Mixing
Langrarian multiplier [closed]
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Consider the following function
$$f(x, y)=x^4-y^2$$
And Set $A=(x,y)in R^2: x^2+y^2=1$
is required.
- find the Lagrangian equation that determines the extreme point of $F$ on $A$ and calculates the solution for this equation.
- characterize the above equation into local maxima and minima.
- using second order condition explain if there are global maxima and minima.
karush-kuhn-tucker
edited Aug 1 at 15:17
Botond
3,8762632
asked Aug 1 at 14:29
RudY the Heck
42
closed as off-topic by amWhy, LinAlg, John Ma, José Carlos Santos, Adrian Keister Aug 2 at 0:16
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РamWhy, LinAlg, John Ma, Jos̩ Carlos Santos, Adrian Keister
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up vote
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Consider the following function
$$f(x, y)=x^4-y^2$$
And Set $A=(x,y)in R^2: x^2+y^2=1$
is required.
- find the Lagrangian equation that determines the extreme point of $F$ on $A$ and calculates the solution for this equation.
- characterize the above equation into local maxima and minima.
- using second order condition explain if there are global maxima and minima.
karush-kuhn-tucker
edited Aug 1 at 15:17
Botond
3,8762632
asked Aug 1 at 14:29
RudY the Heck
42
closed as off-topic by amWhy, LinAlg, John Ma, José Carlos Santos, Adrian Keister Aug 2 at 0:16
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РamWhy, LinAlg, John Ma, Jos̩ Carlos Santos, Adrian Keister
2
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 1 at 14:32
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Consider the following function
$$f(x, y)=x^4-y^2$$
And Set $A=(x,y)in R^2: x^2+y^2=1$
is required.
- find the Lagrangian equation that determines the extreme point of $F$ on $A$ and calculates the solution for this equation.
- characterize the above equation into local maxima and minima.
- using second order condition explain if there are global maxima and minima.
karush-kuhn-tucker
edited Aug 1 at 15:17
Botond
3,8762632
asked Aug 1 at 14:29
RudY the Heck
42
Consider the following function
$$f(x, y)=x^4-y^2$$
And Set $A=(x,y)in R^2: x^2+y^2=1$
is required.
- find the Lagrangian equation that determines the extreme point of $F$ on $A$ and calculates the solution for this equation.
- characterize the above equation into local maxima and minima.
- using second order condition explain if there are global maxima and minima.
karush-kuhn-tucker
edited Aug 1 at 15:17
Botond
3,8762632
asked Aug 1 at 14:29
RudY the Heck
42
edited Aug 1 at 15:17
Botond
3,8762632
edited Aug 1 at 15:17
Botond
3,8762632
edited Aug 1 at 15:17
Botond
3,8762632
3,8762632
asked Aug 1 at 14:29
RudY the Heck
42
asked Aug 1 at 14:29
RudY the Heck
42
asked Aug 1 at 14:29
RudY the Heck
42
42
closed as off-topic by amWhy, LinAlg, John Ma, José Carlos Santos, Adrian Keister Aug 2 at 0:16
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РamWhy, LinAlg, John Ma, Jos̩ Carlos Santos, Adrian Keister
closed as off-topic by amWhy, LinAlg, John Ma, José Carlos Santos, Adrian Keister Aug 2 at 0:16
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РamWhy, LinAlg, John Ma, Jos̩ Carlos Santos, Adrian Keister
2
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 1 at 14:32
add a comment |Â
2
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 1 at 14:32
2
2
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 1 at 14:32
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 1 at 14:32
add a comment |Â
2 Answers
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1
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$nabla (x^4 - y^2) = nabla(lambda(x^2+y^2 - 1))$
$4x^3 = 2lambda x\
-2y = 2lambda y$
From the second equation we get: $lambda = -1 text or y = 0$
if $lambda = -1\
4x^3 = -2x\
x = 0$
Maxima -- $(1,0),(-1,0)$
Minima -- $(0,1),(0,-1)$
answered Aug 1 at 15:13
Doug M
39k31749
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We know that $$nabla f=(4x^3,-2y)$$and$$c(x,y)=x^2+y^2-1=0$$therefore the Lagrangian would be $$L(x,y)=nabla f+lambdanabla c=(4x^3,-2y)+lambda(2x,2y)$$also the Hessian of the function is $$H_f=beginbmatrix12x^2&0\0&-2endbmatrix$$for finding maxima and minima we have $$4x^3=2lambda x\-2y=2lambda y$$if $y=0$ we have $x=pm 1$ and $lambda=2$. The only feasible direction in these points is $vec d=(0,d_2)$ for any $d_2inBbb R-0$ from the other side$$dH_f(pm 1,0)d^T=-2d_2^2<0$$which means that the points $(pm 1,0)$ are local maxima.
