Lagrange multiplier constrictions equality

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if I have $f(x,y,z)=cdots $ and two constrictions $g(x,y,z)$ , $h(x,y,z)$

and I need to calculate the min/max etc.

(1) I know that I can do a Lagrange system with $5$ equations and $5$ variables.

(2) if I do two systems separately (one with $g(x,y,z)$ and the other one with $h(x,y,z)$) will it give me the same results as if I were to do them as point 1?

(3) if my $f(x,y,z)=z$ and let $g(x,y,z)$ be the constraint. Can I still do Lagrange multiplier even though $f_x=0$ and $f_y=0$ ?

multivariable-calculus lagrange-multiplier

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edited Jul 22 at 17:02

Michael Hardy

204k23186462

asked Jul 22 at 15:27

NPLS

1819

add a comment | 

up vote
0
down vote

favorite

if I have $f(x,y,z)=cdots $ and two constrictions $g(x,y,z)$ , $h(x,y,z)$

and I need to calculate the min/max etc.

(1) I know that I can do a Lagrange system with $5$ equations and $5$ variables.

(2) if I do two systems separately (one with $g(x,y,z)$ and the other one with $h(x,y,z)$) will it give me the same results as if I were to do them as point 1?

(3) if my $f(x,y,z)=z$ and let $g(x,y,z)$ be the constraint. Can I still do Lagrange multiplier even though $f_x=0$ and $f_y=0$ ?

multivariable-calculus lagrange-multiplier

share|cite|improve this question

edited Jul 22 at 17:02

Michael Hardy

204k23186462

asked Jul 22 at 15:27

NPLS

1819

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

if I have $f(x,y,z)=cdots $ and two constrictions $g(x,y,z)$ , $h(x,y,z)$

and I need to calculate the min/max etc.

(1) I know that I can do a Lagrange system with $5$ equations and $5$ variables.

(2) if I do two systems separately (one with $g(x,y,z)$ and the other one with $h(x,y,z)$) will it give me the same results as if I were to do them as point 1?

(3) if my $f(x,y,z)=z$ and let $g(x,y,z)$ be the constraint. Can I still do Lagrange multiplier even though $f_x=0$ and $f_y=0$ ?

multivariable-calculus lagrange-multiplier

share|cite|improve this question

edited Jul 22 at 17:02

Michael Hardy

204k23186462

asked Jul 22 at 15:27

NPLS

1819

if I have $f(x,y,z)=cdots $ and two constrictions $g(x,y,z)$ , $h(x,y,z)$

and I need to calculate the min/max etc.

(1) I know that I can do a Lagrange system with $5$ equations and $5$ variables.

(2) if I do two systems separately (one with $g(x,y,z)$ and the other one with $h(x,y,z)$) will it give me the same results as if I were to do them as point 1?

(3) if my $f(x,y,z)=z$ and let $g(x,y,z)$ be the constraint. Can I still do Lagrange multiplier even though $f_x=0$ and $f_y=0$ ?

multivariable-calculus lagrange-multiplier

share|cite|improve this question

edited Jul 22 at 17:02

Michael Hardy

204k23186462

asked Jul 22 at 15:27

NPLS

1819

share|cite|improve this question

share|cite|improve this question

edited Jul 22 at 17:02

Michael Hardy

204k23186462

edited Jul 22 at 17:02

Michael Hardy

204k23186462

edited Jul 22 at 17:02

Michael Hardy

204k23186462

204k23186462

asked Jul 22 at 15:27

NPLS

1819

asked Jul 22 at 15:27

NPLS

1819

asked Jul 22 at 15:27

NPLS

1819

1819

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
0
down vote



accepted

(2) Of course not (in general). Suppose that $f(x,y,z)=x^2+y^2+z^2$, that the first constrain is $x=-1$ and that the second constrain is $y=-1$. If you solve two systems separately, then the solution of the first system will be $(-1,0,0)$, whereas the solution of the second one will be $(0,-1,0)$. None of these points belong to the line $x=y=-1$. The solution will be $(-1,-1,0)$, of course.

