Finding global maxima with Kuhn-Tucker conditions (and distinguish them from other critical points)

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We want to maximize $(x-1)^2 + (y-1)^2$ restricted to $x + y le 2$ and $x, y ge 0$.

I tried the following combinations:

$x gt 0, y gt 0$ This led me to no critical point.

$x gt 0, y = 0$ This led me to these two critical points: $P_1(1, 0), lambda = 0$ and $P_2(2, 0), lambda = -2$.

$x = 0, y gt 0$ This led me to these two critical points: $P_1(0, 1), lambda = 0$ and $P_2(0, 2), lambda = -2$.

$x = 0, y = 0$ This led me to these two critical points: $P_1(0, 0), lambda = 0$ and $P_2(0, 0), lambda lt 2$.

My handbook says the global optima are: $x = 2, y = 0, lambda = -2$ and $x = 0, y = 0, lambda = 0$.

So here I have several questions:

1) How do I distinguish between global and local optima in these cases? The hessian matrix does not seem to be useful in this case. For example, how can I know that $x = 2, y = 0, lambda = -2$ is a global optima, but $x = 1, y = 0, lambda = 0$ is not? Intuition can help, but I want the method.

2) Why would $x = 2, y = 0, lambda = -2$ be a global optima, and $x = 0, y = 2, lambda = -2$ would not?

3) Why would $x = 0, y= 0, lambda = 0$ be a global maxima? It is a bit counterintuitive.

4) Would it be impossible to determine the value of $lambda$ in $P_2$ of $x = 0, y = 0$ combination? (This is the case in which we assume $lambda lt 0$, while in $P_1$ we assumed $lambda = 0$).

5) Feel free to add any information needed to understand why the answer of my handbook is the correct one (which are the global maxima) and not all of above critical points, in case this is not clear with the answers to questions 1, 2, 3 and 4.

Thank you very much for your time.

nonlinear-optimization karush-kuhn-tucker

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asked Jul 25 at 12:00

Ignacio Valdés Zamudio

103

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We want to maximize $(x-1)^2 + (y-1)^2$ restricted to $x + y le 2$ and $x, y ge 0$.

I tried the following combinations:

$x gt 0, y gt 0$ This led me to no critical point.

$x gt 0, y = 0$ This led me to these two critical points: $P_1(1, 0), lambda = 0$ and $P_2(2, 0), lambda = -2$.

$x = 0, y gt 0$ This led me to these two critical points: $P_1(0, 1), lambda = 0$ and $P_2(0, 2), lambda = -2$.

$x = 0, y = 0$ This led me to these two critical points: $P_1(0, 0), lambda = 0$ and $P_2(0, 0), lambda lt 2$.

My handbook says the global optima are: $x = 2, y = 0, lambda = -2$ and $x = 0, y = 0, lambda = 0$.

So here I have several questions:

1) How do I distinguish between global and local optima in these cases? The hessian matrix does not seem to be useful in this case. For example, how can I know that $x = 2, y = 0, lambda = -2$ is a global optima, but $x = 1, y = 0, lambda = 0$ is not? Intuition can help, but I want the method.

2) Why would $x = 2, y = 0, lambda = -2$ be a global optima, and $x = 0, y = 2, lambda = -2$ would not?

3) Why would $x = 0, y= 0, lambda = 0$ be a global maxima? It is a bit counterintuitive.

4) Would it be impossible to determine the value of $lambda$ in $P_2$ of $x = 0, y = 0$ combination? (This is the case in which we assume $lambda lt 0$, while in $P_1$ we assumed $lambda = 0$).

5) Feel free to add any information needed to understand why the answer of my handbook is the correct one (which are the global maxima) and not all of above critical points, in case this is not clear with the answers to questions 1, 2, 3 and 4.

Thank you very much for your time.

nonlinear-optimization karush-kuhn-tucker

share|cite|improve this question

asked Jul 25 at 12:00

Ignacio Valdés Zamudio

103

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

We want to maximize $(x-1)^2 + (y-1)^2$ restricted to $x + y le 2$ and $x, y ge 0$.

I tried the following combinations:

$x gt 0, y gt 0$ This led me to no critical point.

$x gt 0, y = 0$ This led me to these two critical points: $P_1(1, 0), lambda = 0$ and $P_2(2, 0), lambda = -2$.

