How to use fmincon for my function with 2 variables?
2 visualizaciones (últimos 30 días)
Mostrar comentarios más antiguos
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 comentarios
Walter Roberson
el 24 de Jun. de 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?
Respuestas (1)
Walter Roberson
el 24 de Jun. de 2018
Editada: Walter Roberson
el 24 de Jun. de 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.
0 comentarios
Ver también
Categorías
Más información sobre Mathematics and Optimization en Help Center y File Exchange.
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!