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How to use fmincon for my function with 2 variables?

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function [pointoftangency] = Tangentpoint(X,Y,amplitude)
ptf=amplitude*sin(2*pi*x./25)+((2*pi*amplitude/25)*cos(2*pi*x./25)*(X-x))-Y
end
and my constraint equation is
amplitude*sin(2*pi*x/25)=0;
How do I solve this using fmincon??
  2 Commenti
Rik
Rik il 24 Giu 2018
Have a read here and here. It will greatly improve your chances of getting an answer.
You can 'trick' fmincon by entering your variables as a vector into your function.
Walter Roberson
Walter Roberson il 24 Giu 2018
Which are the two variables? You have X, Y, and amplitude as inputs to Tangentpoint, and your code also uses x as well. Your code computs ptf, but your function expects pointoftangency to be output.
What is it that needs to be minimized?

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Risposte (1)

Walter Roberson
Walter Roberson il 24 Giu 2018
Modificato: Walter Roberson il 24 Giu 2018
Assuming that it is ptf that needs to be minimized with X, Y, amplitude constants, then the solution is that ptf becomes arbitrarily small (towards negative infinity) as x approaches positive infinity, assuming the value 2*Pi*amplitude*(X-50*Z)*(1/25)-Y when x = 50*Z with Z being an integer.
ptf becomes arbitrarily small (towards negative infinity) as x approaches negative infinity when x = 50*Z+25 with Z being an integer.

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