How to find the maximum value of two variables of a function in MATLAB

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Hadeel Obaid 2023-5-10

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评论： Matt J 2023-6-20

Hi everyone,

I would like to find the maximum value of \eta and xo in the function below using numerical simulation:

z=1e6*log2(1+(10^(30/10)*4*(3e8/(4*pi*1e12))^2*15^(-4)*exp(-0.0016*15))/10^(-90/10))*(-1/(1e4^(1-0.5)-1))+ 1e6*log2(1+(10^(30/10)*4*(3e8/(4*pi*1e12))^2*x0^(-2)*exp(-0.0016*x0))/10^(-90/10))*((100/eta)^(1-0.5)-1)/(1e4^(1-0.5)-1);
\eta range and xo range are:
eta_range = 0.01:0.01:1;
x0_range = 1:1:100;

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Rik 2023-6-20

I recovered the removed content from the Google cache (something which anyone can do). Editing away your question is very rude. Someone spent time reading your question, understanding your issue, figuring out the solution, and writing an answer. Now you repay that kindness by ensuring that the next person with a similar question can't benefit from this answer.

Matt J 2023-6-20

在 MATLAB Online 中打开

Back-up copy of Hadeel Obaid's question:

Hi everyone,

I would like to find the maximum value of \eta and xo in the function below using numerical simulation:

z=1e6*log2(1+(10^(30/10)*4*(3e8/(4*pi*1e12))^2*15^(-4)*exp(-0.0016*15))/10^(-90/10))*(-1/(1e4^(1-0.5)-1))+ 1e6*log2(1+(10^(30/10)*4*(3e8/(4*pi*1e12))^2*x0^(-2)*exp(-0.0016*x0))/10^(-90/10))*((100/eta)^(1-0.5)-1)/(1e4^(1-0.5)-1);
\eta range and xo range are:
eta_range = 0.01:0.01:1;
x0_range = 1:1:100;

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Answer 1

Matt J 2023-5-10

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https://ww2.mathworks.cn/matlabcentral/answers/1961029-how-to-find-the-maximum-value-of-two-variables-of-a-function-in-matlab#answer_1232509

编辑：Matt J 2023-5-10

在 MATLAB Online 中打开

Your function z is separable and monotonically decreasing in both variables. So, it should come as no surprise that the smallest values of eta and x0 give the maximum. However, you can verify that with the code below:

eta = (0.01:0.01:1)';
x0 = (1:100);
z=1e6.*log2(1+(10.^(30./10).*4.*(3e8./(4.*pi.*1e12)).^2.*15.^(-4).*exp(-0.0016.*15))./10.^(-90./10)).*(-1./(1e4.^(1-0.5)-1))+ 1e6.*log2(1+(10.^(30./10).*4.*(3e8./(4.*pi.*1e12)).^2.*x0.^(-2).*exp(-0.0016.*x0))./10.^(-90./10)).*((100./eta).^(1-0.5)-1)./(1e4.^(1-0.5)-1);
[maxval,k]=max(z,[],'all','linear')
maxval = 1.1152e+07
k = 1
[i,j]=ind2sub(size(z),k);
eta_max=eta(i),
eta_max = 0.0100
x0_max=x0(j),
x0_max = 1

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Hadeel Obaid 2023-5-11

@Walter Robersonbut each one has different results?

Matt J 2023-5-11

@Hadeel Obaid Torsten and I reached the same result. And, as I outlined above, you did not need any code to reach this result. The maximizing point is obvious from the expression for z.

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Answer 2

Torsten 2023-5-10

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在 MATLAB Online 中打开

eta = 0.01:0.01:1;
x0 = (1:1:100).';
z = 1e6*log2(1+(10^(30/10)*4*(3e8/(4*pi*1e12))^2*15^(-4)*exp(-0.0016*15))/10^(-90/10))*(-1/(1e4^(1-0.5)-1))+ 1e6*log2(1+(10^(30/10)*4*(3e8/(4*pi*1e12))^2*x0.^(-2).*exp(-0.0016*x0))/10^(-90/10))*((100./eta).^(1-0.5)-1)/(1e4^(1-0.5)-1);
maximum_z = max(max(z))
maximum_z = 1.1152e+07
[i,j] = find(z==maximum_z)
i = 1
j = 1