Lagrange multiplier method （拉格朗日乘数法）

The Lagrange multiplier method is used to solve the problem that find the extremum of a function z=f(x,y) , given ϕ(x,y)=0 .

simple extremum problem

To solve a simple extremum problem,

z=f(x,y)

we just need to find the (x,y) where

∂z∂x=0∂z∂y=0

By solving the simultaneous equations, we could find the (x,y)s. Then plugging in the (x,y)s into the function z = f(x,y), we find the extremum of z.

extremum problem with restricted condition

The problem is finding the extremum of

z=f(x,y)

where,

ϕ(x,y)=0

By applying Lagrange multiplier method, we define a new function, as:

F(x,y,λ)=f(x,y)+λϕ(x,y)

Then, take the partial derivatives of x,y,λ . We get ∂F∂x,∂F∂y,∂F∂λ . Make them be 0 .

∂F∂x=0∂F∂y=0∂F∂λ=0

Three equations, three unknow variables. The equation set could be solved. By solving the equation set, we get the (x,y) , we care nothing about λ .
Then, plugging the (x,y) , just solve out, in the function z=f(x,y) , we find the extremum of z=f(x,y) with restricted condition ϕ(x,y)=0

explanation

as ϕ(x,y) always be 0 , F(x,y,λ) will always be equal to f(x,y) .

F(x,y,λ)=f(x,y)+λϕ(x,y)ϕ(x,y)=0⇓F(x,y,λ)=f(x,y)

So F(x,y,λ) and f(x,y) have the same 3-d surface, if you thought a=F(x,y,λ) is a function in 3-d space. In fact the function F(x,y,λ) is a function in 4-d space.
What we do is going to find the extremum of the 4-d space function F(x,y,λ) . When F(x,y,λ) get its extremum, the conditions exactly fit the origin problem conditions, occasionally, luckily in fact.
Solving the simple 4-d problem solves the complex 3-d problem. “Why not solve the simple problem?”, Lagrange thought.

[MORE DETAILED EXPLANATION COMES SOON.]

reference

1 [youdaoDic]
特别的，我们学过,怎样找到直线与平面的交点,通过将参数方程带入平面方程。
And, well, we’ve learned in particular how to find where a line intersects a plane by plugging in the parametric equation into the equation of a plane.
2 拉格朗日乘数法_百度百科