Consider the maximization of the area of a box subject to a given perimeter.
max_x,y>=0 xy s.t. x+y<=c
This admits the Lagrangian L(x,y,lambda) = xy + lambda(c-x-y). Notice that this
Lagrangian is CONVEX jointly in (x,y). So why do the first order conditions
define a maximum?
Aside: I KNOW how to argue that the solution finds a maximum and not a minimum.
I must consider the BORDERED HESSIAN determinants. The point of my question is
that I am facing the above counterintuitive where the analogue of the Hessian
determinants seems not to exist, and the standard sufficiency conditions pump
out convexity and not concavity.
Thanks, Lones.
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/ L \ O / N \ E / S \ / S \ M / I \ T / H \
/ `-' `-' `-' `-' `-' `
Lones Smith, Economics Department, M.I.T., E52-252C, Cambridge MA 02139
(617)-253-0914 (work) 253-6915 (fax)
lo...@lones.mit.edu
Your objective function is quasi-concave, which is perfectly consistent
with it being convex, and both strictly so. Also, the constraint is
linear, thus the Arrow-Enthoven result applies. Arrow and Enthoven study
quasi-concavity in a famous article whose particulars of publication I do
not recall offhand. Check out Chiang's or Simon and Blume' mathematical
economics / mathematics for economists textbooks under quasi-concave
programming.
Dimitrios Diamantaras
Department of Economics
Temple University
lo...@lones.mit.edu (Lones A Smith) wrote:
>Consider the maximization of the area of a box subject to a given perimeter.
>
>max_x,y>=0 xy s.t. x+y<=c
>
>This admits the Lagrangian L(x,y,lambda) = xy + lambda(c-x-y). Notice that this
>Lagrangian is CONVEX jointly in (x,y). So why do the first order conditions
>define a maximum?
The objective function is strictly quasiconcave (<=> level curves define a
strictly convex function y=f(x)), while the constraint is linear. It is
therefore the standard textbook case where first-order conditions define a
global maximum. Draw a picture!
Daniel
________________________________
Daniel Royer
University of Geneva (Switzerland)
Daniel Royer (Daniel...@themes.unige.ch) wrote:
: lo...@lones.mit.edu (Lones A Smith) wrote:
: >Consider the maximization of the area of a box subject to a given perimeter.
: >
: >max_x,y>=0 xy s.t. x+y<=c
: >
: >This admits the Lagrangian L(x,y,lambda) = xy + lambda(c-x-y). Notice that
: >this Lagrangian is CONVEX jointly in (x,y). So why do the first order
: >conditions define a maximum?
: The objective function is strictly quasiconcave (<=> level curves define a
: strictly convex function y=f(x)), while the constraint is linear. It is
: therefore the standard textbook case where first-order conditions define a
: global maximum. Draw a picture!
: Daniel
I see that the second paragraph of my post did not appear as I meant it to.
What I had written was basically: Granted that we have other means like
Bordered Hessians and quasiconcavity for dealing with this apparent failure
of concavity, we are OK *in the Euclidean case*. But my problem concerns
a _Hamiltonian_ where the (Arrow or Mangasarian) second order sufficiency
conditions do not help. Now, I would like to have a fallback tool that can
rescue me. For instance, it would be nice to say "But the Hamiltonian is
quasiconcave", or some such. But I cannot find any such equivalent in the texts
I have seen.....
.-. .-. .-. .-. .-. .-.
/ L \ O / N \ E / S \ / S \ M / I \ T / H \
/ `-' `-' `-' `-' `-' `
Lones Smith, Economics Department, M.I.T., E52-252C, Cambridge MA 02139
(617)-253-0914 (work) 253-6915 (fax)
lo...@lones.mit.edu
[Moderator takes this opportunity to insert an apology. Since Monday
a number of posts have been appearing under my signature. This is a
strange side effect of having a new machine at Duke, and recompiled
mailing macros on it. The address @corp.sun.com belongs to the guy who
installed the machine, not me. I hope it is now under control. If the
part of Lones' note that vanished was also a victim of the head-snipper
macro, my apologies to him, too. -dlj.]