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  4. For real values of x and y, the minimum value of f(x, y)=x2+2 x y+2 y2+6 y+10, is

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Q. For real values of $x$ and $y$, the minimum value of $f(x, y)=x^2+2 x y+2 y^2+6 y+10$, is

Complex Numbers and Quadratic Equations

Solution:

$ f(x, y)=(x+y)^2+(y+3)^2+1$
Hence, $f(x, y)$ is minimum if $y=-3$ and $x=3$.
$\therefore f ( x , y )]_{\min }=1 .$