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Find the minimum value of 2x^2 + 2xy + y^2 - 2x + 2y + 4 - x^2 + 4x over all real numbers $x$ and $y.$

 Jun 25, 2022
 #1
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Rearrange as: \( (y^2 + 2y) +(x^2 +2x+ 2xy) + 4 \)

 

Consider this as 2 separate equations (\(x^2 +2x+ 2xy\) and \(y^2 + 2y\)) and a constant. Our goal is to minimize the equations.

 

Start with \(y^2 + 2y\). The minimum value of this is -1, and it occurs when \(y = {-b \over 2a} = {-2 \over 2} = -1\) 

 

Now, subbing in \(y = -1\) into the second equation gives us \(x^2 + 2x - 2x = x^2\)

 

The minimum value of this, quite obviously, is 0, and it occurs when \(x = 0\).

 

This means the minimum value is: \(0 - 1 + 4 = \color{brown}\boxed{3}\)

 Jun 26, 2022

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