# can't find global minimizer from Lagrange Multiplier Rule

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 $$y=2$$

• Is $$2x-y=0$$ given? There is no equality in your post. – Dr. Sonnhard Graubner Jul 13 at 8:39
• @Dr.SonnhardGraubner h is the equality function – Diskun Tsu Jul 13 at 9:32
• sure. I'm attaching it to the question. – Diskun Tsu Jul 13 at 9:33

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 $$\frac{399}{400}. This is also smaller than $$f(0,0)=1$$