I am trying to validate how Lagrange multipliers work.

Looking to maximize $f(x,y)=1-x^2$

along curve $x^2 + y^2 = 1$,

the solutions are $f(0, -1) = f(0,1)=1$.

However, according to Lagrange multipliers, the only $f$ contours that are tanget to this curve go through curve-minima $f(-1, 0)=f(1, 0)=0$.

Does this mean Lagrange multipliers do not catch all extrema? Or is there actually a $\lambda$ that satisfies those two maxima?

  • $\begingroup$ $\lambda=0$ yields the correct maxima $\endgroup$ – user251257 Mar 1 '16 at 23:22
  • $\begingroup$ Right! What if I change the curve to not touch $x=0$, e.g. $(x-2)^2+(y-2)^2=1$? $\endgroup$ – N. J. M. Mar 1 '16 at 23:24
  • $\begingroup$ What do you mean? If $(x,y)$ is an extremum and the gradient of the constraint does not vanish at $(x,y)$ then there exists a Lagrange multiplier $\lambda$ such that the gradient of the Lagrangian vanishes at $(x,y,\lambda)$. $\endgroup$ – user251257 Mar 1 '16 at 23:30
  • $\begingroup$ @user251257 I beg your pardon but I disagree: first, it is better to consider that a Lagrange multiplier cannot be 0, see math.stackexchange.com/q/41534 ; but the main point is that we cannot conclude with this technique because the Hessian is nowhere positive definite in $\mathbb{R}^2$. $\endgroup$ – Jean Marie Mar 2 '16 at 0:40
  • $\begingroup$ @JeanMarie the Lagrangian multiplier might be 0 if a constrained extremum is also a unconstrained extremum. You are mixing sufficient conditions with necessary conditions. Also, it is sufficient that the hessian is definite restricted to the tangent space at the critical point. $\endgroup$ – user251257 Mar 2 '16 at 0:45

Let $L(x,y,\lambda) = 1-x^2 + \lambda(x^2+y^2 -1)$. Then, we have $$ \nabla_{(x,y)} L(x,y,\lambda) = \begin{bmatrix} -2x + 2\lambda x \\ 2\lambda y \end{bmatrix} = 0$$ if and only if

  1. $\lambda=1$ or $x = 0$,
  2. $\lambda=0$ or $y = 0$.

That is, the possible critical points are $(x,0,1)$ for $x\in\mathbb R$, $(0,y,0)$ for $y\in\mathbb R$, and $(0,0,\lambda)$ for $\lambda\notin\{0,1\}$. With the constraint $x^2+y^2=1$ they reduce to $(\pm 1,0,1)$ and $(0, \pm 1, 0)$.

Let us check the necessary and sufficient condition for a maximum. We have $$ H=(x,y,\lambda):=\nabla_{(x,y)}^2 L(x,y,\lambda) = \begin{bmatrix} -2 + 2\lambda \\ & 2\lambda \end{bmatrix} $$ and the tangent space $$ T(x,y) = \{ (h,k) \mid 2xh + 2yk = 0 \} = \{ r(y, -x) \mid r\in\mathbb R \}. $$

  1. Reduced to the tangent space $T(\pm1,0)$, $H(\pm1, 0, 1)$ is positive definite. Thus there is no maximum at $(\pm 1, 0, 1)$. In fact it is a strictly local minimum.
  2. Reduced to the tangent space $T(0,\pm1)$, $H(0,\pm1, 0)$ is negative definite. Thus there is a strictly local maximum at $(\pm 1, 0, 1)$.

Since the gradient of $x^2+y^2-1$ does not vanish for any feasible point, every extremum needs to be critical. Thus, $x = 0, y=\pm1$ are both global maximizers, as they have the same function value $f(0,\pm1)=1$.


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