optimization problem with two variable maxima and minima

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manish kumar
manish kumar am 6 Sep. 2019
Bearbeitet: Bruno Luong am 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 Kommentare
manish kumar
manish kumar am 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
Torsten
Torsten am 6 Sep. 2019
Please show your code.

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Antworten (1)

Catalytic
Catalytic am 6 Sep. 2019
Bearbeitet: Matt J am 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 Kommentare
Torsten
Torsten am 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 am 11 Sep. 2019
Bearbeitet: Bruno Luong am 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
>>

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