(3) Yes, you can.

share|cite|improve this answer

edited Jul 22 at 17:01

answered Jul 22 at 15:32

José Carlos Santos

113k1698176

• 
  
  
  

  
  

  
  
  
  
  
  

  
  In (3): that means that if I found a point is in the form (x,y,z)=(x,y,point_found) or (z)=(point_found) ? I originally asked that question becouse: lets assume I find a possible critical point (z)=(3) and I want to find if it is a max or min extrema , I procede doing the second derivate test ... But wait it doesn't work on f(x,y,z)=z because f_xx =0 = f_yy
  – NPLS
  Jul 22 at 16:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  But that test is for functions defined on an open set. That's not the case here.
  – José Carlos Santos
  Jul 22 at 17:02
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Could you give me an example please? Because I really have trouble understanding the min/max of constrained functions like f(x,y,z)='one variable' , like f(x,y,z)=e^x , f(x,y,z)=y .
  – NPLS
  Jul 22 at 17:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Suppose that your afte that maximum and the minimum of $z$ in the region $x^2+y^2+z^2=1$. Then you solve the system$$left{beginarrayl0=2lambda x\0=2lambda y\1=2lambda z\x^2+y^2+z^2=1.endarrayright.$$This system only has $2$ solutions: $(x,y,z,lambda)=pmleft(0,0,1,frac12right)$. So, you see what's the value of $z$ when $(x,y,z)=(0,0,pm1)$ and you deduce that the maximum and the minimum are attained at $(0,0,1)$ and at $(0,0,-1)$ respectively.
  – José Carlos Santos
  Jul 22 at 17:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  So in order to decide if it is a max or min I only have the 'deduction' method ?
  – NPLS
  Jul 22 at 17:34
  

  
  
  
  

 | 
show 4 more comments

Your Answer

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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

(2) Of course not (in general). Suppose that $f(x,y,z)=x^2+y^2+z^2$, that the first constrain is $x=-1$ and that the second constrain is $y=-1$. If you solve two systems separately, then the solution of the first system will be $(-1,0,0)$, whereas the solution of the second one will be $(0,-1,0)$. None of these points belong to the line $x=y=-1$. The solution will be $(-1,-1,0)$, of course.

(3) Yes, you can.

share|cite|improve this answer

edited Jul 22 at 17:01

answered Jul 22 at 15:32

José Carlos Santos

113k1698176

• 
  
  
  

  
  

  
  
  
  
  
  

  
  In (3): that means that if I found a point is in the form (x,y,z)=(x,y,point_found) or (z)=(point_found) ? I originally asked that question becouse: lets assume I find a possible critical point (z)=(3) and I want to find if it is a max or min extrema , I procede doing the second derivate test ... But wait it doesn't work on f(x,y,z)=z because f_xx =0 = f_yy
  – NPLS
  Jul 22 at 16:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  But that test is for functions defined on an open set. That's not the case here.
  – José Carlos Santos
  Jul 22 at 17:02
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Could you give me an example please? Because I really have trouble understanding the min/max of constrained functions like f(x,y,z)='one variable' , like f(x,y,z)=e^x , f(x,y,z)=y .
  – NPLS
  Jul 22 at 17:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Suppose that your afte that maximum and the minimum of $z$ in the region $x^2+y^2+z^2=1$. Then you solve the system$$left{beginarrayl0=2lambda x\0=2lambda y\1=2lambda z\x^2+y^2+z^2=1.endarrayright.$$This system only has $2$ solutions: $(x,y,z,lambda)=pmleft(0,0,1,frac12right)$. So, you see what's the value of $z$ when $(x,y,z)=(0,0,pm1)$ and you deduce that the maximum and the minimum are attained at $(0,0,1)$ and at $(0,0,-1)$ respectively.
  – José Carlos Santos
  Jul 22 at 17:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  So in order to decide if it is a max or min I only have the 'deduction' method ?
  – NPLS
  Jul 22 at 17:34
  

  
  
  
  

 | 
show 4 more comments

up vote
0
down vote



accepted

(2) Of course not (in general). Suppose that $f(x,y,z)=x^2+y^2+z^2$, that the first constrain is $x=-1$ and that the second constrain is $y=-1$. If you solve two systems separately, then the solution of the first system will be $(-1,0,0)$, whereas the solution of the second one will be $(0,-1,0)$. None of these points belong to the line $x=y=-1$. The solution will be $(-1,-1,0)$, of course.

