optimization problem with two variable maxima and minima

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manish kumar
manish kumar 2019-9-6
编辑： Bruno Luong 2019-9-11
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)
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manish kumar
manish kumar 2019-9-6
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

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回答（1 个）

Catalytic
Catalytic 2019-9-6
编辑：Matt J 2019-9-9
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)
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Torsten
Torsten 2019-9-11
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 2019-9-11
编辑：Bruno Luong 2019-9-11
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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