# Unable to find explicit solution in Lagrangian optimization

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Fabian am 11 Feb. 2024
Kommentiert: Matt J am 11 Feb. 2024
I am trying to find the analytical solution to the following problem:
I tried solving it by coding the Lagrangian by hand and use solve, but Matlab prints the warning: "Unable to find explicit solution".
I used the following code:
syms e1 e2 p1 p2 rho gamma lambda
syms E H(e1,e2)
H(e1,e2) = (e1^rho +e2^rho)^(1/rho)
L(e1, e2, lambda) = p1*e1 +p2*e2 + lambda*(H(e1,e2)-E)
L_e1 = diff(L,e1) == 0
L_e2 = diff(L,e2) == 0
L_lambda = diff(L,lambda) == 0
system = [L_e1,L_e2,L_lambda]
[e1_s,e2_s,lambda_s]=solve(system,[e1 e2 lambda])
Do you know what I could do to solve this? Or is there a different and better way to find an analytical solution?
##### 1 Kommentar-1 ältere Kommentare anzeigen-1 ältere Kommentare ausblenden
Matt J am 11 Feb. 2024
Bearbeitet: Matt J am 11 Feb. 2024
Note that the problem can always be rewritten in the simpler form,
where x=e/E and P=p*E. This is assuming E is a known positive constant.

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### Akzeptierte Antwort

Catalytic am 11 Feb. 2024
Bearbeitet: Catalytic am 11 Feb. 2024
An analytical solution for 0<rho<1 is -
A=[1 0;
0 1;
-1 0;
0 -1]*E;
[fval,i]=min(A*[p1;p2]);
e1=A(i,1);
e2=A(i,2);
##### 2 KommentareKeine anzeigenKeine ausblenden
Catalytic am 11 Feb. 2024
Bearbeitet: Catalytic am 11 Feb. 2024
You can see this graphically by plotting the constrained region. The region always has extreme points at (), so that's where the optimum must lie.
E=1;
for rho=[0.1:0.2:0.9]
fimplicit(@(e1,e2) abs(e1).^rho + abs(e2).^rho - E.^rho, [-1.5,1.5]); hold on
end
Matt J am 11 Feb. 2024
I like it. And, in fact, because the extreme points lie at points where H(e1,e2) is not differentiable, it shows that you will never find the true solution with Lagrange multiplier analysis.

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### Weitere Antworten (1)

Matt J am 11 Feb. 2024
If you make rho explicit, it seems to be able to find solutions. I doubt there would be a closed-form solution for general rho.
rho=2;
syms e1 e2 p1 p2 gamma lambda
syms E H(e1,e2)
H(e1,e2) = (e1^rho +e2^rho)
H(e1, e2) =
L(e1, e2, lambda) = p1*e1 +p2*e2 + lambda*(H(e1,e2)-E^rho)
L(e1, e2, lambda) =
L_e1 = diff(L,e1) == 0
L_e1(e1, e2, lambda) =
L_e2 = diff(L,e2) == 0
L_e2(e1, e2, lambda) =
L_lambda = diff(L,lambda) == 0
L_lambda(e1, e2, lambda) =
system = [L_e1,L_e2,L_lambda]
system(e1, e2, lambda) =
[e1_s,e2_s,lambda_s]=solve(system,[e1 e2 lambda])
e1_s =
e2_s =
lambda_s =
##### 4 Kommentare2 ältere Kommentare anzeigen2 ältere Kommentare ausblenden
Matt J am 11 Feb. 2024
Even when it can be explicitly solved, the result isn't nice:
rho=sym(1/4);
syms e1 e2 p1 p2 gamma lambda
syms H(e1,e2)
H(e1,e2) = (e1^rho +e2^rho);
L(e1, e2, lambda) = p1*e1 +p2*e2 + lambda*(H(e1,e2)-1);
L_e1 = diff(L,e1) == 0;
L_e2 = diff(L,e2) == 0;
L_lambda = diff(L,lambda) == 0;
system = [L_e1,L_e2,L_lambda];
[e1_s,e2_s,lambda_s]=solve(system,[e1 e2 lambda])
Warning: Possibly spurious solutions.
e1_s =
e2_s =
lambda_s =
Walter Roberson am 11 Feb. 2024
You can eliminate the root() constructs, but the result is confusing.
rho=sym(1/4);
syms e1 e2 p1 p2 gamma lambda
syms H(e1,e2)
H(e1,e2) = (e1^rho +e2^rho);
L(e1, e2, lambda) = p1*e1 +p2*e2 + lambda*(H(e1,e2)-1);
L_e1 = diff(L,e1) == 0;
L_e2 = diff(L,e2) == 0;
L_lambda = diff(L,lambda) == 0;
system = [L_e1,L_e2,L_lambda];
[e1_s,e2_s,lambda_s]=solve(system,[e1 e2 lambda], 'maxdegree', 3)
Warning: Possibly spurious solutions.
e1_s =
e2_s =
lambda_s =

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