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