Which optimization tool to use? And how to get the accurate result?

function Z=igx(x,y)
igx1=((x*y^3)/12)+(x*y)*((((x*y)*(150-(y/2))+113664+6000)/((x*y)+1536+1200))-(150-(y/2)))^2;
igx2=2097152+1536*((((x*y)*(150-(y/2))+113664+6000)/((x*y)+1536+1200))-74)^2;
igx3=10000+1200*((((x*y)*(150-(y/2))+113664+6000)/((x*y)+1536+1200))-5)^2;
z=igx1+igx2+igx3;
Z=-z;
end

Above is the function handle. There are 2 variables which are (x,y). 'x' bounds are 13 to 120 and 'y' bounds are 1 to 12. I want to maximize "igx". Which global optimisation to use and how to use the bounds.

[Kevin Chng]

In your command window type :

optimtool

then in the Objective Function

put @igx

then in the Bounds there

[13 1] to [120 12]

then give some starts point. here you go.

[Kiran HR]

It's not getting solved, any other suggestion.

[Torsten]

Objective function:

@(x)igx(x(1),x(2))

[Kevin Chng]

or try change your function to

function Z=igx(x)
 %within it, all x change to x(1), all y change to x(2)
end

[Kiran HR]

Thanks, it worked, Can you please explain why it should be given as @(x)igx(x(1),x(2)), or please suggest me where to look in order understand it completely.

[Kevin Chng]

I started optimization with this video : https://www.mathworks.com/videos/tips-and-tricks-getting-started-using-optimization-with-matlab-81594.html?elqsid=1544016061314&potential_use=Commercial

Answer 1

[Kevin Chng]

n your command window type :

optimtool

then in the Objective Function

put @igx

then in the Bounds there

[13 1] to [120 12]

then give some starts point.

You may need to change your function to

function Z=igx(x)
 %within it, all x change to x(1), all y change to x(2)
end

or try what Torsten recommended to do.