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myl5865115

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[求助] 函数最值问题!!!!

小虫,刚接触MATLAB,需求一函数的最值,但约束条件有些复杂,顿时凌乱,望高手指导!
以下是我的文件内容:
function f = fun(x);
x0=[1 1 1 1 10*pi/10 10*pi/180 10*pi/180 4*pi/180 8*pi/180 80*pi/180 175*pi/180];
T=1300;
M=11500;
R=10;
U=0.16;
fun=@(x)1/1.6900e-007*abs(abs(((T/R)*(x(1)*tan(x(5))-x(2)*tan(x(6))+x(3)*tan(x(7))*cos(x(9))-x(4)*tan(x(8)))-(2*M/R^2)*x(3)*sin(x(9)))*(sin(x(11))...
-U*cos(x(11)))+((2*M/R^2)*(x(1)-x(2)+cos(x(9))-x(4))+(T/R)*x(3)*tan(x(7))*sin(x(9)))*(cos(x(10))+U*sin(x(11)))./(1.0256*sin(x(11)-x(10))))...
-abs(((2*M/R^2)*(x(1)-x(2)+cos(x(9))-x(4))+(T/R)*x(3)*tan(x(7))*sin(x(9)))))+1/2.0331e-006*(abs(abs(((T/R)*(x(1)*tan(x(5))-x(2)*tan(x(6))+x(3)*tan(x(7))*cos(x(9))-x(4)*tan(x(8)))-(2*M/R^2)*x(3)*sin(x(9)))*(sin(x(10))...
-U*cos(x(10)))+((2*M/R^2)*(x(1)-x(2)+cos(x(9))-x(4))+(T/R)*x(3)*tan(x(7))*sin(x(9)))*(cos(x(10))+U*sin(x(10)))./((1+U^2)*sin(x(10)-x(11))))...
-abs((T/R)*(x(1)*x(5)-x(2)*x(6)+x(3)*tan(x(7))*cos(x(9))-x(4)*tan(x(8)))-(2*M/R^2)*x(3)*sin(x(9)))))


A=[];
B=[];
Aeq=[1 1 1 1 0 0 0 0 0 0 0];
Beq=[10];
lb=[1,1,1,1,5*pi/180,8*pi/180,8*pi/180,10*pi/180,8*pi/180,85*pi/180,185*pi/180];
ub=[5,5,5,5,16*pi/180,14*pi/180,14*pi/180,14*pi/180,12*pi/180,95*pi/180,200*pi/180];
[x,fval]= fmincon(fun,x0,A,B,Aeq,Beq,lb,ub)
c=(T/R)*(x(1)*tan(x(5))-x(2)*tan(x(6))+x(3)*tan(x(7))*cos(x(9))-x(4)*tan(x(8)))-(2*M/R^2)*x(3)*sin(x(9))
d=(2*M/R^2)*(x(1)-x(2)+cos(x(9))-x(4))+(T/R)*x(3)*tan(x(7))*sin(x(9))
p=((T/R)*(x(1)*tan(x(5))-x(2)*tan(x(6))+x(3)*tan(x(7))*cos(x(9))-x(4)*tan(x(8)))-(2*M/R^2)*x(3)*sin(x(9)))*(sin(x(11))...
-U*cos(x(11)))+((2*M/R^2)*(x(1)-x(2)+cos(x(9))-x(4))+(T/R)*x(3)*tan(x(7))*sin(x(9)))*(cos(x(10))+U*sin(x(11)))./((1+U^2)*sin(x(11)-x(10)))
w=((T/R)*(x(1)*tan(x(5))-x(2)*tan(x(6))+x(3)*tan(x(7))*cos(x(9))-x(4)*tan(x(8)))-(2*M/R^2)*x(3)*sin(x(9)))*(sin(x(10))-U*cos(x(10)))+((2*M/R^2)*(x(1)-x(2)+cos(x(9))-x(4))+(T/R)*x(3)*tan(x(7))*sin(x(9)))*(cos(x(10))+U*sin(x(10)))./((1+U^2)*sin(x(10)-x(11)))
其中要使c,d大于0;p,w小于0,不知道如何添加约束,还望高手指导》》》
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