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Why $λ$ in Lagrange-multiplier changes when constraint is scaled?

Simple Lagrange Multiplier Problem, not working outLagrange multiplier or not?Optimization of utility function with Lagrange multiplierLagrange multiplier expressionCan lagrange multiplier(Kuhn tucker multipliers?) change in corner solution?Maximum of $x+y+z$ subject to $frac{a}{x} + frac{b}{y} + frac{c}{z} = 1$ via Lagrange multiplierHow to solve the following optimzation use lagrange multiplier method?When is there a symmetry between constraint and objective function in Lagrange multipliers?Lagrange multiplier for two constraints?Why is the Lagrange multiplier considered to be a variable?

2

$begingroup$

Consider the problem

$$begin{array}{ll} text{maximize} & x^2+y^2 \ text{subject to} & dfrac{x^2}{25} + dfrac{y^2}{9} = 1end{array}$$

Solving this using the Lagrange multiplier method, I get

$$x = pm5, qquad y = 0, qquad lambda = 25$$

However, if I rewrite the constraint as $9x^2+25y^2=225$, I get a different value of $lambda$, namely, $lambda = frac 19$. I am unclear about why this should happen: the constraint is exactly the same, only rewritten after cross multiplication — why should that affect the value of the multiplier? What am I missing?

optimization lagrange-multiplier qcqp

share|cite|improve this question

edited 7 hours ago

user21820

40.1k544161

asked 13 hours ago

PGuptaPGupta

1745

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That shouldn't be the case - maybe check for an arithmetic error?
  $endgroup$
  – Vasting
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Double checked, doesn't seem to be an error.
  $endgroup$
  – PGupta
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Why wouldn't the multiplier's value change? There's no reason that it shouldn't.
  $endgroup$
  – littleO
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Because the constrained is the same. Relaxing the constraint by a small unit should still have the same effect on the value function, shouldn't it?
  $endgroup$
  – PGupta
  12 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

Consider the problem

$$begin{array}{ll} text{maximize} & x^2+y^2 \ text{subject to} & dfrac{x^2}{25} + dfrac{y^2}{9} = 1end{array}$$

Solving this using the Lagrange multiplier method, I get

$$x = pm5, qquad y = 0, qquad lambda = 25$$

However, if I rewrite the constraint as $9x^2+25y^2=225$, I get a different value of $lambda$, namely, $lambda = frac 19$. I am unclear about why this should happen: the constraint is exactly the same, only rewritten after cross multiplication — why should that affect the value of the multiplier? What am I missing?

optimization lagrange-multiplier qcqp

share|cite|improve this question

edited 7 hours ago

user21820

40.1k544161

asked 13 hours ago

PGuptaPGupta

1745

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That shouldn't be the case - maybe check for an arithmetic error?
  $endgroup$
  – Vasting
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Double checked, doesn't seem to be an error.
  $endgroup$
  – PGupta
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Why wouldn't the multiplier's value change? There's no reason that it shouldn't.
  $endgroup$
  – littleO
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Because the constrained is the same. Relaxing the constraint by a small unit should still have the same effect on the value function, shouldn't it?
  $endgroup$
  – PGupta
  12 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Consider the problem

$$begin{array}{ll} text{maximize} & x^2+y^2 \ text{subject to} & dfrac{x^2}{25} + dfrac{y^2}{9} = 1end{array}$$

Solving this using the Lagrange multiplier method, I get

$$x = pm5, qquad y = 0, qquad lambda = 25$$

However, if I rewrite the constraint as $9x^2+25y^2=225$, I get a different value of $lambda$, namely, $lambda = frac 19$. I am unclear about why this should happen: the constraint is exactly the same, only rewritten after cross multiplication — why should that affect the value of the multiplier? What am I missing?

optimization lagrange-multiplier qcqp

share|cite|improve this question

edited 7 hours ago

user21820

40.1k544161

asked 13 hours ago

PGuptaPGupta

1745

$endgroup$

share|cite|improve this question

edited 7 hours ago

user21820

40.1k544161

asked 13 hours ago

PGuptaPGupta

1745

share|cite|improve this question

edited 7 hours ago

user21820

40.1k544161

asked 13 hours ago

PGuptaPGupta

1745

edited 7 hours ago

user21820

40.1k544161

edited 7 hours ago

user21820

40.1k544161

asked 13 hours ago

PGuptaPGupta

1745

asked 13 hours ago

PGuptaPGupta

1745

3 Answers
3

active

oldest

votes

2

$begingroup$

Let $f(x,y)=x^2+y^2+lambda(9x^2+25y^2-225).$

Thus, from $$frac{partial f}{partial x}=2x+18lambda x=0$$ and
$$frac{partial f}{partial y}=2y+50lambda y=0$$ we obtain two possibilities: $lambda=-frac{1}{9}$ or $lambda=-frac{1}{25}.$