If $lambda=-1$ we obtain $$4x^3=-2x$$ the only possible candidates are $(0,pm 1)$ that both are local minima similarly. Here is a sketch:
answered Aug 1 at 18:32
Mostafa Ayaz
8,5183630
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
$nabla (x^4 - y^2) = nabla(lambda(x^2+y^2 - 1))$
$4x^3 = 2lambda x\
-2y = 2lambda y$
From the second equation we get: $lambda = -1 text or y = 0$
if $lambda = -1\
4x^3 = -2x\
x = 0$
Maxima -- $(1,0),(-1,0)$
Minima -- $(0,1),(0,-1)$
answered Aug 1 at 15:13
Doug M
39k31749
add a comment |Â
up vote
1
down vote
$nabla (x^4 - y^2) = nabla(lambda(x^2+y^2 - 1))$
$4x^3 = 2lambda x\
-2y = 2lambda y$
From the second equation we get: $lambda = -1 text or y = 0$
if $lambda = -1\
4x^3 = -2x\
x = 0$
Maxima -- $(1,0),(-1,0)$
Minima -- $(0,1),(0,-1)$
answered Aug 1 at 15:13
Doug M
39k31749
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$nabla (x^4 - y^2) = nabla(lambda(x^2+y^2 - 1))$
$4x^3 = 2lambda x\
-2y = 2lambda y$
From the second equation we get: $lambda = -1 text or y = 0$
if $lambda = -1\
4x^3 = -2x\
x = 0$
Maxima -- $(1,0),(-1,0)$
Minima -- $(0,1),(0,-1)$
answered Aug 1 at 15:13
Doug M
39k31749
$nabla (x^4 - y^2) = nabla(lambda(x^2+y^2 - 1))$
$4x^3 = 2lambda x\
-2y = 2lambda y$
From the second equation we get: $lambda = -1 text or y = 0$
if $lambda = -1\
4x^3 = -2x\
x = 0$
Maxima -- $(1,0),(-1,0)$
Minima -- $(0,1),(0,-1)$
answered Aug 1 at 15:13
Doug M
39k31749
answered Aug 1 at 15:13
Doug M
39k31749
answered Aug 1 at 15:13
Doug M
39k31749
answered Aug 1 at 15:13
Doug M
39k31749
39k31749
add a comment |Â
add a comment |Â
up vote
0
down vote
We know that $$nabla f=(4x^3,-2y)$$and$$c(x,y)=x^2+y^2-1=0$$therefore the Lagrangian would be $$L(x,y)=nabla f+lambdanabla c=(4x^3,-2y)+lambda(2x,2y)$$also the Hessian of the function is $$H_f=beginbmatrix12x^2&0\0&-2endbmatrix$$for finding maxima and minima we have $$4x^3=2lambda x\-2y=2lambda y$$if $y=0$ we have $x=pm 1$ and $lambda=2$. The only feasible direction in these points is $vec d=(0,d_2)$ for any $d_2inBbb R-0$ from the other side$$dH_f(pm 1,0)d^T=-2d_2^2<0$$which means that the points $(pm 1,0)$ are local maxima.
If $lambda=-1$ we obtain $$4x^3=-2x$$ the only possible candidates are $(0,pm 1)$ that both are local minima similarly. Here is a sketch:
answered Aug 1 at 18:32
Mostafa Ayaz
8,5183630
add a comment |Â
up vote
0
down vote
We know that $$nabla f=(4x^3,-2y)$$and$$c(x,y)=x^2+y^2-1=0$$therefore the Lagrangian would be $$L(x,y)=nabla f+lambdanabla c=(4x^3,-2y)+lambda(2x,2y)$$also the Hessian of the function is $$H_f=beginbmatrix12x^2&0\0&-2endbmatrix$$for finding maxima and minima we have $$4x^3=2lambda x\-2y=2lambda y$$if $y=0$ we have $x=pm 1$ and $lambda=2$. The only feasible direction in these points is $vec d=(0,d_2)$ for any $d_2inBbb R-0$ from the other side$$dH_f(pm 1,0)d^T=-2d_2^2<0$$which means that the points $(pm 1,0)$ are local maxima.
If $lambda=-1$ we obtain $$4x^3=-2x$$ the only possible candidates are $(0,pm 1)$ that both are local minima similarly. Here is a sketch:
answered Aug 1 at 18:32
Mostafa Ayaz
8,5183630
add a comment |Â
up vote
0
down vote
up vote
0
down vote
We know that $$nabla f=(4x^3,-2y)$$and$$c(x,y)=x^2+y^2-1=0$$therefore the Lagrangian would be $$L(x,y)=nabla f+lambdanabla c=(4x^3,-2y)+lambda(2x,2y)$$also the Hessian of the function is $$H_f=beginbmatrix12x^2&0\0&-2endbmatrix$$for finding maxima and minima we have $$4x^3=2lambda x\-2y=2lambda y$$if $y=0$ we have $x=pm 1$ and $lambda=2$. The only feasible direction in these points is $vec d=(0,d_2)$ for any $d_2inBbb R-0$ from the other side$$dH_f(pm 1,0)d^T=-2d_2^2<0$$which means that the points $(pm 1,0)$ are local maxima.
If $lambda=-1$ we obtain $$4x^3=-2x$$ the only possible candidates are $(0,pm 1)$ that both are local minima similarly. Here is a sketch:
answered Aug 1 at 18:32
Mostafa Ayaz
8,5183630
We know that $$nabla f=(4x^3,-2y)$$and$$c(x,y)=x^2+y^2-1=0$$therefore the Lagrangian would be $$L(x,y)=nabla f+lambdanabla c=(4x^3,-2y)+lambda(2x,2y)$$also the Hessian of the function is $$H_f=beginbmatrix12x^2&0\0&-2endbmatrix$$for finding maxima and minima we have $$4x^3=2lambda x\-2y=2lambda y$$if $y=0$ we have $x=pm 1$ and $lambda=2$. The only feasible direction in these points is $vec d=(0,d_2)$ for any $d_2inBbb R-0$ from the other side$$dH_f(pm 1,0)d^T=-2d_2^2<0$$which means that the points $(pm 1,0)$ are local maxima.
If $lambda=-1$ we obtain $$4x^3=-2x$$ the only possible candidates are $(0,pm 1)$ that both are local minima similarly. Here is a sketch:
answered Aug 1 at 18:32
Mostafa Ayaz
8,5183630
answered Aug 1 at 18:32
Mostafa Ayaz
8,5183630
answered Aug 1 at 18:32
Mostafa Ayaz
8,5183630
answered Aug 1 at 18:32
Mostafa Ayaz
8,5183630
8,5183630
add a comment |Â
add a comment |Â
2
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 1 at 14:32