How to find the x value at the maximum?

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sophp
sophp 2018-2-7
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
Jan
Jan 2018-2-7
编辑：Jan 2018-2-7
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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采纳的回答

Birdman
Birdman 2018-2-7
编辑：Birdman 2018-2-7
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');

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