How to correct the code?

I have to maximize the objective function

b*Eby-s*Esl+gb*(Esl-Eby)

subjected to constraints

gb<=s;b<=gs; % where gb=3;gs=8;

By solving this equation using KKT condition,

L(b,s,lambda1,lambda2)=b*Eby-s*Esl+gb*(Esl-Eby)+lambda1(gb-s)+lambda2(b-gs)

solution obtained is,

case1 : lambda1=0;lambda2=0

s*=3.302;b*=6.928

case1 : lambda1=0;lambda2>0

s*=3.302;b*=8

case1 : lambda1>0;lambda2=0

s*=3;b*=6.928

case1 : lambda1>0;lambda2>0

s*=3;b*=8

I solving this function using fmincon ,but didn't got the correct solution.

gb=3;gs=8;
ES=4.5;is=20;EB=1.5;ib=40;
b=8;s=3;Ns=1;Nb=1;
Esl=(ES)-(is/s)+Ns;
Eby=(ib/b)-(EB)-Nb;
f = @(s,b)(b*Eby-s*Esl+gb*(Esl-Eby)); 
[sb,fv] = fmincon(@(sb) -f(sb(1),sb(2)), rand(2,1), [],[],[],[],[3 -Inf],[inf 8])

Problem appears unbounded.

fmincon stopped because the objective function value is less than
the value of the objective function limit and constraints 
are satisfied to within the value of the constraint tolerance.

sb = 2×1

1.0e+20 * 3.2933 -0.0000

fv = -3.8422e+20

b=sb(2)

b = -1.4635e+13

s=sb(1)

s = 3.2933e+20

Is this code is correct? Is it possible to get a solution between 3 and 8.That is case 1 have to be excecute.How to solve this problem?Is there any other method? Please help me to correct this code.

NB:values of ES,EB,is,ib are changable variables.

Answer 1

[Torsten]

gb=3;gs=8;
ES=4.5;is=20;EB=1.5;ib=40;
b=8;s=3;Ns=1;Nb=1;
Esl=@(s)(ES)-(is/s)+Ns

Esl = function_handle with value:
    @(s)(ES)-(is/s)+Ns

Eby=@(b)(ib/b)-(EB)-Nb

Eby = function_handle with value:
    @(b)(ib/b)-(EB)-Nb

f = @(s,b)(b*Eby(b)-s*Esl(s)+gb*(Esl(s)-Eby(b))); 
[sb,fv] = fmincon(@(sb) -f(sb(1),sb(2)), [5 5], [],[],[],[],[3 -Inf],[inf 8])

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.

sb = 1×2

3.3029 6.9282

fv = -13.0272