Minimize $f(x,y)=x^2+y^2$ subject to constraint $(x-1)^3-y^2=0$.

Question

Minimize $f(x,y)=x^2+y^2$ subject to constraint $(x-1)^3-y^2=0$. This was presented as an example for which the method of Lagrange multipliers fails to identify the minima at $(1,0)$. My issue is without reference to the plot, how would you find and classify the point $(1,0)$?

Answer 1

For the condition $(x-1)^3=y^2$ we have $x\geq1$ and $(1,0)$ can not be a point of a local extreme,

but in $(1,0)$ our $f$ gets a minimal value.

The condition was $(x-1)^2=y^3$.

For this condition:

Let $g(x)=x^2+\sqrt[3]{(x-1)^2}.$

Thus, $g'(x)=2x+\frac{2}{3\sqrt[3]{x-1}}.$

We see that $g'(x)\rightarrow+\infty$ for $x\rightarrow1^+$ and $g'(x)\rightarrow-\infty$ for $x\rightarrow1^-$.

Id est, $(1,0)$ is a local minimum point of $g$.

Answer 2

The Lagrange multiplier's technique can be successfully applied when the involved functions are class $\mathcal{C}^1$. The function used as restriction

$$ g(x,y)=(x-1)^3-y^2=0 $$ does not pertain to $\mathcal{C}^1$ because it's implicit derivative is discontinuous at $x = 1$ as can be easily verified because

$$ y' = \frac 32\sqrt{x-1} $$

then at this point can occur problems with the method.

Concluding, the assertion " This was presented as a counter example to the method of Lagrange multipliers" is not correct because the methot cannot be applied.