How do i use fminsearch to find the minimum or maximum of a function . x.^4-3.*x.*y+2.*y.^2

I trying to use fmin search for a function of 2 variable

[Matt J]

Ok, show us.

[Ibraheem]

fun = @(x) x.^4-3.*x.*y +2.*y.^2;

x1 =x;

x2 =y;

x0 = [-0.5,0.5];

x = fminsearch(fun,x0)

I don't really know what to do to find the local minimum

Answer 1

[James Tursa]

You need to have your function handle accept a vector and return a scalar. I.e., the x argument to the function handle is a vector of two elements representing your original x and y variables. Assuming x(1) and x(2) are your intended original x and y variables, that would mean something like this:

fun = @(x) x(1).^4-3.*x(1).*x(2) +2.*x(2).^2;
x0 = [-0.5,0.5];
x = fminsearch(fun,x0)

x = 1×2

-0.7500 -0.5625

[Ibraheem]

Thank you very much, but why are the answers negative but when i calculated it i got 3/4 ,9/16

[James Tursa]

format longg
fun = @(x) x(1).^4-3.*x(1).*x(2) +2.*x(2).^2;
x0 = [-0.5,0.5];
x = fminsearch(fun,x0)

x = 1×2

-0.749979058202537 -0.562478134566371

fun(x)

ans = 
        -0.316406248937402

fun([3/4,9/16])

ans = 
               -0.31640625

Also note that fminsearch( ) can only find local minimums, of which there can be more than one depending on the function. So different starting points can result in different answers.

x = fminsearch(fun,[3/4+0.1,9/16-0.1])

x = 1×2

0.749992146496559 0.562470511151394

fun(x)

ans = 
        -0.316406248747428

[the cyclist]

The function is invariant in the transformation (x,y) --> (-x,-y)

Answer 2

[the cyclist]

Did you try reading the documentation for fminsearch? The very first example is exactly like your problem.

fun = @(x)(x(1)^4 - 3*x(1)*x(2) + 2*x(2)^2);
x0 = [-0.5, 0.5];
x = fminsearch(fun,x0)

x = 1×2

-0.7500 -0.5625