Suppose you started with a function myfun(x) that inputs a vector x and outputs a vector, for example:
myfun = @(x) x.^2-1;
x0 = [-1.5 1.5];
x1 = fsolve(myfun,x0)
x1 =
-1.0000 1.0000
To incorporate your constraints, you could do this:
lb = zeros(size(x0)); %lower bound of zero
ub = Inf*ones(size(x0));
x2 = lsqnonlin(myfun,x0,lb,ub)
x2 =
1.0000 1.0000
As John D'Errico implies, the minimization may give you an answer that is not a root, so you then need to check your answer:
myfun(x2)
ans =
1.0e-08 *
0.3056 0.0030
(Note: answer rewritten completely.)
