Find argmin of function with two input variables

Hello, I want to find the arguments of a function La(x1, x2) that minimize this function by calculating the diff of the function in order of x1 and x2, equal to 0 and solving them. MATLAB keeps returning Empty sym: 0-by-1 and I don't understand why. Any help?

Also, if the derivative of the function in order of one of the variables was equal to 1, and then MATLAB tried to calculate 1 == 0, what would it return?

syms x1 x2
assume(x1 > 0.1)
assumeAlso(x1 < 2.0)
assumeAlso(x2 > 0.1)
assumeAlso(x2 < 2.5)
La(x1, x2) = x1^2 + x1 + x2^2 + 5*x2;
eqn1 = [diff(La, x1)==0];
solx1 = solve(eqn1, x1);
eqn2 = [diff(La, x2)==0];
solx2 = solve(eqn2, x2);

Answer 1

[Matt J]

MATLAB keeps returning Empty sym: 0-by-1 and I don't understand why. Any help?

The solution to diff(La,x1)==0 is x1 = -0.5, which does not satisfy your assumption x1>0.1

and then MATLAB tried to calculate 1 == 0,

It would return no solution, as it should.

syms x

solve(diff(x)==0,x)

 
ans =
 
Empty sym: 0-by-1
 

[ABCDEFG HIJKLMN]

It makes total sense, I just experimented with diferent boundaries for the input variables and it worked, thank you for your time.

[ABCDEFG HIJKLMN]

However, I would like the code to return to me the minimum possible for the limits imposed. For instance, in the original limits of x1 and x2, what would be the pair that lead to the minimum value of La? Is there a way to obtain one of the solutions when there are multiple? @Matt J

[Matt J]

It makes total sense, I just experimented with diferent boundaries for the input variables and it worked, thank you for your time.

You're welcome, but please accept-click the answer to indicate that your question is resolved.

For instance, in the original limits of x1 and x2, what would be the pair that lead to the minimum value of La? Is there a way to obtain one of the solutions when there are multiple?

You can use fminbnd.

fminbnd(@(x1) x1^2+x1,  0.1,2)

ans = 0.1000

[ABCDEFG HIJKLMN]

But fminbnd only works if the function only had one variable, right? I would need the coordinates (x1, x2) instead! @Matt J

[Matt J]

Your function is additively separable in this example, so the terms dependent on x2 can be minimized independently of x1.

fminbnd(@(x2) x2^2+5*x2,  0.1,2.5)

ans = 0.1001

If you have to do deal with non-separable functions, you can use fmincon().