Lagrange multiplier $f(x,y)=x^2+2xy^2+y^2$. Show that $(-1,1)$ is the minimal point of $f$.

Question

Given the function $f(x,y)=x^2+2xy^2+y^2$ with the restriction $x+2y-1=0$ I can't justify why the point (-1,1) is the minimum point of $f$.

I used the lagrange multiplier and found the following maximum and minimum candidates: $(-1,1), (-1/2,2/3)$, and $(0,2/3)$.

Answer 1

The solution (-1,1) is not a global min. The problem has no global min, it is unbounded to $-\infty$.

To see that make the substitution $x=1-2y$ in the objective function to have now an unconstrained problem in $y$:

$$\underset{y\in \mathbb{R}}{\text{min}}\ \ f(y)=(1-2y)^2+2(1-2y)y^2+y^2=1-4y+7y^2-4y^3$$

Notice that $f(y)\rightarrow -\infty$ as $y\rightarrow \infty$.