The second gives a minimal value, wile the first gives a maximal value:
$$f(x,y)=x^2+y^2-frac{1}{9}(9x^2+25y^2-225)=25-frac{16}{9}y^2leq25,$$ where the equality occurs for $y=0.$

If we consider $f(x,y)=x^2+y^2+lambdaleft(frac{x^2}{25}+frac{y^2}{9}-1right)$ so we'll get $lambda=-25$

and we'll get the same answer of course.

It happens because $-frac{1}{9}cdot225=-25$ and
$$x^2+y^2-frac{1}{9}(9x^2+25y^2-225)=x^2+y^2-25left(frac{x^2}{25}+frac{y^2}{9}-1right).$$

share|cite|improve this answer

edited 12 hours ago

answered 12 hours ago

Michael RozenbergMichael Rozenberg

110k1896201

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, this is what I am doing. My question is, why does the multiplier's value change when I just divide the constraint by 225.
  $endgroup$
  – PGupta
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @PGupta I added something. See now.
  $endgroup$
  – Michael Rozenberg
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This makes sense, thank you.
  $endgroup$
  – PGupta
  12 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

If $x^star$ minimizes $f(x)$ subject to the constraint that $g(x)=0$, then under mild assumptions there exists a Lagrange multiplier $lambda$ that satisfies
$$
tag{1} nabla f(x^star) = lambda nabla g(x^star).
$$
If $g$ is replaced with $c g$, then $x^star$ is still a minimizer, but of course $lambda$ no longer satisfies (1). We must correspondingly multiply $lambda$ by $1/c$ in order for (1) to remain true.

Different but equivalent constraints have different Lagrange multipliers. The way you write the constraint matters.

share|cite|improve this answer

answered 12 hours ago

littleOlittleO

30.4k648111

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Right, this makes it clear. Could you also shed some light on how to interpret this? The multiplier tells us how the value function changes with respect to a small change in the constraint. The maximised value remains the same no matter how we write the constraint. So, relaxing the constraint should change the maximixed value by the same amount--how is rescaling affecting this?
  $endgroup$
  – PGupta
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @PGupta Perturbing the constraint $g(x) = 0$ by a small amount $delta$ is equivalent to perturbing the constraint $cg(x) =0$ by $cdelta$. The change in the optimal value is the same amount $epsilon$ in either case. But the rate of change is $lambda = epsilon/delta$ in the first case, and $lambda/c = epsilon/(c delta)$ in the second case.
  $endgroup$
  – littleO
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Great, got it. Thanks a lot!
  $endgroup$
  – PGupta
  10 hours ago
  

  
  
  
  

add a comment | 

0

$begingroup$

Writing the equation $$frac{x^2}{25}+frac{y^2}{9}=1$$ in the form
$$y^2=9-frac{9}{25}x^2$$ you will have the objective function
$$f(x)=frac{16}{25}x^2+9$$

share|cite|improve this answer

edited 12 hours ago

answered 13 hours ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.7k42867

$endgroup$

add a comment | 

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2

$begingroup$

Let $f(x,y)=x^2+y^2+lambda(9x^2+25y^2-225).$

Thus, from $$frac{partial f}{partial x}=2x+18lambda x=0$$ and
$$frac{partial f}{partial y}=2y+50lambda y=0$$ we obtain two possibilities: $lambda=-frac{1}{9}$ or $lambda=-frac{1}{25}.$

The second gives a minimal value, wile the first gives a maximal value:
$$f(x,y)=x^2+y^2-frac{1}{9}(9x^2+25y^2-225)=25-frac{16}{9}y^2leq25,$$ where the equality occurs for $y=0.$

If we consider $f(x,y)=x^2+y^2+lambdaleft(frac{x^2}{25}+frac{y^2}{9}-1right)$ so we'll get $lambda=-25$

and we'll get the same answer of course.

