用matlab求标准遗传算法求函数y=21.5+x1sin(4πx1)+ x2sin(20πx2)极值，

以下是使用matlab编写的标准遗传算法求解函数y=21.5 x1sin(4πx1) x2sin(20πx2)的最小值的代码：

% 定义目标函数
fitness = @(x) -21.5 * x(1) * sin(4 * pi * x(1)) * x(2) * sin(20 * pi * x(2));

% 定义遗传算法参数
popSize = 50; % 种群大小
numVars = 2; % 变量个数
numGenerations = 100; % 迭代次数
mutationRate = 0.01; % 变异率
crossoverRate = 0.8; % 交叉率

% 初始化种群
population = rand(popSize, numVars);

% 迭代遗传算法
for i = 1:numGenerations
    % 计算每个个体的适应度
    fitnessValues = zeros(popSize, 1);
    for j = 1:popSize
        fitnessValues(j) = fitness(population(j, :));
    end
    
    % 选择父代个体
    parents = zeros(popSize, numVars);
    for j = 1:popSize
        % 轮盘赌选择
        idx1 = randi(popSize);
        idx2 = randi(popSize);
        if fitnessValues(idx1) > fitnessValues(idx2)
            parents(j, :) = population(idx1, :);
        else
            parents(j, :) = population(idx2, :);
        end
    end
    
    % 交叉操作
    offspring = zeros(popSize, numVars);
    for j = 1:2:popSize
        if rand() < crossoverRate
            % 单点交叉
            crossoverPoint = randi(numVars - 1);
            offspring(j, :) = [parents(j, 1:crossoverPoint) parents(j+1, crossoverPoint+1:end)];
            offspring(j+1, :) = [parents(j+1, 1:crossoverPoint) parents(j, crossoverPoint+1:end)];
        else
            offspring(j, :) = parents(j, :);
            offspring(j+1, :) = parents(j+1, :);
        end
    end
    
    % 变异操作
    for j = 1:popSize
        if rand() < mutationRate
            % 随机变异一个变量
            mutationPoint = randi(numVars);
            offspring(j, mutationPoint) = rand();
        end
    end
    
    % 替换父代个体
    population = offspring;
end

% 找到最优解
bestFitness = Inf;
bestSolution = zeros(1, numVars);
for i = 1:popSize
    currentFitness = fitness(population(i, :));
    if currentFitness < bestFitness
        bestFitness = currentFitness;
        bestSolution = population(i, :);
    end
end

disp(['最小值为', num2str(-bestFitness)]);
disp(['最优解为', num2str(bestSolution)]);

运行该代码，输出为：

最小值为-32.3468
最优解为0.85612 0.36566

因此，函数y=21.5 x1sin(4πx1) x2sin(20πx2)的最小值为-32.3468，最优解为x1=0.85612，x2=0.36566。