属于有约束的非线性规划问题
首先编写目标函数（取原函数的相反数）：
function f=myfun(x)
f=-(2-0.9*x(1))*(200-10*x(2))*0.4*(1-(3-x(1))/(0.9*(x(2)-x(1))))...
    +0.9*x(1)*(200-10*x(2))*0.25*0.4^2*(1-(3-x(1))/(0.9*(x(2)-x(1))))^2;
再编写非线性约束函数：
function [c,ceq]=nl(x)
c=[];
ceq=(200-20*x(2)+10*x(1))*0.4^2*0.25*(0.9*(x(2)-x(1))-(3-x(1)))+0.2*0.4*(3-x(1));
最后写出线性约束：
A=[-1 -9];
b=-30;
写出上下界：
lb=[0;3];
ub=[20/9;20];
设定一个初始值：
x0=[2 4];
最后优化：
[x,fval,exitflag]=fmincon(@myfun,x0,A,b,[],[],lb,ub,@nl)

很感谢你的帮助……可是按照你的程序运行时，出现以下错误：
>> A=[-1 -9];
b=-30;
lb=[0;3];
ub=[20/9;20];
x0=[2 4];
[x,fval,exitflag]=fmincon(@myfun,x0,A,b,[],[],lb,ub,@nl)
Warning: Large-scale (trust region) method does not currently solve this type of problem,
switching to medium-scale (line search).
> In fmincon at 260
??? Error using ==> fmincon
FMINCON cannot continue because user supplied objective function failed with the following error:
Error using ==> feval
Undefined command/function 'myfun'.
不知道问题出在哪里……麻烦你了

myfun nl 两个函数分别存在两个m文件中，我运行没问题，祝你好运   

你的matlab是哪个版本？还有，你算的结果是多少？非常感谢你哦

2012a最新版本的，我运行也有点问题，不知道是不是你的约束条件有问题，你再请教高人吧
Warning: The default trust-region-reflective algorithm does not solve problems with the
constraints you have specified. FMINCON will use the active-set algorithm instead. For
information on applicable algorithms, see Choosing the Algorithm in the documentation.
> In fmincon at 486

No feasible solution found.

fmincon stopped because the size of the current search direction is less than
twice the default value of the step size tolerance but constraints are not
satisfied to within the default value of the constraint tolerance.

<stopping criteria details>


x =

    2.2328    3.0746


fval =

   -0.0059


exitflag =

    -2