It happens because $-frac{1}{9}cdot225=-25$ and
$$x^2+y^2-frac{1}{9}(9x^2+25y^2-225)=x^2+y^2-25left(frac{x^2}{25}+frac{y^2}{9}-1right).$$

share|cite|improve this answer

edited 12 hours ago

answered 12 hours ago

Michael RozenbergMichael Rozenberg

110k1896201

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, this is what I am doing. My question is, why does the multiplier's value change when I just divide the constraint by 225.
  $endgroup$
  – PGupta
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @PGupta I added something. See now.
  $endgroup$
  – Michael Rozenberg
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This makes sense, thank you.
  $endgroup$
  – PGupta
  12 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

Let $f(x,y)=x^2+y^2+lambda(9x^2+25y^2-225).$

Thus, from $$frac{partial f}{partial x}=2x+18lambda x=0$$ and
$$frac{partial f}{partial y}=2y+50lambda y=0$$ we obtain two possibilities: $lambda=-frac{1}{9}$ or $lambda=-frac{1}{25}.$

The second gives a minimal value, wile the first gives a maximal value:
$$f(x,y)=x^2+y^2-frac{1}{9}(9x^2+25y^2-225)=25-frac{16}{9}y^2leq25,$$ where the equality occurs for $y=0.$

If we consider $f(x,y)=x^2+y^2+lambdaleft(frac{x^2}{25}+frac{y^2}{9}-1right)$ so we'll get $lambda=-25$

and we'll get the same answer of course.

It happens because $-frac{1}{9}cdot225=-25$ and
$$x^2+y^2-frac{1}{9}(9x^2+25y^2-225)=x^2+y^2-25left(frac{x^2}{25}+frac{y^2}{9}-1right).$$

share|cite|improve this answer

edited 12 hours ago

answered 12 hours ago

Michael RozenbergMichael Rozenberg

110k1896201

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, this is what I am doing. My question is, why does the multiplier's value change when I just divide the constraint by 225.
  $endgroup$
  – PGupta
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @PGupta I added something. See now.
  $endgroup$
  – Michael Rozenberg
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This makes sense, thank you.
  $endgroup$
  – PGupta
  12 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Let $f(x,y)=x^2+y^2+lambda(9x^2+25y^2-225).$

Thus, from $$frac{partial f}{partial x}=2x+18lambda x=0$$ and
$$frac{partial f}{partial y}=2y+50lambda y=0$$ we obtain two possibilities: $lambda=-frac{1}{9}$ or $lambda=-frac{1}{25}.$

The second gives a minimal value, wile the first gives a maximal value:
$$f(x,y)=x^2+y^2-frac{1}{9}(9x^2+25y^2-225)=25-frac{16}{9}y^2leq25,$$ where the equality occurs for $y=0.$

If we consider $f(x,y)=x^2+y^2+lambdaleft(frac{x^2}{25}+frac{y^2}{9}-1right)$ so we'll get $lambda=-25$

and we'll get the same answer of course.

It happens because $-frac{1}{9}cdot225=-25$ and
$$x^2+y^2-frac{1}{9}(9x^2+25y^2-225)=x^2+y^2-25left(frac{x^2}{25}+frac{y^2}{9}-1right).$$

share|cite|improve this answer

edited 12 hours ago

answered 12 hours ago

Michael RozenbergMichael Rozenberg

110k1896201

$endgroup$

share|cite|improve this answer

edited 12 hours ago

answered 12 hours ago

Michael RozenbergMichael Rozenberg

110k1896201

answered 12 hours ago

Michael RozenbergMichael Rozenberg

110k1896201

answered 12 hours ago

Michael RozenbergMichael Rozenberg

110k1896201

4

$begingroup$

If $x^star$ minimizes $f(x)$ subject to the constraint that $g(x)=0$, then under mild assumptions there exists a Lagrange multiplier $lambda$ that satisfies
$$
tag{1} nabla f(x^star) = lambda nabla g(x^star).
$$
If $g$ is replaced with $c g$, then $x^star$ is still a minimizer, but of course $lambda$ no longer satisfies (1). We must correspondingly multiply $lambda$ by $1/c$ in order for (1) to remain true.

