Show that Lagrange multiplier method fails to solve $\min(x^2+y^2)$ subject to $(x-1)^3-y^2=0$

Question

Ok, so I am doing some homework from textbook. The task is:

In the (a) I found out that set of feasible points from g(x) = 0 is: and x is more or equal 1. (and the min of f(x) is in $x^*$ = (1, 0))

In (b) we need to show that Langange multipliers method fails. However,

L(.) = $x^2 + y^2 + \lambda[(x - 1)^3 - y^2]$

Second and third FOCs give us:

(2): $2y(1-\lambda) = 0$

(3): $(x - 1)^3 - y^2 = 0$

If y = 0 from (2), then x = 1 from (3). And this is the point of minimum. So method of Lagrange does not fail?

Answer 1

First, the method of Lagrange does not apply, it is not that it fails. The gradient of the constraint is zero at the solution. The constraint qualification for the method to apply is that the active constraint gradients are linearly independent at a solution.

The $x$ component of the derivative of the Lagrangian with respect to $x$ is $2x + 3 \lambda (x-1)^2$, and at the minimum, this has value $1$. So it cannot be zero in this case.

Answer 2

The Lagrange multiplier method also requires that $$ \frac{\partial\mathcal{L}}{\partial x} = 0, $$ which is not true for the point $(1,0)$.

The method fails because it assumes that $\nabla g(x,y) \neq (0,0)$, but $\nabla g(1,0) = (0,0)$.