://math.stackexchange.com/questions/2872669/find-the-least-value-of-the-expression-x22xy2y24y7

Check out more info there!

CalTheGreat  Dec 31, 2020

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