How to find the x value at the maximum?

Question

sophp 2018-2-7

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/381299-how-to-find-the-x-value-at-the-maximum

编辑： Jan 2018-2-7

Here is a MATLAB code to find the maximum of the Y_B (which is a function of

T and t) against t.
     t = 0.2:0.1:20;
     T = 600:2:850;
     for i = 1:numel(T)
        k1 = 1e7 .* exp(-12700 ./ T(i));
        k2 = 5e4 .* exp(-10800 ./ T(i));
        k3 = 7e7 .* exp(-15000 ./ T(i));
        for j = 1:numel(t)
           Y_B(i,j) = (k1/(k2-k1-k3))*(exp(-(k1+k3)*t(j))-exp(-k2*t(j)));
          end
       end
    plot(t, max(Y_B, [], 1));

How do I find the corresponding value of T at the Y_Bmax and plot it against t?

1 个评论
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Jan 2018-2-7

编辑：Jan 2018-2-7

在 MATLAB Online 中打开

As mentioned in your other thread, this is more efficient and nicer without a loop:

t   = 0.2:0.1:20;
T   = (600:50:850).';
k1  = 1e7 .* exp(-12700 ./ T);
k2  = 5e4 .* exp(-10800 ./ T);
k3  = 7e7 .* exp(-15000 ./ T);
Y_B = (k1 ./ (k2-k1-k3)) .* (exp(-(k1+k3) * t) - exp(-k2*t));  % >= R2016

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Answer 1

Birdman 2018-2-7

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/381299-how-to-find-the-x-value-at-the-maximum#answer_303892

编辑：Birdman 2018-2-7

在 MATLAB Online 中打开

       t = 0.2:0.1:20;
       T = 600:2:850;
       for i = 1:numel(T)
          k1 = 1e7 .* exp(-12700 ./ T(i));
          k2 = 5e4 .* exp(-10800 ./ T(i));
          k3 = 7e7 .* exp(-15000 ./ T(i));
          for j = 1:numel(t)
             Y_B(i,j) = (k1/(k2-k1-k3))*(exp(-(k1+k3)*t(j))-exp(-k2*t(j)));
            end
         end
       [val,idx]=max(Y_B);
      plot(T(idx), val,'-o');