引用matlab linprog的例子
  http://www.mathworks.com/help/toolbox/optim/ug/linprog.html
Find x that minimizes
  f(x) = –5*x1 – 4*x2 –6*x3,
subject to
  x1 – x2 + x3 ≤ 20
  3x1 + 2x2 + 4x3 ≤ 42
  3x1 + 2x2 ≤ 30
  0 ≤ x1, 0 ≤ x2, 0 ≤ x3.

%%%%%matlab linprog code
f = [-5; -4; -6];
A =  [1 -1  1
      3  2  4
      3  2  0];
b = [20; 42; 30];
lb = zeros(3,1);

[x,fval,exitflag,output,lambda] = linprog(f,A,b,[],[],lb);

%%%%cvx toolbox
cvx_begin
  variables x1 x2 x3;
  minimize(-5*x1 - 4*x2 -6*x3 );
  subject to
   x1- x2 + x3 <= 20
   3*x1 + 2*x2 + 4*x3 <= 42
   3*x1 + 2*x2 <= 30
   0 <= x1; 0 <= x2; 0 <= x3;
cvx_end

Calling sedumi: 6 variables, 3 equality constraints
   For improved efficiency, sedumi is solving the dual problem.
------------------------------------------------------------
SeDuMi 1.21 by AdvOL, 2005-2008 and Jos F. Sturm, 1998-2003.
Alg = 2: xz-corrector, Adaptive Step-Differentiation, theta = 0.250, beta = 0.500
eqs m = 3, order n = 7, dim = 7, blocks = 1
nnz(A) = 11 + 0, nnz(ADA) = 9, nnz(L) = 6
it :     b*y       gap    delta  rate   t/tP*  t/tD*   feas cg cg  prec
  0 :            4.35E+002 0.000
  1 :  5.10E+001 1.44E+002 0.000 0.3314 0.9000 0.9000   1.84  1  1  3.0E+000
  2 :  6.89E+001 3.89E+001 0.000 0.2695 0.9000 0.9000   2.50  1  1  4.9E-001
  3 :  7.62E+001 9.96E+000 0.000 0.2563 0.9000 0.9000   0.99  1  1  1.4E-001
  4 :  7.80E+001 4.04E-001 0.000 0.0405 0.9900 0.9900   1.03  1  1  
iter seconds digits       c*x               b*y
  4      0.1  15.7  7.8000000000e+001  7.8000000000e+001
|Ax-b| =  0.0e+000, [Ay-c]_+ =  5.3E-015, |x|= 1.9e+000, |y|= 1.5e+001

Detailed timing (sec)
   Pre          IPM          Post
1.092E-001    1.404E-001    3.120E-002   
Max-norms: ||b||=6, ||c|| = 42,
Cholesky |add|=0, |skip| = 0, ||L.L|| = 1.82429.
------------------------------------------------------------
Status: Solved
Optimal value (cvx_optval): -78

谢谢你epoh，因为是唯一的答案，当然是最佳答案啦。这个问题我后来问了助教，才发现这段code在她的matlab里是可以运行出结果的，只是问题过于简单直接解方程即可。。。可能是我的matlab版本问题吧