请帮我看一段matlab程序

问题：
Min  10X2+10X3+40
Sub. to  1≤X1≤4
1≤X2≤4
1≤X3≤4
1≤X4≤4
-(6.9402131+0.252469*x(1)-0.276517*x(2)+0.207394*x(3)+0.314560*x(4)+0.039335*x(1)*x(1)+0.039335*x(2)*x(2)-0.003563*x(3)*x(3)-0.006647*x(4)*x(4)-0.006296*x(1)*x(2)-0.034093*x(1)*x(3)-0.059668*x(1)*x(4)+0.010462*x(2)*x(3)+0.005134*x(2)*x(4)-0.008807*x(3)*x(4))+8.62≤0
6.9402131+0.252469*x(1)-0.276517*x(2)+0.207394*x(3)+0.314560*x(4)+0.039335*x(1)*x(1)+0.039335*x(2)*x(2)-0.003563*x(3)*x(3)-0.006647*x(4)*x(4)-0.006296*x(1)*x(2)-0.034093*x(1)*x(3)-0.059668*x(1)*x(4)+0.010462*x(2)*x(3)+0.005134*x(2)*x(4)-0.008807*x(3)*x(4)-9.17≤0
0.6730551+0.301146*x(1)-0.078675*x(2)+0.011658*x(3)+0.178452*x(4)+0.009664*x(1)*x(1)+0.009664*x(2)*x(2)+0.001284*x(3)*x(3)-0.002713*x(4)*x(4)-0.012422*x(1)*x(2)-0.023157*x(1)*x(3)-0.036011*x(1)*x(4)+0.024744*x(2)*x(3)-0.020623*x(2)*x(4)-0.002149*x(3)*x(4)-1.5≤0
-(0.6730551+0.301146*x(1)-0.078675*x(2)+0.011658*x(3)+0.178452*x(4)+0.009664*x(1)*x(1)+0.009664*x(2)*x(2)+0.001284*x(3)*x(3)-0.002713*x(4)*x(4)-0.012422*x(1)*x(2)-0.023157*x(1)*x(3)-0.036011*x(1)*x(4)+0.024744*x(2)*x(3)-0.020623*x(2)*x(4)-0.002149*x(3)*x(4)) ≤0
-(51.8277931+6.142814*x(1)+1.235285*x(2)-1.799889*x(3)+4.118358*x(4)-0.434893*x(1)*x(1)-0.434893*x(2)*x(2)+0.033447*x(3)*x(3)+0.076468*x(4)*x(4)-0.201259*x(1)*x(2)-0.652022*x(1)*x(3)+0.820377*x(1)*x(4)+1.839250*x(2)*x(3)-1.451273*x(2)*x(4)-0.023900*x(3)*x(4))+95≤0
51.8277931+6.142814*x(1)+1.235285*x(2)-1.799889*x(3)+4.118358*x(4)-0.434893*x(1)*x(1)-0.434893*x(2)*x(2)+0.033447*x(3)*x(3)+0.076468*x(4)*x(4)-0.201259*x(1)*x(2)-0.652022*x(1)*x(3)+0.820377*x(1)*x(4)+1.839250*x(2)*x(3)-1.451273*x(2)*x(4)-0.023900*x(3)*x(4)-100≤0

