I have differnt coodinate after use fmincon

i want to get a coordinate for maximum distance which form is distance =10.361+0.40125.*a-0.34875.*c-0.78375.*a.*c i plot this surface, and i have (1,-1) is the best coordinate and the maximum is 11.8947 but when i do this for fmincon, x0=[0 0]; lowerbound=[-1 -1]; upperbound =[1 1]; function f = distance(x) A=x(1); C=x(2); f= (-1).*(10.361+0.40125.*A-0.34875.*C-0.78375.*A.*C); end function [c ceq]=const(x) A=x(1); C=x(2); c=A+C-2; ceq=[]; end the answer is x=1 y=1 , not a x=1 y=-1!!!!!! plz help me

[Torsten]

upperbound = lowerbound? Must be wrong.

[jong eun kim]

the range of A and B is (-1<=A<=1) (-1<=B<=1)

[Torsten]

And you set

-1 <= A <= -1 , -1 <= C <= -1.

[jong eun kim]

nono i mistake the question i set lower=[-1 -1] upper=[1 1]

[Torsten]

Both Matt J and I already changed the settings.

And the result is as you expect: A = 1 and C = -1.

Answer 1

[Matt J]

x0=[0 0]; lb=[-1 -1]; ub =-lb;

fmincon(@distance,x0,[1,1],2,[],[],lb,ub)

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.

ans = 1×2

1.0000 -1.0000

function f = distance(x)
 A=x(1); C=x(2); f= (-1).*(10.361+0.40125.*A-0.34875.*C-0.78375.*A.*C);
end

[jong eun kim]

my question was mistake plz check my changed question!!

[Matt J]

I don't see any changes in your question other than those that have already been corrected in the code I've presented.

[jong eun kim]

i reupload my question by more specificly. i think u dont understand my question. but thank you for your help

[jong eun kim]

plz check my new question sir

Answer 2

[Torsten]

x0=[0 0]; 
lowerbound =[-1 -1]; 
upperbound = [1 1]; 
sol = fmincon(@distance,x0,[1 1],2,[],[],lowerbound,upperbound)

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.

sol = 1×2

1.0000 -1.0000

function f = distance(x); 
  A=x(1); 
  C=x(2); 
  f= (-1).*(10.361+0.40125.*A-0.34875.*C-0.78375.*A.*C);
end