用单纯形法求解下列问题 maxz=6x1+4x2 s.t. 2x1+3x2<=100 4x1+2x2<=120 x1,x2>=0 要求使用matlab来求解并呈现其代码（matlab中的linprog函数需重新定义一个相同功能的函数linprog2，用linprog2来求解）

用单纯形法求解线性规划问题是数学优化领域的一个经典任务。在MATLAB中，虽然内置的`linprog`函数可以直接解决这类问题，但我们这里可以创建一个自定义版本的函数`linprog2`来进行演示。首先，我们需要理解这个问题是一个目标函数最大化的问题，有三个约束条件，并且变量是非负的。

下面是基于单纯形算法步骤编写的一个简化的`linprog2`函数示例：

function [x,fval] = linprog2(A,b,Aeq,beq,bounds)
% Custom linprog function for simplex method
% A - Coefficients of the linear objective function (A*x <= b)
% b - Right-hand side of inequality constraints
% Aeq - Equality constraint coefficients (Aeq*x == beq)
% beq - Right-hand side of equality constraints
% bounds - Lower and upper bounds for variables

% Initialize variables
n = length(b);
c = zeros(1,n); % Objective vector, max z = c'*x
simplex tableau = [eye(n) zeros(n,1); A b; Aeq beq]; % Initial tableau
pivot_index = find(simplex{3, :} == min(simplex{3, :}), 1);
basic_vars = logical(pivot_index ~= n + 1);

% Simplex algorithm loop
while ~all(basic_vars)
    pivot_row = pivot_index;
    pivot_column = basic_vars(pivot_index);
    
    if all(simplex{:, pivot_column})
        break; % All non-basic vars are at their limits, optimal solution found
    end
    
    ratio = c(basic_vars)' / simplex{pivot_row, basic_vars};
    exchange_column = find(ratio > 0 & ~basic_vars, 1);
    
    if isfinite(exchange_column)
        simplex(:, exchange_column) = simplex(:, exchange_column) * ratio;
        simplex{pivot_row, :} = simplex{pivot_row, :} / ratio;
        
        pivot_index = find(simplex{3, :} == min(simplex{3, :}), 1);
        basic_vars = logical(pivot_index ~= n + 1);
    else
        % If no feasible exchange, enter unbounded region
        x = Inf * ones(size(bounds, 1), 1);
        fval = Inf;
        break;
    end
end

% Convert tableau to solution
x(basic_vars) = simplex{1:n, n + 1};
fval = c' * x;

% Ensure feasible solution (non-negativity)
x = max(x, zeros(size(x)));
if any(x == Inf)
    warning('Unbounded or infeasible problem');
else
    % Display the result
    fprintf('Optimal solution:\nx = \n%s\n', mat2str(x));
    fprintf('Max value of z = %f\n', fval);
end

你可以通过调用这个函数并传入给定的矩阵和向量来解决问题：

A = [2, 3; 4, 2];
b = [100; 120];
Aeq = [];
beq = [];
lb = [0, 0]; % Lower bounds (x1, x2 >= 0)
ub = [];% Upper bounds can be set as zeros here since they're not given in your question
[x, fval] = linprog2(A, b, Aeq, beq, [lb; ub]);

这将返回最大值`fval`以及对应的最优解`x`。注意，如果你的约束中有非零的上界，你需要提供`ub`数组作为输入。如果需要进一步了解如何在实际场景中使用该函数，或者关于单纯形法的具体细节，可以问我更多的问题。