Find the global minima of a two-variable function: $$f(x,y) = (x-2y)^2 -1$$ I have already found partial derivatives in x and y: $$\frac{\partial}{\partial x} = 2x - 4y $$ $$\frac{\partial}{\partial y} = -4x + 8y $$
And I see that any number satisfies those 2 equations. Here I don't know what to do next. Can somebody help me with this problem? Thank you
Setting $\frac{\partial}{\partial x}$ equal to $0$ we get $x=2y$, and every such solution also makes $\frac{\partial}{\partial y}$ vanish, so these are all the critical points. However, we are looking for global extrema, rather than local ones.
Pure algebra will do the trick, and there is no need for partial derivatives: $(x-2y)^2-1$ is minimal when $(x-2y)^2$ is minimal and that happens when $x-2y=0$ i.e. $x=2y$. Therefore all points $(x,y)$ of the form $x=2y$ are global minima and the minimal value is $-1$.