optimization problem with two variable maxima and minima

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)

[Matt J]

Which variable is the unknown? You only have 1 equation, so only one of the variables can be.

[manish kumar]

here x(1) and x(2) are unkown you can use other equation as

1=x(1)*x(2)

[Torsten]

Differentiate Y with respect to x(1) and x(2), set the derivatives to 0 and solve for x(1) and x(2).

help diff

help solve

[manish kumar]

if i am differentiating the equation with respect to x(1) then the term under square root x(2) is coming

please guide under this matter

[Torsten]

I don't understand what you mean.

[manish kumar]

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]

Please show your code.

Answer 1

[Catalytic]

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]

i am unable to run your code

plesae provide the code for x(1)' and x(2)'

[Matt J]

You probably need to remove all the spaces in the expression for fun.

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]

it says

Undefined function or variable 'initial_guess'.

[Walter Roberson]

You should have used

initial_guess = [-42, pi];

[manish kumar]

can you please provide combined code in a single script

as the data provided by you is not working

[Torsten]

It's still not clear to me what you are trying to do.

Do you want to solve the two equations

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)
1=x(1)*x(2)

for x(1) and x(2) (for a given value of Y)

or do you want to determine maximum and minimum of

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)

or ...

[manish kumar]

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]

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]

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

>>