Why does the Lagrange multiplier method not work to minimize $f(x,y) = x$ subject to $x^3=y^2+x^4$?

Question

Why does the Lagrange multiplier method not work to minimize $f(x,y) = x$ subject to the equality constraint $x^3=y^2+x^4$? How does one use the 2nd derivative test to classify the critical point? And how can you use the definition of the derivative to classify the extrema?

Answer 1

The Lagrange multipliers method gives what it can, namely, points $(a,0)$ at which there can be an extremum. Further, it refuses to help and you need to move yourself. But it is quite simple.

$x^3=y^2+x^4\Rightarrow x^3(1-x)=y^2\geqslant 0\Rightarrow 0\leqslant x \leqslant 1.$

So the minimum of the function $f(x,y)=x$ is zero - it achived at point $(0;0)$