Question

1. show that $f(x,y)=y^{2}+y^{4}$ has infinitely many critical points (all along $y = 0$) and that although it is clear that each one of them is a global minimum, the second - derivative test is inconclusive for each one.

Explanation:

Step1: Find the first - order partial derivatives

First, find $\frac{\partial f}{\partial x}=0$ and $\frac{\partial f}{\partial y}=2y + 4y^{3}=2y(1 + 2y^{2})$. Setting $\frac{\partial f}{\partial x}=0$ (since there is no $x$ in the function, this derivative is always 0), and setting $\frac{\partial f}{\partial y}=0$. We have $2y(1 + 2y^{2}) = 0$. Since $1+2y^{2}>0$ for all real $y$, then $y = 0$. So the critical points are of the form $(x,0)$ for all real $x$.

Step2: Find the second - order partial derivatives

Calculate $f_{xx}=0$, $f_{yy}=2 + 12y^{2}$, $f_{xy}=0$. Then the discriminant $D=f_{xx}f_{yy}-(f_{xy})^{2}=0\times(2 + 12y^{2})-0^{2}=0$ for all critical points $(x,0)$. So the second - derivative test is inconclusive.

Step3: Analyze the function value

For $f(x,y)=y^{2}+y^{4}$, when $y = 0$, $f(x,0)=0$. And for any real $y$, $y^{2}\geq0$ and $y^{4}\geq0$, so $f(x,y)=y^{2}+y^{4}\geq0$. Thus, each point of the form $(x,0)$ is a global minimum.

Answer:

We have shown that $f(x,y)=y^{2}+y^{4}$ has infinitely many critical points of the form $(x,0)$ for all real $x$, the second - derivative test is inconclusive for these points, and each of them is a global minimum since $f(x,y)\geq0$ for all $(x,y)$ and $f(x,0) = 0$.