Minimum of a function

I am having a hard time figuring out how to call the file-functions. I want to make a nonlinear contraint , and then another file function to call the initial approximations x0(1) и х0(2), then call the the equtaion below

f = @(х) (х(1) - 2).^2 + (х(2) - 4)

Below are my limits:

c=[-x(1);

-x(2);

2*x(1).^2+x(2).^2-34;

2*x(1)+3*x(2).^2-18];

So far I got to here

function f = fun( x )

f = @(х) (х(1) - 2).^2 + (х(2) - 4)

end

function [ с,сequ ] = mycon( x )

с=[-x(1);

x(2);

-x(3);

2*x(1)+2+(x)+3*x(3).^2-34; %% I'm not sure if this is even written correctly.

сequ=[]; end

This speaks nothing to me and I've been trying every option under the moon to make it run but it doesn't work. How can I even write down the last limit as a MATLAB function? I need a slight boost, please. I read the fmincon documentation but this function is giving me a hard time.

Answer 1

[Thiago Henrique Gomes Lobato]

You had some typos in your constrain function and also you don't need to create a function file for your function since you already use an annonymous one. Use this and it should work:

f = @(x) (x(1) - 2).^2 + (x(2) - 4);

x0 = [1,1]; % Here you put your x0 function you said you had
nonlcon = @mycon;
x = fmincon(f,x0,[],[],[],[],[],[],nonlcon)

% Another file
function [ c,cequ ] = mycon( x ) 
 c=[-x(1);
-x(2); 
2*x(1).^2+x(2).^2-34;
2*x(1)+3*x(2).^2-18]; %% I'm not sure if this is even written correctly.
cequ=[];
end 


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.

<stopping criteria details>

x =

    2.0000    0.0000

[John D'Errico]

I would only add that the first two inequalities are just bound constraints on x(1) and x(2). so they could be put in directly as bound constraints, rather than needing a nonlinear inequality constraint. If there were 3 variables, then LB would be just

LB = [0,0,-inf];