(3) Yes, you can.

share|cite|improve this answer

edited Jul 22 at 17:01

answered Jul 22 at 15:32

José Carlos Santos

113k1698176

• 
  
  
  

  
  

  
  
  
  
  
  

  
  In (3): that means that if I found a point is in the form (x,y,z)=(x,y,point_found) or (z)=(point_found) ? I originally asked that question becouse: lets assume I find a possible critical point (z)=(3) and I want to find if it is a max or min extrema , I procede doing the second derivate test ... But wait it doesn't work on f(x,y,z)=z because f_xx =0 = f_yy
  – NPLS
  Jul 22 at 16:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  But that test is for functions defined on an open set. That's not the case here.
  – José Carlos Santos
  Jul 22 at 17:02
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Could you give me an example please? Because I really have trouble understanding the min/max of constrained functions like f(x,y,z)='one variable' , like f(x,y,z)=e^x , f(x,y,z)=y .
  – NPLS
  Jul 22 at 17:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Suppose that your afte that maximum and the minimum of $z$ in the region $x^2+y^2+z^2=1$. Then you solve the system$$left{beginarrayl0=2lambda x\0=2lambda y\1=2lambda z\x^2+y^2+z^2=1.endarrayright.$$This system only has $2$ solutions: $(x,y,z,lambda)=pmleft(0,0,1,frac12right)$. So, you see what's the value of $z$ when $(x,y,z)=(0,0,pm1)$ and you deduce that the maximum and the minimum are attained at $(0,0,1)$ and at $(0,0,-1)$ respectively.
  – José Carlos Santos
  Jul 22 at 17:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  So in order to decide if it is a max or min I only have the 'deduction' method ?
  – NPLS
  Jul 22 at 17:34
  

  
  
  
  

 | 
show 4 more comments

up vote
0
down vote



accepted

up vote
0
down vote



accepted

(2) Of course not (in general). Suppose that $f(x,y,z)=x^2+y^2+z^2$, that the first constrain is $x=-1$ and that the second constrain is $y=-1$. If you solve two systems separately, then the solution of the first system will be $(-1,0,0)$, whereas the solution of the second one will be $(0,-1,0)$. None of these points belong to the line $x=y=-1$. The solution will be $(-1,-1,0)$, of course.

(3) Yes, you can.

share|cite|improve this answer

edited Jul 22 at 17:01

answered Jul 22 at 15:32

José Carlos Santos

113k1698176

(2) Of course not (in general). Suppose that $f(x,y,z)=x^2+y^2+z^2$, that the first constrain is $x=-1$ and that the second constrain is $y=-1$. If you solve two systems separately, then the solution of the first system will be $(-1,0,0)$, whereas the solution of the second one will be $(0,-1,0)$. None of these points belong to the line $x=y=-1$. The solution will be $(-1,-1,0)$, of course.

(3) Yes, you can.

share|cite|improve this answer

edited Jul 22 at 17:01

answered Jul 22 at 15:32

José Carlos Santos

113k1698176

share|cite|improve this answer

share|cite|improve this answer

edited Jul 22 at 17:01

edited Jul 22 at 17:01

edited Jul 22 at 17:01

answered Jul 22 at 15:32

José Carlos Santos

113k1698176

answered Jul 22 at 15:32

José Carlos Santos

113k1698176

answered Jul 22 at 15:32

José Carlos Santos

113k1698176

113k1698176

• 
  
  
  

  
  

  
  
  
  
  
  

