非线性规划问题（数学建模 lingo）,求指教

比如说x

min=((x2-x1^2)^2+(1-x1)^2+90*(x4-x3^2)^2+(1-x3)^2+10.1*((1-x2)^2+(1-x4)^2))+19.8*(x2-1)*(x4-1);
@bnd(-10,x1,10);
@bnd(-10,x2,10);
@bnd(-10,x3,10);
@bnd(-10,x4,10);
x1*x2-x1-x2+1.5=0;
x1*x2*x3*x4

sets:
r/1..5/:a;
endsets
n=@size(r);
b=@sum(r:a)/5;
@sum(r:(a-b)^2)/n

This topic research nonlinear programming optimal solution algorithm new method. Nonlinear programming in the military, economy, management, production process automation, engineering design and product optimization design all have the important applications. But nonlinear programming research is still not mature, many questions need to be further perfected. Nonlinear programming not as linear programming have unified algorithm for different questions need, with different algorithm processing, present different algorithm has certain limitation, only for various algorithm to correct it, can effectively solve people in daily production, life throws optimization problem, make the best decisions.
This paper is mainly to the existing algorithm, looking for a test by path dependence, convergence faster optimal algorithms. First introduced the nonlinear programming research background and research status, then discusses the selection process, key scheme of experimental process, main is traced to various nonlinear optimal calculation method using matlab software programming in engineering, and presents an optimal function of typical examples, with a lot of testing, and gives the analysis of the results. Finally gives the whole experiment summary and the outlook for the future.
Keywords: nonlinear programming; The optimal decision; Path dependence

要限制k+1在范围内
Min=@sum(Time(i,j):t(i,j)*@sum(order(k)|k#lt#@size(order):a(i,k)*a(j,k+1)));

令x^2=u>0,则
目标函数为u+y+4z求最小值
约束条件如下:
4u+y +z>=5
y-z=0,y>=2,z>=0

方便。
虽然动态规划主要用于求解以时间划分阶段的动态过程的优化问题，但是一些与时
间无关的静态规划（如线性规划、非线性规划），只要人为地引进时间因

主要问题：
1、关于Trust-region-reflective算法的警告
该类方法不适用于存在非线性约束的情况,所以用警告的方式提示你使用别的方法.如果不介意看到警告信息,直接忽略无妨；如果不想看到警告,主程序可以这样调用：
opt=optimset('Algorithm','active-set',);
x0=[2.5 21 7];
[x,FVAL] = fmincon(@myfun,x0,[4 -1 0;-18 1 0],[0;0],[],[],[2.2;16;3],[3.5;28;Inf],@mycon,opt)
2、程序现在最大的问题是,你给出的初始解不在可行域内,也就是说,不满足你的所有约束条件（很容易验证,不满足非线性不等式约束c(x)