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Saddle point of the Lagrangian

Generalizing Lagrange multipliers to use the subdifferential?Constrained optimization problems resulting in equal variable assignmentsDerive the solution to the Lagrangian $ mathcal L= y(x)sqrt1+y'(x)^2$Minimize and maximize the function $g(a,b) = a + b$ given the constraint $h(a,b) = frac1a + frac1b = 1$.Find the maximum of $U (x,y) = x^alpha y^beta$ subject to $I = px + qy$Constructing a function with saddle pointsApplying log transform to find critical points when there is a constraintAmbiguity about the Lagrangian method of optimizationSolving system of polynomial equations stemming from a constrained optimization problemSolving difficult Lagrangian

0

$begingroup$

Consider a standard constrained optimisation problem with an equality constraint.

Maximize $f(x,y)$ s.t. $g(x,y)=c$.

We construct the Lagrangian to solve this problem.

$mathcalL(x,y)=f(x,y)-lambda(g(x,y)-c)$

The first order conditions give us the stationary point of the Lagrangian.

$mathcalL_x=0$

$mathcalL_y=0$

$g(x,y)=c$

Say the vector $(x*,y*,lambda*)$ solves this set of equations. Is this point a saddle point of the Lagrangian?

Further, does $(x*,y*,lambda*)$ maximize $mathcalL(x,y)$ as well?

multivariable-calculus

share|cite|improve this question

edited Apr 2 at 7:02

PGupta

asked Apr 1 at 17:03

PGuptaPGupta

1746

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If the functional is stricly concave, the stationary point cannot be a saddle point: you can grasp some intuition of it thinking about concave function of real variables
  $endgroup$
  – Gabriele Cassese
  Apr 1 at 17:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Then how do we establish that the function is maximixed?
  $endgroup$
  – PGupta
  Apr 2 at 2:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  A function is not maximized on a saddle point
  $endgroup$
  – Gabriele Cassese
  Apr 2 at 5:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  intuitively, a saddle point is when the function is max in along one variable and min along the other variable (just like the saddle on the horseback)
  $endgroup$
  – farruhota
  Apr 2 at 6:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @farruhota I understand what is a saddle point. What I am trying to ask is about the relation (if any exists) between the optimum of the constrained problem and the stationary point of the Lagrangian we construct to solve the said problem. Is the optimum of the constrained problem also the maximum of the Lagrangian?
  $endgroup$
  – PGupta
  Apr 2 at 6:56
  

  
  
  
  

 | 
show 1 more comment

0

$begingroup$

Consider a standard constrained optimisation problem with an equality constraint.

Maximize $f(x,y)$ s.t. $g(x,y)=c$.

We construct the Lagrangian to solve this problem.

$mathcalL(x,y)=f(x,y)-lambda(g(x,y)-c)$

The first order conditions give us the stationary point of the Lagrangian.

$mathcalL_x=0$

$mathcalL_y=0$

$g(x,y)=c$

Say the vector $(x*,y*,lambda*)$ solves this set of equations. Is this point a saddle point of the Lagrangian?

Further, does $(x*,y*,lambda*)$ maximize $mathcalL(x,y)$ as well?

multivariable-calculus

share|cite|improve this question

edited Apr 2 at 7:02

PGupta

asked Apr 1 at 17:03

PGuptaPGupta

1746

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If the functional is stricly concave, the stationary point cannot be a saddle point: you can grasp some intuition of it thinking about concave function of real variables
  $endgroup$
  – Gabriele Cassese
  Apr 1 at 17:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Then how do we establish that the function is maximixed?
  $endgroup$
  – PGupta
  Apr 2 at 2:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  A function is not maximized on a saddle point
  $endgroup$
  – Gabriele Cassese
  Apr 2 at 5:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  intuitively, a saddle point is when the function is max in along one variable and min along the other variable (just like the saddle on the horseback)
  $endgroup$
  – farruhota
  Apr 2 at 6:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @farruhota I understand what is a saddle point. What I am trying to ask is about the relation (if any exists) between the optimum of the constrained problem and the stationary point of the Lagrangian we construct to solve the said problem. Is the optimum of the constrained problem also the maximum of the Lagrangian?
  $endgroup$
  – PGupta
  Apr 2 at 6:56
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

Consider a standard constrained optimisation problem with an equality constraint.

Maximize $f(x,y)$ s.t. $g(x,y)=c$.

We construct the Lagrangian to solve this problem.

$mathcalL(x,y)=f(x,y)-lambda(g(x,y)-c)$

The first order conditions give us the stationary point of the Lagrangian.

$mathcalL_x=0$

$mathcalL_y=0$

$g(x,y)=c$

Say the vector $(x*,y*,lambda*)$ solves this set of equations. Is this point a saddle point of the Lagrangian?

Further, does $(x*,y*,lambda*)$ maximize $mathcalL(x,y)$ as well?

multivariable-calculus

share|cite|improve this question

edited Apr 2 at 7:02

PGupta

asked Apr 1 at 17:03

PGuptaPGupta

1746

$endgroup$

share|cite|improve this question

edited Apr 2 at 7:02

PGupta

asked Apr 1 at 17:03

PGuptaPGupta

1746

share|cite|improve this question

edited Apr 2 at 7:02

PGupta

asked Apr 1 at 17:03

PGuptaPGupta

1746

asked Apr 1 at 17:03

PGuptaPGupta

1746

asked Apr 1 at 17:03

PGuptaPGupta

1746