Equivalence between minimization and solving system of equations

Question

Why is solving the system of equations $$1+x-y^2=0$$ $$y-x^2=0$$ the same as minimizing $$f(x,y)=(1+x-y^2)^2 + (y-x^2)^2$$

Originally I thought it was because if you take the partial derivatives of $f(x,y)$ and set them equal to zero that is what you are doing in the system. But when I worked out the partial derivatives it was not clear that that is what was going on.

Can someone clarify why they are equivalent?

Answer 1

Note that $$ f(x,y)=(1+x-y^2)^2 + (y-x^2)^2$$ is sum of two squares which is always non-negative.

The minimum value of $f(x,y)$ is zero which is attained when both squares are zero.

Your system of equations are simply making the squares equal zero and finding the points at which the minimum is attained.

Answer 2

We can say $f(x,y)=g(x,y)^2+h(x,y)^2$. It is clear that being the sum of two square terms, $f(x,y)\geq 0$. So, the minimum value of $f(x,y)$ (which is $0$) comes about when $g(x,y)=0$ and $h(x,y)=0$.