\(C=x^2-4x+y^2-6y+70\)
\(=\left(x^2-4x+4\right)+\left(y^2-6y+9\right)+57\)
\(=\left(x-2\right)^2+\left(y-3\right)^2+57\)
\(C\ge57\forall x,y\in R\)
Dấu = xảy ra khi: \(\left\{{}\begin{matrix}\left(x-2\right)^2=0\\\left(y-3\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
Vậy GTNN của C = 57 khi x = 2, y = 3.