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:

$$F(x,y,lambda)=f(x,y)+lambdavarphi(x,y)$$

Then solve this equation：

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

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

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 $y=y_0$):

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

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:

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

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

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

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:

$$begin{cases} f_x^{'}(x_0,y_0)-lambdacdotvarphi_x^{'}(x_0,y_0)=0\ &\ f_y^{'}(x_0,y_0)-lambdacdotvarphi_y^{'}(x_0,y_0)=0\ &\ end{cases}$$

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

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

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