Filter löschen
Filter löschen

How to use fmincon for my function with 2 variables?

1 Ansicht (letzte 30 Tage)
Sanjal C C
Sanjal C C am 23 Jun. 2018
Bearbeitet: Walter Roberson am 24 Jun. 2018
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 Kommentare
Rik
Rik am 24 Jun. 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 am 24 Jun. 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?

Melden Sie sich an, um zu kommentieren.

Antworten (1)

Walter Roberson
Walter Roberson am 24 Jun. 2018
Bearbeitet: Walter Roberson am 24 Jun. 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.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by