我的程序如下：
m文件
function [c,ceq]=mycon(x)
c=[-(6.9402131+0.252469*x(1)-0.276517*x(2)+0.207394*x(3)+0.314560*x(4)+0.039335*x(1)*x(1)+0.039335*x(2)*x(2)-0.003563*x(3)*x(3)-0.006647*x(4)*x(4)-0.006296*x(1)*x(2)-0.034093*x(1)*x(3)-0.059668*x(1)*x(4)+0.010462*x(2)*x(3)+0.005134*x(2)*x(4)-0.008807*x(3)*x(4))+8.62; 
    6.9402131+0.252469*x(1)-0.276517*x(2)+0.207394*x(3)+0.314560*x(4)+0.039335*x(1)*x(1)+0.039335*x(2)*x(2)-0.003563*x(3)*x(3)-0.006647*x(4)*x(4)-0.006296*x(1)*x(2)-0.034093*x(1)*x(3)-0.059668*x(1)*x(4)+0.010462*x(2)*x(3)+0.005134*x(2)*x(4)-0.008807*x(3)*x(4)-9.17;
    0.6730551+0.301146*x(1)-0.078675*x(2)+0.011658*x(3)+0.178452*x(4)+0.009664*x(1)*x(1)+0.009664*x(2)*x(2)+0.001284*x(3)*x(3)-0.002713*x(4)*x(4)-0.012422*x(1)*x(2)-0.023157*x(1)*x(3)-0.036011*x(1)*x(4)+0.024744*x(2)*x(3)-0.020623*x(2)*x(4)-0.002149*x(3)*x(4)-1.5; 
    -(0.6730551+0.301146*x(1)-0.078675*x(2)+0.011658*x(3)+0.178452*x(4)+0.009664*x(1)*x(1)+0.009664*x(2)*x(2)+0.001284*x(3)*x(3)-0.002713*x(4)*x(4)-0.012422*x(1)*x(2)-0.023157*x(1)*x(3)-0.036011*x(1)*x(4)+0.024744*x(2)*x(3)-0.020623*x(2)*x(4)-0.002149*x(3)*x(4)); 
    -(51.8277931+6.142814*x(1)+1.235285*x(2)-1.799889*x(3)+4.118358*x(4)-0.434893*x(1)*x(1)-0.434893*x(2)*x(2)+0.033447*x(3)*x(3)+0.076468*x(4)*x(4)-0.201259*x(1)*x(2)-0.652022*x(1)*x(3)+0.820377*x(1)*x(4)+1.839250*x(2)*x(3)-1.451273*x(2)*x(4)-0.023900*x(3)*x(4))+95; 
    51.8277931+6.142814*x(1)+1.235285*x(2)-1.799889*x(3)+4.118358*x(4)-0.434893*x(1)*x(1)-0.434893*x(2)*x(2)+0.033447*x(3)*x(3)+0.076468*x(4)*x(4)-0.201259*x(1)*x(2)-0.652022*x(1)*x(3)+0.820377*x(1)*x(4)+1.839250*x(2)*x(3)-1.451273*x(2)*x(4)-0.023900*x(3)*x(4)-100];
ceq=[];

主程序：
>> fun='x(1)*0+x(2)*10+10+x(3)*10+30+x(4)*0';
>> x0=[1 1 1 1];
>> A=[];
>> b=[];
>> Aeq=[];
>> beq=[];
>> lb=[1 1 1 1];
>> ub=[4 4 4 4];
>> [x,fval,exitflag,output,lambda,grad,hessian]=fmincon(fun,x0,A,b,Aeq,beq,lb,ub,@mycon)


运行结果：
[b]Warning: Large-scale (trust region) method does not currently solve this type of problem,
switching to medium-scale (line search).
> In fmincon at 260
Maximum number of function evaluations exceeded;
increase OPTIONS.MaxFunEvals.[/b]

x =

    4.2343    3.1404    4.3459    4.3459


fval =

  114.8638


exitflag =

     0


output = 

       iterations: 66
        funcCount: 401
         stepsize: 1
        algorithm: 'medium-scale: SQP, Quasi-Newton, line-search'
    firstorderopt: 10.0000
     cgiterations: []
          message: [1x79 char]


lambda = 

         lower: [4x1 double]
         upper: [4x1 double]
         eqlin: [0x1 double]
      eqnonlin: [0x1 double]
       ineqlin: [0x1 double]
    ineqnonlin: [6x1 double]


grad =

         0
   10.0000
   10.0000
         0


hessian =

    0.5552    0.1233    0.0784   -0.4841
    0.1233    0.0670    0.0061   -0.1023
    0.0784    0.0061    0.0198   -0.0586
   -0.4841   -0.1023   -0.0586    0.4462

请问是什么原因出错了？谢谢 

X1、X2、X3、X4的范围明明是
1≤X1≤4
1≤X2≤4
1≤X3≤4
1≤X4≤4
但是它给出的值为
x =    4.2343    3.1404    4.3459    4.3459
X1、X3、X4值均超过4了，怎么会事？