Why does the Lagrange multiplier $lambda$ change when the equality constraint is scaled?
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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 constraints qcqp
$endgroup$
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 constraints qcqp
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$begingroup$
That shouldn't be the case - maybe check for an arithmetic error?
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– Vasting
2 days ago
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Double checked, doesn't seem to be an error.
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– PGupta
2 days ago
1
$begingroup$
Why wouldn't the multiplier's value change? There's no reason that it shouldn't.
$endgroup$
– littleO
2 days 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
2 days 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 constraints qcqp
$endgroup$
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 constraints qcqp
optimization lagrange-multiplier constraints qcqp
edited 2 days ago
Rodrigo de Azevedo
13.2k41962
13.2k41962
asked Apr 7 at 6:29
PGuptaPGupta
1746
1746
$begingroup$
That shouldn't be the case - maybe check for an arithmetic error?
$endgroup$
– Vasting
2 days ago
$begingroup$
Double checked, doesn't seem to be an error.
$endgroup$
– PGupta
2 days ago
1
$begingroup$
Why wouldn't the multiplier's value change? There's no reason that it shouldn't.
$endgroup$
– littleO
2 days 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
2 days ago
add a comment |
$begingroup$
That shouldn't be the case - maybe check for an arithmetic error?
$endgroup$
– Vasting
2 days ago
$begingroup$
Double checked, doesn't seem to be an error.
$endgroup$
– PGupta
2 days ago
1
$begingroup$
Why wouldn't the multiplier's value change? There's no reason that it shouldn't.
$endgroup$
– littleO
2 days 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
2 days ago
$begingroup$
That shouldn't be the case - maybe check for an arithmetic error?
$endgroup$
– Vasting
2 days ago
$begingroup$
That shouldn't be the case - maybe check for an arithmetic error?
$endgroup$
– Vasting
2 days ago
$begingroup$
Double checked, doesn't seem to be an error.
$endgroup$
– PGupta
2 days ago
$begingroup$
Double checked, doesn't seem to be an error.
$endgroup$
– PGupta
2 days ago
1
1
$begingroup$
Why wouldn't the multiplier's value change? There's no reason that it shouldn't.
$endgroup$
– littleO
2 days ago
$begingroup$
Why wouldn't the multiplier's value change? There's no reason that it shouldn't.
$endgroup$
– littleO
2 days 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
2 days 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
2 days ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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
$$x^2+y^2-frac{1}{9}(9x^2+25y^2-225)=x^2+y^2-25left(frac{x^2}{25}+frac{y^2}{9}-1right).$$
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$begingroup$
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
2 days ago
$begingroup$
@PGupta I added something. See now.
$endgroup$
– Michael Rozenberg
2 days ago
$begingroup$
This makes sense, thank you.
$endgroup$
– PGupta
2 days 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.
$endgroup$
$begingroup$
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
2 days 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.
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– littleO
2 days ago
$begingroup$
Great, got it. Thanks a lot!
$endgroup$
– PGupta
2 days ago
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
$$f(x)=frac{16}{25}x^2+9$$
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
$$x^2+y^2-frac{1}{9}(9x^2+25y^2-225)=x^2+y^2-25left(frac{x^2}{25}+frac{y^2}{9}-1right).$$
$endgroup$
$begingroup$
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
2 days ago
$begingroup$
@PGupta I added something. See now.
$endgroup$
– Michael Rozenberg
2 days ago
$begingroup$
This makes sense, thank you.
$endgroup$
– PGupta
2 days 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
$$x^2+y^2-frac{1}{9}(9x^2+25y^2-225)=x^2+y^2-25left(frac{x^2}{25}+frac{y^2}{9}-1right).$$
$endgroup$
$begingroup$
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
2 days ago
$begingroup$
@PGupta I added something. See now.
$endgroup$
– Michael Rozenberg
2 days ago
$begingroup$
This makes sense, thank you.
$endgroup$
– PGupta
2 days 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
$$x^2+y^2-frac{1}{9}(9x^2+25y^2-225)=x^2+y^2-25left(frac{x^2}{25}+frac{y^2}{9}-1right).$$
$endgroup$
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
$$x^2+y^2-frac{1}{9}(9x^2+25y^2-225)=x^2+y^2-25left(frac{x^2}{25}+frac{y^2}{9}-1right).$$
edited 2 days ago
answered 2 days ago
Michael RozenbergMichael Rozenberg
110k1896201
110k1896201
$begingroup$
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
2 days ago
$begingroup$
@PGupta I added something. See now.
$endgroup$
– Michael Rozenberg
2 days ago
$begingroup$
This makes sense, thank you.
$endgroup$
– PGupta
2 days ago
add a comment |
$begingroup$
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
2 days ago
$begingroup$
@PGupta I added something. See now.
$endgroup$
– Michael Rozenberg
2 days ago
$begingroup$
This makes sense, thank you.
$endgroup$
– PGupta
2 days ago
$begingroup$
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
2 days ago
$begingroup$
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
2 days ago
$begingroup$
@PGupta I added something. See now.
$endgroup$
– Michael Rozenberg
2 days ago
$begingroup$
@PGupta I added something. See now.
$endgroup$
– Michael Rozenberg
2 days ago
$begingroup$
This makes sense, thank you.
$endgroup$
– PGupta
2 days ago
$begingroup$
This makes sense, thank you.
$endgroup$
– PGupta
2 days 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.
$endgroup$
$begingroup$
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
2 days 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
2 days ago
$begingroup$
Great, got it. Thanks a lot!
$endgroup$
– PGupta
2 days 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.
$endgroup$
$begingroup$
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
2 days 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
2 days ago
$begingroup$
Great, got it. Thanks a lot!
$endgroup$
– PGupta
2 days 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.
$endgroup$
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.
answered 2 days ago
littleOlittleO
30.5k649111
30.5k649111
$begingroup$
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
2 days 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
2 days ago
$begingroup$
Great, got it. Thanks a lot!
$endgroup$
– PGupta
2 days ago
add a comment |
$begingroup$
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
2 days 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
2 days ago
$begingroup$
Great, got it. Thanks a lot!
$endgroup$
– PGupta
2 days ago
$begingroup$
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
2 days ago
$begingroup$
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
2 days ago
1
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
2 days ago
$begingroup$
@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
2 days ago
$begingroup$
Great, got it. Thanks a lot!
$endgroup$
– PGupta
2 days ago
$begingroup$
Great, got it. Thanks a lot!
$endgroup$
– PGupta
2 days ago
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
$$f(x)=frac{16}{25}x^2+9$$
$endgroup$
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
$$f(x)=frac{16}{25}x^2+9$$
$endgroup$
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
$$f(x)=frac{16}{25}x^2+9$$
$endgroup$
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
$$f(x)=frac{16}{25}x^2+9$$
edited 2 days ago
answered 2 days ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.8k42867
78.8k42867
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$begingroup$
That shouldn't be the case - maybe check for an arithmetic error?
$endgroup$
– Vasting
2 days ago
$begingroup$
Double checked, doesn't seem to be an error.
$endgroup$
– PGupta
2 days ago
1
$begingroup$
Why wouldn't the multiplier's value change? There's no reason that it shouldn't.
$endgroup$
– littleO
2 days 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
2 days ago