Calculus

Completing the squares: $f(x, y) = (x^2 - 4x + 4) + 2(y^2 - 2y + 1) + 10 - 4 - 2$ $f(x, y) = (x - 2)^2 + 2(y - 1)^2 + 4$ Since $(x - 2)^2 \ge 0$ and $2(y - 1)^2 \ge 0$, the minimum value is $4$, occurring at $(x, y) = (2, 1)$.