simple fmincon problem, optimization toolbox

I have an optimization problem in that Matlab solves depending on the initial value given. First, I give X0 = [10 10 10] as the initial value, and ok. But then I give X0 = [0 0 0] as the initial value, which is not a minimum, Matlab gives it as the minimum. Why?

FUN = @(x)-prod(x);

X0 = [10 10 10];

A = [-1 -2 -3;
      1  2  3];
  
B = [0 ; 72];

options = optimset('Display','iter');

X = fmincon(FUN, X0, A, B, [], [], [], [], [], options)

                                            First-order      Norm of
 Iter F-count            f(x)  Feasibility   optimality         step
    0       4   -1.000000e+03    0.000e+00    9.622e+01
    1       8   -2.137831e+03    0.000e+00    1.817e+02    7.782e+00
    2      12   -2.198098e+03    0.000e+00    8.705e+01    3.238e+00
    3      17   -2.281344e+03    0.000e+00    2.825e+01    6.225e+00
    4      21   -2.303595e+03    0.000e+00    2.554e+00    1.755e+00
    5      25   -2.303998e+03    0.000e+00    1.707e-01    2.181e-01
    6      29   -2.304000e+03    0.000e+00    1.713e-02    2.213e-02
    7      33   -2.304000e+03    0.000e+00    1.381e-03    1.352e-03
    8      37   -2.304000e+03    0.000e+00    7.437e-06    8.647e-05

Local minimum found that satisfies the constraints.

Optimization completed because the objective function is non-decreasing in 
feasible directions, to within the value of the optimality tolerance,
and constraints are satisfied to within the value of the constraint tolerance.

X = 1×3

24.0000 12.0000 8.0000

FUN = @(x)-prod(x);

X0 = [0 0 0];

A = [-1 -2 -3;
      1  2  3];
  
B = [0 ; 72];

options = optimset('Display','iter');

X = fmincon(FUN, X0, A, B, [], [], [], [], [], options)

                                            First-order      Norm of
 Iter F-count            f(x)  Feasibility   optimality         step
    0       4   -0.000000e+00    0.000e+00    0.000e+00

Initial point is a local minimum that satisfies the constraints.

Optimization completed because at the initial point, the objective function is non-decreasing 
in feasible directions to within the value of the optimality tolerance, and 
constraints are satisfied to within the value of the constraint tolerance.

X = 1×3

0 0 0

Answer 1

[Matt J]

The point X=[0,0,0] is a first order local minimum. It is a point of zero-gradient, and it satisfies the constraints. You made an unlucky initial guess.

Incidentally, if you are looking for solutions with positive x(i) it would probably be better, numerically, to use the equivalent objective function

FUN=@(x)-sum(log(x))

as well as to impose lower bounds lb=[0;0;0]