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

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;

[Rik]

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]

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;

Answer 1

[Matt J]

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

[Walter Roberson]

Notice that @Matt J and @Torsten both combined row vectors with column vectors and vectorized operations such as .^2 instead of ^2 . When those kinds of operations are used, you get a 2D array of results with all pairwise combinations of the eta and x0 values.

[Hadeel Obaid]

@Walter Robersonbut each one has different results?

[Matt J]

@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.

Answer 2

[Torsten]

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