# In Lagrange Multiplier, why level curves of $f$ and $g$ are tangent to each other?

In Lagrange multiplier method, e.g. optimize a function $f(x_1, \dots, x_n)$ under a constraint $g(x_1, \dots, x_n) = 0$. There is a fact that $\nabla f$ is parallel to $\nabla g$ which is given rise from the level curves of $f$ and $g$ are tangent to each other (i.e. there tangent lines are parallel, then because gradient and tangent of level curve are orthogonal implies the fact above) at the points when $f$ optimized under constraint $g$.

The only part I don't have intuitive understanding is that why level curves of $f$ and $g$ are tangent to each other at where $f$ optimized under $g$.

The level curves of $f$ represent single values of $f$ that increase in a direction parallel to the gradient. This means that, given a level curve that does not represent a local maximum, there is another level curve nearby whose value for $f$ is greater than the first curve.

Imagine $g$ as a curve that cuts through a level curve of $f$ at a point $p$. Since $g$ cuts the curve, there are level curves of $f$ on either side of $p$ that also intersect with $g$. Therefore, we can choose another level curve with a greater value for $f$ than the one that contains $p$ and so the maximum cannot occur on that curve.

Therefore, to maximize $f$, we choose level curves in the direction of increase until we can go no further which will occur when the level curve of $f$ is tangent to $g$.

The thing is like in the following picture. of wikipedia.

For $f=d$ you increment $d$ until you touch $g=c$. In the moment of contact you take a minimum. If you go on, just before $f=d$ leaves the contact, you take the maximum.

Thinking it well it is like parking! Really, the idea is so productive that is the base of the Morse's theory.

Parametrize the curve $g(x) = 0$ with $c(t)$ s.t $c(0) = p$ where $p$ is the local extrema of $f, c'(0) \not = 0$. Then you know that $f(c(t))$ has local min/max when $t = 0$ i.e;

$$\frac{d}{dt} f(c(t)) |_{t=0} = \nabla f(p) \cdot c'(0) = 0$$

We also know that $\nabla g(p) \cdot c'(0) = 0$ and so there exists a non-zero scalar $\lambda$ s.t;

$$\nabla f(p) = \lambda \nabla g(p)$$

Assume you want to optimize the system $f(x_1,x_2)$ subject to $g(x_1,x_2)=c$. To assure that the constraing holds $$dg=\frac{\partial g}{\partial x_1}dx_1+\frac{\partial g}{\partial x_2}dx_2=0$$ $$\Rightarrow dx_2=-\frac{\frac{\partial g}{\partial x_1}}{\frac{\partial g}{\partial x_2}}dx_1$$ By definition $f$ is optimized when $$df=\frac{\partial f}{\partial x_1}dx_1+\frac{\partial f}{\partial x_2}dx_2=0$$ Replacing $dx_2$ $$df=\frac{\partial f}{\partial x_1}dx_1-\frac{\partial f}{\partial x_2}\frac{\frac{\partial g}{\partial x_1}}{\frac{\partial g}{\partial x_2}}dx_1=0$$ $$df=\bigg(\frac{\partial f}{\partial x_1}-\frac{\partial f}{\partial x_2}\frac{\frac{\partial g}{\partial x_1}}{\frac{\partial g}{\partial x_2}}\bigg)dx_1=0$$ $$\Rightarrow \frac{\partial f}{\partial x_1}-\frac{\partial f}{\partial x_2}\frac{\frac{\partial g}{\partial x_1}}{\frac{\partial g}{\partial x_2}}=0$$ $$\Rightarrow \frac{\frac{\partial f}{\partial x_1}}{\frac{\partial g}{\partial x_1}}=\frac{\frac{\partial f}{\partial x_2}}{\frac{\partial g}{\partial x_2}}=\lambda$$ Now by using the dot product we can check the angle between $\nabla f$ and $\nabla g$ $$\nabla f=(\frac{\partial f}{\partial x_1} \ \ \frac{\partial f}{\partial x_2})\ \ \ \nabla g=(\frac{\partial g}{\partial x_1} \ \ \frac{\partial g}{\partial x_2})$$ $$\cos(\theta)=\frac{\nabla f \cdot \nabla g}{\Vert \nabla f\Vert \quad \Vert \nabla g\Vert}$$ $$\cos(\theta)=\frac{\frac{\partial f}{\partial x_1}\frac{\partial g}{\partial x_1}+\frac{\partial f}{\partial x_2}\frac{\partial g}{\partial x_2}}{\sqrt{\frac{\partial f}{\partial x_1}^2+\frac{\partial f}{\partial x_2}^2}\sqrt{\frac{\partial g}{\partial x_1}^2+\frac{\partial g}{\partial x_2}^2}}$$ $$\cos(\theta)=\frac{\lambda \frac{\partial g}{\partial x_1}\frac{\partial g}{\partial x_1}+\lambda \frac{\partial g}{\partial x_2}\frac{\partial g}{\partial x_2}}{\sqrt{\lambda^2 \frac{\partial g}{\partial x_1}^2+\lambda^2 \frac{\partial g}{\partial x_2}^2}\sqrt{\frac{\partial g}{\partial x_1}^2+\frac{\partial g}{\partial x_2}^2}}$$ $$\cos(\theta)=\frac{\lambda \left(\frac{\partial g}{\partial x_1}^2 + \frac{\partial g}{\partial x_2}^2\right)}{\lambda \left(\frac{\partial g}{\partial x_1}^2 + \frac{\partial g}{\partial x_2}^2\right)}=1$$ Since $\cos(\theta)=1$ $\nabla f$ and $\nabla g$ are paralel.

• "By definition $f$ is optimized when $df=\frac{\partial f}{\partial x_1}dx_1+\frac{\partial f}{\partial x_2}dx_2=0$" - umm, no, this is true only when the extremum is free, i.e. not constrained. Here, though, it is constrained. What you say would be true only in local coordinates on the submanifold $g = 0$. – Alex M. Apr 12 at 14:35