Different but equivalent constraints have different Lagrange multipliers. The way you write the constraint matters.

share|cite|improve this answer

answered 12 hours ago

littleOlittleO

30.4k648111

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Right, this makes it clear. Could you also shed some light on how to interpret this? The multiplier tells us how the value function changes with respect to a small change in the constraint. The maximised value remains the same no matter how we write the constraint. So, relaxing the constraint should change the maximixed value by the same amount--how is rescaling affecting this?
  $endgroup$
  – PGupta
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @PGupta Perturbing the constraint $g(x) = 0$ by a small amount $delta$ is equivalent to perturbing the constraint $cg(x) =0$ by $cdelta$. The change in the optimal value is the same amount $epsilon$ in either case. But the rate of change is $lambda = epsilon/delta$ in the first case, and $lambda/c = epsilon/(c delta)$ in the second case.
  $endgroup$
  – littleO
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Great, got it. Thanks a lot!
  $endgroup$
  – PGupta
  10 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

If $x^star$ minimizes $f(x)$ subject to the constraint that $g(x)=0$, then under mild assumptions there exists a Lagrange multiplier $lambda$ that satisfies
$$
tag{1} nabla f(x^star) = lambda nabla g(x^star).
$$
If $g$ is replaced with $c g$, then $x^star$ is still a minimizer, but of course $lambda$ no longer satisfies (1). We must correspondingly multiply $lambda$ by $1/c$ in order for (1) to remain true.

Different but equivalent constraints have different Lagrange multipliers. The way you write the constraint matters.

share|cite|improve this answer

answered 12 hours ago

littleOlittleO

30.4k648111

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Right, this makes it clear. Could you also shed some light on how to interpret this? The multiplier tells us how the value function changes with respect to a small change in the constraint. The maximised value remains the same no matter how we write the constraint. So, relaxing the constraint should change the maximixed value by the same amount--how is rescaling affecting this?
  $endgroup$
  – PGupta
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @PGupta Perturbing the constraint $g(x) = 0$ by a small amount $delta$ is equivalent to perturbing the constraint $cg(x) =0$ by $cdelta$. The change in the optimal value is the same amount $epsilon$ in either case. But the rate of change is $lambda = epsilon/delta$ in the first case, and $lambda/c = epsilon/(c delta)$ in the second case.
  $endgroup$
  – littleO
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Great, got it. Thanks a lot!
  $endgroup$
  – PGupta
  10 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

If $x^star$ minimizes $f(x)$ subject to the constraint that $g(x)=0$, then under mild assumptions there exists a Lagrange multiplier $lambda$ that satisfies
$$
tag{1} nabla f(x^star) = lambda nabla g(x^star).
$$
If $g$ is replaced with $c g$, then $x^star$ is still a minimizer, but of course $lambda$ no longer satisfies (1). We must correspondingly multiply $lambda$ by $1/c$ in order for (1) to remain true.

Different but equivalent constraints have different Lagrange multipliers. The way you write the constraint matters.

share|cite|improve this answer

answered 12 hours ago

littleOlittleO

30.4k648111

$endgroup$

share|cite|improve this answer

answered 12 hours ago

littleOlittleO

30.4k648111

answered 12 hours ago

littleOlittleO

30.4k648111

answered 12 hours ago

littleOlittleO

30.4k648111

0

$begingroup$

Writing the equation $$frac{x^2}{25}+frac{y^2}{9}=1$$ in the form
$$y^2=9-frac{9}{25}x^2$$ you will have the objective function
$$f(x)=frac{16}{25}x^2+9$$

share|cite|improve this answer

edited 12 hours ago

answered 13 hours ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.7k42867

$endgroup$

add a comment | 

0

$begingroup$

Writing the equation $$frac{x^2}{25}+frac{y^2}{9}=1$$ in the form
$$y^2=9-frac{9}{25}x^2$$ you will have the objective function
$$f(x)=frac{16}{25}x^2+9$$

share|cite|improve this answer

edited 12 hours ago

answered 13 hours ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.7k42867

$endgroup$

add a comment | 

$begingroup$

Writing the equation $$frac{x^2}{25}+frac{y^2}{9}=1$$ in the form
$$y^2=9-frac{9}{25}x^2$$ you will have the objective function
$$f(x)=frac{16}{25}x^2+9$$

share|cite|improve this answer

edited 12 hours ago

answered 13 hours ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.7k42867

$endgroup$

share|cite|improve this answer

edited 12 hours ago

answered 13 hours ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.7k42867

answered 13 hours ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.7k42867

answered 13 hours ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.7k42867