【求助】使用lsqnonlin函数大规模计算如何设置函数最大值？？？急急急

大海龟

x = lsqnonlin(fun,x0) starts at the point x0 and finds a minimum to the sum of
squares of the functions described in fun. fun should return a vector of values and
not the sum-of-squares of the values. (fun(x) is summed and squared implicitly in the
algorithm.)
  
x = lsqnonlin(fun,x0,lb,ub) defines a set of lower and upper bounds on the design
variables, x, so that the solution is always in the range lb <= x <= ub.
  
x = lsqnonlin(fun,x0,lb,ub,options) minimizes with the optimization parameters
specified in the structure options. Pass empty matrices for lb an ub if no bounds
exist.
  
x = lsqnonlin(fun,x0,lb,ub,options,P1,P2,...) passes the problem-dependent
parameters P1, P2, etc., directly to the function fun. Pass an empty matrix for options
to use the default values for options.
  
[x,resnorm] = lsqnonlin(...) returns the value of the squared 2-norm of the residual
at x: sum(fun(x).^2).
  
[x,resnorm,residual] = lsqnonlin(...) returns the value of the residual, fun(x), at
the solution x.
  
[x,resnorm,residual,exitflag] = lsqnonlin(...) returns a value exitflag that
describes the exit condition.
  
[x,resnorm,residual,exitflag,output] = lsqnonlin(...) returns a structure output
that contains information about the optimization.
  
[x,resnorm,residual,exitflag,output,lambda] = lsqnonlin(...) returns a structure
lambda whose fields contain the Lagrange multipliers at the solution x.
  
[x,resnorm,residual,exitflag,output,lambda,jacobian] = lsqnonlin(...) returns
the Jacobian of fun at the solution x.
  
英文的，自己看。