$x = 0, y gt 0$ This led me to these two critical points: $P_1(0, 1), lambda = 0$ and $P_2(0, 2), lambda = -2$.

$x = 0, y = 0$ This led me to these two critical points: $P_1(0, 0), lambda = 0$ and $P_2(0, 0), lambda lt 2$.

My handbook says the global optima are: $x = 2, y = 0, lambda = -2$ and $x = 0, y = 0, lambda = 0$.

So here I have several questions:

1) How do I distinguish between global and local optima in these cases? The hessian matrix does not seem to be useful in this case. For example, how can I know that $x = 2, y = 0, lambda = -2$ is a global optima, but $x = 1, y = 0, lambda = 0$ is not? Intuition can help, but I want the method.

2) Why would $x = 2, y = 0, lambda = -2$ be a global optima, and $x = 0, y = 2, lambda = -2$ would not?

3) Why would $x = 0, y= 0, lambda = 0$ be a global maxima? It is a bit counterintuitive.

4) Would it be impossible to determine the value of $lambda$ in $P_2$ of $x = 0, y = 0$ combination? (This is the case in which we assume $lambda lt 0$, while in $P_1$ we assumed $lambda = 0$).

5) Feel free to add any information needed to understand why the answer of my handbook is the correct one (which are the global maxima) and not all of above critical points, in case this is not clear with the answers to questions 1, 2, 3 and 4.

Thank you very much for your time.

nonlinear-optimization karush-kuhn-tucker

share|cite|improve this question

asked Jul 25 at 12:00

Ignacio Valdés Zamudio

103

We want to maximize $(x-1)^2 + (y-1)^2$ restricted to $x + y le 2$ and $x, y ge 0$.

I tried the following combinations:

$x gt 0, y gt 0$ This led me to no critical point.

$x gt 0, y = 0$ This led me to these two critical points: $P_1(1, 0), lambda = 0$ and $P_2(2, 0), lambda = -2$.

$x = 0, y gt 0$ This led me to these two critical points: $P_1(0, 1), lambda = 0$ and $P_2(0, 2), lambda = -2$.

$x = 0, y = 0$ This led me to these two critical points: $P_1(0, 0), lambda = 0$ and $P_2(0, 0), lambda lt 2$.

My handbook says the global optima are: $x = 2, y = 0, lambda = -2$ and $x = 0, y = 0, lambda = 0$.

So here I have several questions:

1) How do I distinguish between global and local optima in these cases? The hessian matrix does not seem to be useful in this case. For example, how can I know that $x = 2, y = 0, lambda = -2$ is a global optima, but $x = 1, y = 0, lambda = 0$ is not? Intuition can help, but I want the method.

2) Why would $x = 2, y = 0, lambda = -2$ be a global optima, and $x = 0, y = 2, lambda = -2$ would not?

3) Why would $x = 0, y= 0, lambda = 0$ be a global maxima? It is a bit counterintuitive.

4) Would it be impossible to determine the value of $lambda$ in $P_2$ of $x = 0, y = 0$ combination? (This is the case in which we assume $lambda lt 0$, while in $P_1$ we assumed $lambda = 0$).

5) Feel free to add any information needed to understand why the answer of my handbook is the correct one (which are the global maxima) and not all of above critical points, in case this is not clear with the answers to questions 1, 2, 3 and 4.

Thank you very much for your time.

nonlinear-optimization karush-kuhn-tucker

share|cite|improve this question

asked Jul 25 at 12:00

Ignacio Valdés Zamudio

103

share|cite|improve this question

share|cite|improve this question

asked Jul 25 at 12:00

Ignacio Valdés Zamudio

103

asked Jul 25 at 12:00

Ignacio Valdés Zamudio

103

asked Jul 25 at 12:00

Ignacio Valdés Zamudio

103

103

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
0
down vote



accepted

Graphically, refer to the graph:

$hspace0.5cm$

The green area is the feasible region.

The red circle center $C$ and passing through the points $A$, $B$ and $O$ is the largest contour curve. Hence, the global maxima is at $A(0,2)$, $B(2,0)$ and $O(0,0)$.

The red point $C$ is the smallest contour curve. Hence, the global minima is at $C(1,1)$.

