求MATLAB大神给指出程序错误!function y = objfun(x)y = 1/x(1) + 2/x(2) + 4/x(3) + 6/x(4)function [c,ceq]=confun(x)c = [x(1)+x(2)+3*x(3)+3*x(4)-2;x(1)+2*x(2)+2*x(3)+x(4)-1.8];ceq = []x0 = [0,0,0,0]lb = [0.2,0.2,0.2,0.2]ub = []options = optimse

求MATLAB大神给指出程序错误!function y = objfun(x)y = 1/x(1) + 2/x(2) + 4/x(3) + 6/x(4)function [c,ceq]=confun(x)c = [x(1)+x(2)+3*x(3)+3*x(4)-2;x(1)+2*x(2)+2*x(3)+x(4)-1.8];ceq = []x0 = [0,0,0,0]lb = [0.2,0.2,0.2,0.2]ub = []options = optimse
求MATLAB大神给指出程序错误!
function y = objfun(x)
y = 1/x(1) + 2/x(2) + 4/x(3) + 6/x(4)
function [c,ceq]=confun(x)
c = [x(1)+x(2)+3*x(3)+3*x(4)-2;x(1)+2*x(2)+2*x(3)+x(4)-1.8];
ceq = []
x0 = [0,0,0,0]
lb = [0.2,0.2,0.2,0.2]
ub = []
options = optimset('largescale' ,'off');
[x,fval] = fmincon('objfun',x0,[],[],[],[],lb,ub,'confun',options)
[c,ceq] = confun(x)
为什么没法运行,出现如下错误
>> confun
Error using confun (line 2)
Not enough input arguments.

求MATLAB大神给指出程序错误!function y = objfun(x)y = 1/x(1) + 2/x(2) + 4/x(3) + 6/x(4)function [c,ceq]=confun(x)c = [x(1)+x(2)+3*x(3)+3*x(4)-2;x(1)+2*x(2)+2*x(3)+x(4)-1.8];ceq = []x0 = [0,0,0,0]lb = [0.2,0.2,0.2,0.2]ub = []options = optimse
function y = objfun(x)
y = 1/x(1) + 2/x(2) + 4/x(3) + 6/x(4)；
%这个程序独立保存为objfun.m
function [c,ceq]=confun(x)
c = [x(1)+x(2)+3*x(3)+3*x(4)-2;x(1)+2*x(2)+2*x(3)+x(4)-1.8];
ceq = [];
%这个程序独立保存为objfun.m;
clc
clear
x0 = [0,0,0,0]
lb = [0.2,0.2,0.2,0.2]
ub = []
options = optimset('largescale' , 'off','Algorithm','active-set');%Algorithm = 'active-set' 
[x,fval] = fmincon('objfun',x0,[],[],[],[],lb,ub,'confun',options)
[c,ceq] = confun(x)
%这个程序独立保存为myfmin.m;运行本主程序就得到下面结果

运行结果

x0 =

     0     0     0     0

lb =

    0.2000    0.2000    0.2000    0.2000

ub =

     []

Local minimum possible. Constraints satisfied.

fmincon stopped because the predicted change in the objective function

is less than the default value of the function tolerance and constraints 

were satisfied to within the default value of the constraint tolerance.

<stopping criteria details>

Active inequalities (to within options.TolCon = 1e-006):

  lower      upper     ineqlin   ineqnonlin

    1                                1

x =

    0.2000    0.2791    0.2279    0.2791

fval =

   51.2177

c =

         0

   -0.3070

ceq =

     []