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 件のコメント
Walter Roberson
2018 年 6 月 24 日
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?
回答 (1 件)
Walter Roberson
2018 年 6 月 24 日
編集済み: Walter Roberson
2018 年 6 月 24 日
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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