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I have an equality constraint function h and a function f to minimize. \begin{array}{c}{h : \mathbb{R}^{2} \rightarrow \mathbb{R}, (x,y) \mapsto 2 x-y} \\ {f : \mathbb{R}^{2} \rightarrow \mathbb{R},(x, y) \mapsto 100\left(y-x^{2}\right)^{2}+(1-x)^{2}}\end{array} By applying Lagrange Multiplier Rule, only single solution x=1 and y =2 is found to satisfy first order necessary condition. But when I check x=0 and y = 0, it also satisfy equality constraint h and even get smaller function value than solution(x=1 and y=2) from Lagrange Multiplier Rule. Does this mean that we can't rely on Lagrange Multiplier Rule to find global minimizer?

The Solution is as follow:

since $\nabla h(x) \neq 0, \nabla h(x)$ is linear independent and thus every $x \in \mathbb{R}^{2}$ with $h(x)=0$ is a regular point. Assume that $x$ satisfies the necessary first order condition for a local mimizer of $f$ subject to the constraint $h(x)=0 .$ Then

$$ \nabla f(x)+\lambda \nabla h(x)=0 $$

for a λ ∈ R. Because $$ \nabla f(x)=\left(\begin{array}{c}{-400 x\left(y-x^{2}\right)+2 x-2} \\ {200\left(y-x^{2}\right)}\end{array}\right) $$

this implies

$$ 0=\left(\begin{array}{c}{-400 x\left(y-x^{2}\right)+2 x-2+2 \lambda} \\ {200\left(y-x^{2}\right)-\lambda}\end{array}\right) $$

Hence $$ \lambda=200\left(y-x^{2}\right) $$ This implies, together with the previous equation, $$ 0=-2 \lambda x+2 x-2+2 \lambda=2\left(x(1-\lambda)-(1-\lambda)\right) $$ Hence $$ x=1 $$ since $h(x)=0,$ this implies $y=2$

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  • $\begingroup$ Is $$2x-y=0$$ given? There is no equality in your post. $\endgroup$ – Dr. Sonnhard Graubner Jul 13 at 8:39
  • $\begingroup$ @Dr.SonnhardGraubner h is the equality function $\endgroup$ – Diskun Tsu Jul 13 at 9:32
  • $\begingroup$ sure. I'm attaching it to the question. $\endgroup$ – Diskun Tsu Jul 13 at 9:33
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You forgot another possible case. From $$0=2(x(1-\lambda)-(1-\lambda))=2(x-1)(1-\lambda),$$ it does not necessarily follow that $x=1$. Namely, if $\lambda=1$, we can have $x\neq 1$. We then get $$y-x^2=\frac{1}{200},\hspace{0.5cm} 2x-y=0$$ and thus $$x^2+\frac{1}{200}=2x\Leftrightarrow (x-1)^2=\frac{199}{200}\Leftrightarrow x=1\pm\sqrt{\frac{199}{200}}$$ and therefore $$y=2\pm\sqrt{\frac{199}{50}}.$$

Plugging these two pairs of values into $f$ yields (in both cases) a value of $\frac{399}{400}$, which is the global minimizer since $\frac{399}{400}<f(1,2)=100$. This is also smaller than $f(0,0)=1$

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  • $\begingroup$ Thank you so much for the correction! $\endgroup$ – Diskun Tsu Jul 13 at 12:56

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