solving a function equal to zero

i am trying to run this code to obtain the last value to the function y=0

format loose
format compact
format long
m=[ 4000 50 ] ; 
s= [400 5 ] ; 

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
  temp =(-diff(Y,x(i)));   
  ls=[ls, temp];
  end
  
  for i=1:n  
    if i<n
        xi=m(i);
    disp(x);
    disp(i);
    else
        last=4000 ;
        xi = fzero(Y,last) ; 
    end   
  end   
  

i am trying to define the last value of the function y=0 using initial guess but i am getting this error

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

Answer 1

[dpb]

fsolve does the work for you...if you define the functional correctly--

fnY= @(x) (29-(6*x(1))-(18*x(2)));
opt= optimoptions('fsolve','algorithm','levenberg-marquardt');
>> fsolve(Y,[0 0],opt)

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the default value of the function tolerance, and
the problem appears regular as measured by the gradient.

<stopping criteria details>

ans =
    0.4833    1.4500
>>

Of course, there are an infinite number of possible solutions; pick a value for one or the other of the two X and solve for the other.

[sarra aloui]

thank you but the value of x1 is given already as 4000 i am trying to look just for x2

[dpb]

That's simply

>> x1=4000;
>> x2=(29-(6*x1))/18
x2 =
  -1.3317e+03
>>

But, you can still use fsolve if must...there's just one variable to solve for, however...

>> Y= @(x) (29-(6*x1)-(18*x));
>> x2=fsolve(Y,[ 0],opt)

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the default value of the function tolerance, and
the problem appears regular as measured by the gradient.

<stopping criteria details>

x2 =
  -1.3317e+03
>>