Lagrange Multiplier Method

Today, I learned a method of finding the conditional extremum of the two-variables function, called the Lagrange multiplier method.

How to use Lagrange Multiplier Method

When we need to know the unconditional extremum of a two-variables, calculating partial derivatives and second derivatives will be useful.

But what if I add a condition $\varphi(x,y)=0$?

To solve the problem about the extremum of $f(x,y)$ under $\varphi(x,y)=0$, we can construct the Lagrange Function:

$$F(x,y,\lambda)=f(x,y)+\lambda\varphi(x,y)$$

Then solve this equation：

$$\begin{cases} \tag{*} \dfrac{\partial F}{\partial x}=\dfrac{\partial f}{\partial x}+\lambda\dfrac{\partial \varphi}{\partial x}=0\\ &\\ \dfrac{\partial F}{\partial y}=\dfrac{\partial f}{\partial y}+\lambda\dfrac{\partial \varphi}{\partial y}=0\\ &\\ \dfrac{\partial F}{\partial \lambda}=\varphi(x,y)=0\\ \end{cases}$$

All x and y meet with this equation are extreme points meeting $\varphi(x,y)=0$.

Proof of Lagrange Multiplier Method

Now consider the function $y=y(x)$ confirmed by $\varphi(x,y)=0$, we plug this into $z=f(x,y)$:

$$z=f(x,y(x))\tag{1}$$

Then the two-variables problem is transformed into the single variable problem. Suppose $x_0$ is the extreme point of z, and $y_0=y(x_0)$. Obviously when $x=x_0$(while $y=y_0$):

$$\frac{dz}{dx}\bigg|_{x=x_0}=f_x^{'}(x_0,y(x_0))+f_y^{'}(x_0,y(x_0))\cdot y^{'}(x_0)=0\tag{2}$$

By using formula of derivation calculus for hidden function we know:

$$y^{'}(x_0)=-\frac{\varphi_x^{'}(x_0,y_0)}{\varphi_y^{'}(x_0,y_0)}\tag{3}$$

Plug $y_0=y(x_0)$ and (3) into (2), and simplify it:

$$\frac{dz}{dx}\bigg|_{x=x_0}=f_x^{'}(x_0,y_0)-\frac{f_y^{'}(x_0,y_0)}{\varphi_y^{'}(x_0,y_0)}\cdot\varphi_x^{'}(x_0,y_0)=0\tag{4}$$

If we consider $x=x(y)$confirmed by $\varphi(x,y)=0$. By the similar way, we can get:

$$\frac{dz}{dy}\bigg|_{y=y_0}=f_y^{'}(x_0,y_0)-\frac{f_x^{'}(x_0,y_0)}{\varphi_x^{'}(x_0,y_0)}\cdot\varphi_y^{'}(x_0,y_0)=0\tag{5}$$

Let $\lambda=-\dfrac{f’_y(x_0,y_0)}{\varphi’_y(x_0,y_0)}=-\dfrac{f’_x(x_0,y_0)}{\varphi’_x(x_0,y_0)}$. Plug this into (4) and (5). Then we have the following equation:

$$\begin{cases} f_x^{'}(x_0,y_0)-\lambda\cdot\varphi_x^{'}(x_0,y_0)=0\\ &\\ f_y^{'}(x_0,y_0)-\lambda\cdot\varphi_y^{'}(x_0,y_0)=0\\ &\\ \end{cases}$$

Obviously $x=x_0$ and $y=y_0$ are the solutions of following equation:

$$\begin{cases} \dfrac{\partial f}{\partial x}+\lambda\dfrac{\partial \varphi}{\partial x}=0\\ &\\ \dfrac{\partial f}{\partial y}+\lambda\dfrac{\partial \varphi}{\partial y}=0\\ &\\ \varphi(x,y)=0\\ \end{cases}$$

Let $F(x,y,\lambda)=f(x,y)+\lambda\varphi(x,y)$, we can get (*).
Q.E.D.

If you are puzzled by why $-\dfrac{f’_y(x_0,y_0)}{\varphi’_y(x_0,y_0)}=-\dfrac{f’_x(x_0,y_0)}{\varphi’_x(x_0,y_0)}$, you can contact me by email umbrellalalalala@qq.com . Actually it is a nature of point of tangency, so I didn’t explain it here.