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answered Jul 25 at 13:54

farruhota

13.6k2632

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you very much. This graph really helped me to understand the problem.
  – Ignacio Valdés Zamudio
  Jul 25 at 14:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are welcome. If possible, the graphical method is easier to use than algebraic (KKK) method. Good luck.
  – farruhota
  Jul 26 at 10:57
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote



accepted

Graphically, refer to the graph:

$hspace0.5cm$

The green area is the feasible region.

The red circle center $C$ and passing through the points $A$, $B$ and $O$ is the largest contour curve. Hence, the global maxima is at $A(0,2)$, $B(2,0)$ and $O(0,0)$.

The red point $C$ is the smallest contour curve. Hence, the global minima is at $C(1,1)$.

share|cite|improve this answer

answered Jul 25 at 13:54

farruhota

13.6k2632

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you very much. This graph really helped me to understand the problem.
  – Ignacio Valdés Zamudio
  Jul 25 at 14:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are welcome. If possible, the graphical method is easier to use than algebraic (KKK) method. Good luck.
  – farruhota
  Jul 26 at 10:57
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

Graphically, refer to the graph:

$hspace0.5cm$

The green area is the feasible region.

The red circle center $C$ and passing through the points $A$, $B$ and $O$ is the largest contour curve. Hence, the global maxima is at $A(0,2)$, $B(2,0)$ and $O(0,0)$.

The red point $C$ is the smallest contour curve. Hence, the global minima is at $C(1,1)$.

share|cite|improve this answer

answered Jul 25 at 13:54

farruhota

13.6k2632

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you very much. This graph really helped me to understand the problem.
  – Ignacio Valdés Zamudio
  Jul 25 at 14:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are welcome. If possible, the graphical method is easier to use than algebraic (KKK) method. Good luck.
  – farruhota
  Jul 26 at 10:57
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

up vote
0
down vote



accepted

Graphically, refer to the graph:

$hspace0.5cm$

The green area is the feasible region.

The red circle center $C$ and passing through the points $A$, $B$ and $O$ is the largest contour curve. Hence, the global maxima is at $A(0,2)$, $B(2,0)$ and $O(0,0)$.

The red point $C$ is the smallest contour curve. Hence, the global minima is at $C(1,1)$.

share|cite|improve this answer

answered Jul 25 at 13:54

farruhota

13.6k2632

Graphically, refer to the graph:

$hspace0.5cm$

The green area is the feasible region.

The red circle center $C$ and passing through the points $A$, $B$ and $O$ is the largest contour curve. Hence, the global maxima is at $A(0,2)$, $B(2,0)$ and $O(0,0)$.

The red point $C$ is the smallest contour curve. Hence, the global minima is at $C(1,1)$.

share|cite|improve this answer

answered Jul 25 at 13:54

farruhota

13.6k2632

share|cite|improve this answer

share|cite|improve this answer

answered Jul 25 at 13:54

farruhota

13.6k2632

answered Jul 25 at 13:54

farruhota

13.6k2632

answered Jul 25 at 13:54

farruhota

13.6k2632

13.6k2632

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you very much. This graph really helped me to understand the problem.
  – Ignacio Valdés Zamudio
  Jul 25 at 14:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are welcome. If possible, the graphical method is easier to use than algebraic (KKK) method. Good luck.
  – farruhota
  Jul 26 at 10:57
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you very much. This graph really helped me to understand the problem.
  – Ignacio Valdés Zamudio
  Jul 25 at 14:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are welcome. If possible, the graphical method is easier to use than algebraic (KKK) method. Good luck.
  – farruhota
  Jul 26 at 10:57
  

  
  
  
  

Thank you very much. This graph really helped me to understand the problem.
– Ignacio Valdés Zamudio
Jul 25 at 14:12

Thank you very much. This graph really helped me to understand the problem.
– Ignacio Valdés Zamudio
Jul 25 at 14:12

You are welcome. If possible, the graphical method is easier to use than algebraic (KKK) method. Good luck.
– farruhota
Jul 26 at 10:57

You are welcome. If possible, the graphical method is easier to use than algebraic (KKK) method. Good luck.
– farruhota
Jul 26 at 10:57

add a comment | 

 

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