MATLAB Answers

optimization problem with two variable maxima and minima

2 views (last 30 days)
manish kumar
manish kumar on 6 Sep 2019
Edited: Bruno Luong on 11 Sep 2019
can any one give me the solution or help me out in solving this equation mathamatically
Y=2x(1)^2 + 23.08x(2)^2 +4(6+x(1))^2 +24+14(x(1)^2 +x(2)^2)^0.5 +3(x(1)^2 + x(2)^2)
the other equation is 1=x(1)*x(2)

  7 Comments

Show 4 older comments
manish kumar
manish kumar on 6 Sep 2019
first step :
by differentiating y with respect to x(1)
then by putting it equal to zero the term x(2) is coming due to square root term
how to solve this
and if we are putting x(2)=1/x(1) then complex term is coming
can you help me out

Sign in to comment.

Answers (1)

Catalytic
Catalytic on 6 Sep 2019
Edited: Matt J on 9 Sep 2019
fun=@(x) [2*x(1)^2+23.08*x(2)^2+4*(6+x(1))^2+24+14*(x(1)^2 +x(2)^2)^0.5+3*(x(1)^2+x(2)^2)-Y;...
prod(x)-1];
x=fsolve(fun,initial_guess)

  9 Comments

Show 6 older comments
manish kumar
manish kumar on 11 Sep 2019
i want to minimize
Y(x(1),x(2)) = 2x(1)^2 + 23.08x(2)^2 +4(6+x(1))^2 +24+14(x(1)^2 +x(2)^2)^0.5 +3(x(1)^2 + x(2)^2)
under the constraint
1 = x(1)*x(2)
Torsten
Torsten on 11 Sep 2019
fun= @(x)2*x.^2+23.08*(1./x).^2+4*(6+x).^2+24+14*(x.^2+(1./x).^2).^0.5+3*(x.^2+(1./x).^2)
x0 = 1.0;
xmin = fminsearch(fun,x0)
Bruno Luong
Bruno Luong on 11 Sep 2019
Careful on local minimum
>> xmin = fminsearch(fun,1), fun(xmin) % not global minimum
xmin =
0.9418
ans =
270.4623
>> xmin = fminsearch(fun,-2), fun(xmin)
xmin =
-2.2066
ans =
142.7984
>>

Sign in to comment.