已知目标函数和约束条件,用MATLAB怎么求最大值
发布网友
发布时间：2022-04-24 04:29
共2个回答
热心网友
时间：2023-10-28 03:59
clc;clear;
syms x y z t%定义自变量x,y,z，拉格朗日乘数t
f(x,y,z)=x+2*y+3*z;%设需要求最大值的表达式x+2*y+3*z
g=x^2+y^2+z^2-4;%设约束条件x^2+y^2+z^2-4=0
L=f-t*g;
sln=solve(diff(L,x)==0,diff(L,y)==0,diff(L,z)==0,g==0);%解拉格朗日数乘法的方程组
eval(f(sln.x,sln.y,sln.z))%把解带回f，求出条件极值
运行结果如下：
ans =
7.4833
-7.4833
即得到x+2*y+3*z在x^2+y^2+z^2-4=0条件下的最大值7.4833，最小值-7.4833。
热心网友
时间：2023-10-28 03:59
clc;clear;
syms x y z t%定义自变量x,y,z，拉格朗日乘数t
f(x,y,z)=x+2*y+3*z;%设需要求最大值的表达式x+2*y+3*z
g=x^2+y^2+z^2-4;%设约束条件x^2+y^2+z^2-4=0
L=f-t*g;
sln=solve(diff(L,x)==0,diff(L,y)==0,diff(L,z)==0,g==0);%解拉格朗日数乘法的方程组
eval(f(sln.x,sln.y,sln.z))%把解带回f，求出条件极值
运行结果如下：
ans =
7.4833
-7.4833
即得到x+2*y+3*z在x^2+y^2+z^2-4=0条件下的最大值7.4833，最小值-7.4833。
热心网友
时间：2023-10-28 04:00
>> a = [-1 0 0;0 -1 0;0 0 -1;3 6 12]; 约束条件
>> b = [0 0 0 20];
>> x0 = [1 1 1]; % 迭代初始值
>> [x,val] = fmincon(f,x0,a,b,[],[],[],[],[],optimset('Algorithm','interior-point'))
Local minimum found that satisfies the constraints.
Optimization completed because the objective function is non-decreasing in
feasible directions, to within the default value of the function tolerance,
and constraints were satisfied to within the default value of the constraint tolerance.
<stopping criteria details>
x =
0.0000 3.3333 0.0000
val =
-11.1231
即x1=0，x2=3.333，x3=0时，原目标函数取得最大值11.1231
热心网友
时间：2023-10-28 04:00
>> a = [-1 0 0;0 -1 0;0 0 -1;3 6 12]; 约束条件
>> b = [0 0 0 20];
>> x0 = [1 1 1]; % 迭代初始值
>> [x,val] = fmincon(f,x0,a,b,[],[],[],[],[],optimset('Algorithm','interior-point'))
Local minimum found that satisfies the constraints.
Optimization completed because the objective function is non-decreasing in
feasible directions, to within the default value of the function tolerance,
and constraints were satisfied to within the default value of the constraint tolerance.
<stopping criteria details>
x =
0.0000 3.3333 0.0000
val =
-11.1231
即x1=0，x2=3.333，x3=0时，原目标函数取得最大值11.1231
