solving a multi variables function equal to zero uzing fzero

i was trying to solve this code

format loose
format compact
format long
m=[ 4000 50 ] ; 
s= [400 5 ] ; 
r=[1 0.5 , 0.5 1 ];
q=[1 1 ] ;
ls=[];

n=length(m) ;
for i = 1 : n
    eval(sprintf('syms x%i,',i));
    eval(sprintf('x(%i) = x%i;', i, i));
  
end
Y= @(x1, x2) 29-(6*x(1))-(18*x(2));

  
  for i=1:n  
    if i<n
        xi=m(i);
    disp(x);
    disp(i);
    else
        last=m(n)
        xi = fzero(Y,last) 
   %=>>>>>>>>>>>>>> 5
    end   
  end   
  

i was trying to obtain the value of the last varible in the function Y=0 but i am getting this error :

Error using fzero (line 328)
Function value at starting guess must be finite and real. 

Answer 1

[John D'Errico]

As much as you want it to do so, fzero is not designed to solve multi-variable problems. Software cannot be made to do what it was not written to do.

You can instead use a tool like fsolve. Or solve. you could even make fminsearch do it for you. But fzero? Not really an option. (Yes, with a fairly great deal of effort, you might be able to cobble something up, using nested calls. Pray tell, why would you even want to do so?)

Use the correct tool for a problem. Would you use a hammer to cut a piece of wood in half? A saw makes more sense.

[sarra aloui]

thank you i am a beginneer in matlab can you please give me more details about the solution . but i want to explaine something : while my function i am looking just for the last variables all the other variables are already knowen

and thank you

[Walter Roberson]

http://www.mathworks.com/help/matlab/math/parameterizing-functions.html

[John D'Errico]

So then it is not a multi-variable problem? I'm trying to read your code, but it is highly confusing. You create symbolic variables for no reason at all.

Is all that you want to do is solve the function Y, with respect to x(2), while fixing x(1) to be m(1)? Forgetting that the problem is trivial, and that you could solve it using solve anyway, but I'll assume your real problem is more complicated.

Y = @(x1,x2) 29 - 6*x1 - 18*x2;

So y is a function of TWO variables, x1, and x2. Now, we wish to fix x1 to have some value, while solveing for the value of x2. Note there was no need to do anything silly, like create symbolic variables using eval. No symbolic variables will be created here.

Assume the desired value for x1 is m(1).

m=[ 4000 50 ] ;
x2 = fzero(@(x2) Y(m(1),x2),-1000)

x2 =
         -1331.72222222222

In fact, we know the solution would be:

(29 - 6*m(1))/18
ans =
         -1331.72222222222

As you can see, fzero found it too.