solve function with two variables by ga toolbox

hello,

I have a function with two variables ang i want to optimize the function by using ga. I used the toolbox long time ago and i forgot how to use it. Can any one help me please.

my function is

y= 0.25-4.27*x(1)+0.61*x(2)+13.34*x(1)*x(2)-4.69*x(2).^2;

and the range of the variables are

lb = [0.001,0.01];
ub = [0.045,0.1];;

I'm trying to use

numberOfVariables = 2;
[x,fval] = ga(FitnessFunction,numberOfVariables,lb,ub)

Thanks

Answer 1

[Walter Roberson]

format long g
FitnessFunction = @(x) 0.25-4.27*x(1)+0.61*x(2)+13.34*x(1)*x(2)-4.69*x(2).^2;
lb = [0.001,0.01];
ub = [0.045,0.1];
numberOfVariables = 2;
A = []; b = [];
Aeq = []; beq = [];
[x,fval] = ga(FitnessFunction, numberOfVariables, A, b, Aeq, beq, lb, ub)

Optimization terminated: average change in the fitness value less than options.FunctionTolerance.

x = 1×2

0.045 0.01

fval = 
                  0.069484

%cross-check
[xf, fvalf] = fmincon(FitnessFunction, randn(numberOfVariables, 1), A, b, Aeq, beq, 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.

xf = 2×1

0.0449999033023298 0.0100003582625156

fvalf = 
        0.0694847999986172

[nadia nadi]

Thank you very much

Answer 2

[Sulaymon Eshkabilov]

Here is how it can be solved with GA:

lb = [0.001, 0.01];
ub = [0.045, 0.1];
numberOfVariables = 2;
y=@(x) 0.25-4.27*x(1)+0.61*x(2)+13.34*x(1)*x(2)-4.69*x(2).^2;
options = optimoptions('ga','PlotFcn', @gaplotbestf);
[x,fval] = ga(y,numberOfVariables,[],[],[],[],lb,ub, [], options)

Optimization terminated: average change in the fitness value less than options.FunctionTolerance.

x = 1×2

0.0450 0.0100

fval = 0.0695

[nadia nadi]

Thanks it is nice of you to send me this code.