# optimization problem with two variable maxima and minima

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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)

Torsten on 6 Sep 2019
I don't understand what you mean.
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
Torsten on 6 Sep 2019

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)

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 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 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
>>