The specific method is as follows:
Let a given binary function z=? (x, y) and the additional condition φ(x, y)=0, in order to find z=? For the extreme point of (x, y) under additional conditions, the Lagrange function L(x, y)=? (x, y)+λφ(x, y), where λ is the parameter.
Find the first partial derivative of L(x, y) to x, y to make it equal to zero, and combine with additional conditions, namely
L'x(x，y)=？ x(x，y)+λφ'x(x，y)=0
L'y(x，y)=？ y(x，y)+λφ'y(x，y)=0
φ(x，y)=0
Applied to microeconomics: Set the utility function U(Qx, Qy), and in order to obtain the extreme value under the constraint conditions, first establish the Lagrange function: L=U(Qx, Qy)+λ( I-Px? Qx-Py？ Qy), λ is the parameter.
Find the first partial derivative of L(x, y) relative to x, y to make it equal to zero, and the additional conditions are established.
that is
L/？ Qx=？ U/？ Qx-λPx=0 ( 1)
L/？ Qy=？ U/？ Qy-λPy=0 (2)
I-Px？ Qx-Py？ Qy=0 (3)
Divide equation (1) by equation (2) to obtain:
U/？ Qx =Px means MUx = MUy.
U/？ Qy =Py
Therefore, if consumers want to maximize the utility of two commodities, the ratio of marginal utility should be equal to the ratio of price.
The above is about X and Y goods. Does it also apply to many commodities? The answer is yes. If consumers choose among n kinds of goods, the principle of consumer equilibrium can be expressed as:
MU 1=MU2 =MU3 = …=MUn
P 1= P2= P3=...= Pn
This conclusion can also be proved by Lagrange multiplier method.
Can Lagrange multiplier method be extended to find n-variable function? Conditional extreme value of (x 1, x2, …, xn) under m additional conditions φ(x 1, x2, …, xn).
The method is as follows:
(1) Lagrange function L(x 1, x2, …, xn)=? (x 1，x2，…，xn)+ ∑λiφi(x 1，…x2)；
(2) Find the partial derivative of L(x 1, …xn) about x 1, …xn, make it equal to zero, and make it stand with additional conditions, namely
L'xi==？ xi+ ∑λiφ'i=0，i= 1，2，…，n
φk(x 1，x2，…，xn)=0，k= 1，2，…，n
By solving this set of equations, the extreme point can be obtained.
Back to our question, let's set the utility function u (qx 1, qx2, ... qxn). In order to obtain the extreme value under the constraint conditions, we first establish the Lagrange function:
L=U(Qx 1，Qx2，…Qxn )+λ(I-Px 1？ Qx 1-P2？ Qy2-…-Pxn？ Qxn), λ is the parameter. Find the first partial derivative of L(x 1, x2, …xn) to x 1, …, xn, make them equal to zero, and establish with additional conditions.
that is
L/？ Qx 1=？ U/？ Qx 1-λPx 1=0 ( 1)
L/？ Qx2=？ U/？ Qx2-λPx2=0 (2)
…… …
L/？ Qxn=？ U/？ Qxn-λPxn=0 (n)
I-Px 1？ Qx 1-P2？ Qy2-…-Pxn？ Qxn
The equation (1) is divided by (n),
MUx 1 = MUx2 =…=MUxn
Px 1 =Px2 =...=Pn
Therefore, if consumers want to maximize the utility of N commodities, the ratio of marginal utility should be equal to the ratio of price.
Extended data:
This method transforms an optimization problem with n variables and k constraints into an extreme value problem of a system of equations with n+k variables, and its variables are not subject to any constraints.
This method introduces a new scalar unknown, that is, Lagrange multiplier: the coefficient of each vector in the linear combination of gradient of constraint equation.
Let a given binary function z=? (x, y) and the additional condition φ(x, y)=0, in order to find z=? For the extreme point of (x, y) under additional conditions, do Lagrange function first.
f(x，y，λ)=f(x，y)+λφ(x，y)
Where λ is a parameter.
Let the first partial derivative of F(x, y, λ) to x and y, λ be equal to zero, that is
F'x=？ x(x，y)+λφ'x(x，y)=0？ [ 1]?
F'y=？ y(x，y)+λφ'y(x，y)=0
F'λ=φ(x，y)=0
Solve x, y, λ from the above equation, and the (x, y) thus obtained is the function z=? Possible extreme points of (x, y) under the additional condition φ(x, y)=0.
If there is only one such point, it can be directly determined from the actual problem.
The conditional extreme point of a function under constraint conditions should be the solution of a set of equations.
The so-called Lagrangian function (real number is called Lagrangian multiplier) is introduced, and the above equations are equations.
Therefore, there are usually three methods to solve conditional extremum:
1) direct method is to solve the problem from the system of equations (1) and express it as a function with variables, and transform the problem into an unconditional extreme value problem of the function;
2) In general, the equations (1) are difficult or even impossible to solve, so the above solving methods are often not feasible.
The Lagrange multiplier method, which is usually adopted, avoids the difficulty of solving the equation (1), transforms the conditional extreme value problem into the stable point problem of the Lagrange function below, and then determines which stable points are extreme values according to the characteristics of the practical problems discussed.
3) Under given conditions, if the unknown quantity can be replaced or solved, the conditional extreme value can be transformed into unconditional extreme value, thus avoiding the trouble of introducing Lagrange multiplier.
Note: The points with φ (x, y, z)=0 and φ(x, y, z) = 0 will not be calculated in this way, therefore, when seeking the maximum or minimum value, these points should be listed and calculated separately.
References:
Baidu encyclopedia-Lagrange multiplier method