  
  In (3): that means that if I found a point is in the form (x,y,z)=(x,y,point_found) or (z)=(point_found) ? I originally asked that question becouse: lets assume I find a possible critical point (z)=(3) and I want to find if it is a max or min extrema , I procede doing the second derivate test ... But wait it doesn't work on f(x,y,z)=z because f_xx =0 = f_yy
  – NPLS
  Jul 22 at 16:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  But that test is for functions defined on an open set. That's not the case here.
  – José Carlos Santos
  Jul 22 at 17:02
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Could you give me an example please? Because I really have trouble understanding the min/max of constrained functions like f(x,y,z)='one variable' , like f(x,y,z)=e^x , f(x,y,z)=y .
  – NPLS
  Jul 22 at 17:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Suppose that your afte that maximum and the minimum of $z$ in the region $x^2+y^2+z^2=1$. Then you solve the system$$left{beginarrayl0=2lambda x\0=2lambda y\1=2lambda z\x^2+y^2+z^2=1.endarrayright.$$This system only has $2$ solutions: $(x,y,z,lambda)=pmleft(0,0,1,frac12right)$. So, you see what's the value of $z$ when $(x,y,z)=(0,0,pm1)$ and you deduce that the maximum and the minimum are attained at $(0,0,1)$ and at $(0,0,-1)$ respectively.
  – José Carlos Santos
  Jul 22 at 17:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  So in order to decide if it is a max or min I only have the 'deduction' method ?
  – NPLS
  Jul 22 at 17:34
  

  
  
  
  

 | 
show 4 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  In (3): that means that if I found a point is in the form (x,y,z)=(x,y,point_found) or (z)=(point_found) ? I originally asked that question becouse: lets assume I find a possible critical point (z)=(3) and I want to find if it is a max or min extrema , I procede doing the second derivate test ... But wait it doesn't work on f(x,y,z)=z because f_xx =0 = f_yy
  – NPLS
  Jul 22 at 16:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  But that test is for functions defined on an open set. That's not the case here.
  – José Carlos Santos
  Jul 22 at 17:02
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Could you give me an example please? Because I really have trouble understanding the min/max of constrained functions like f(x,y,z)='one variable' , like f(x,y,z)=e^x , f(x,y,z)=y .
  – NPLS
  Jul 22 at 17:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Suppose that your afte that maximum and the minimum of $z$ in the region $x^2+y^2+z^2=1$. Then you solve the system$$left{beginarrayl0=2lambda x\0=2lambda y\1=2lambda z\x^2+y^2+z^2=1.endarrayright.$$This system only has $2$ solutions: $(x,y,z,lambda)=pmleft(0,0,1,frac12right)$. So, you see what's the value of $z$ when $(x,y,z)=(0,0,pm1)$ and you deduce that the maximum and the minimum are attained at $(0,0,1)$ and at $(0,0,-1)$ respectively.
  – José Carlos Santos
  Jul 22 at 17:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  So in order to decide if it is a max or min I only have the 'deduction' method ?
  – NPLS
  Jul 22 at 17:34
  

  
  
  
  

In (3): that means that if I found a point is in the form (x,y,z)=(x,y,point_found) or (z)=(point_found) ? I originally asked that question becouse: lets assume I find a possible critical point (z)=(3) and I want to find if it is a max or min extrema , I procede doing the second derivate test ... But wait it doesn't work on f(x,y,z)=z because f_xx =0 = f_yy
– NPLS
Jul 22 at 16:53

In (3): that means that if I found a point is in the form (x,y,z)=(x,y,point_found) or (z)=(point_found) ? I originally asked that question becouse: lets assume I find a possible critical point (z)=(3) and I want to find if it is a max or min extrema , I procede doing the second derivate test ... But wait it doesn't work on f(x,y,z)=z because f_xx =0 = f_yy
– NPLS
Jul 22 at 16:53

But that test is for functions defined on an open set. That's not the case here.
– José Carlos Santos
Jul 22 at 17:02

But that test is for functions defined on an open set. That's not the case here.
– José Carlos Santos
Jul 22 at 17:02

Could you give me an example please? Because I really have trouble understanding the min/max of constrained functions like f(x,y,z)='one variable' , like f(x,y,z)=e^x , f(x,y,z)=y .
– NPLS
Jul 22 at 17:17

Could you give me an example please? Because I really have trouble understanding the min/max of constrained functions like f(x,y,z)='one variable' , like f(x,y,z)=e^x , f(x,y,z)=y .
– NPLS
Jul 22 at 17:17

Suppose that your afte that maximum and the minimum of $z$ in the region $x^2+y^2+z^2=1$. Then you solve the system$$left{beginarrayl0=2lambda x\0=2lambda y\1=2lambda z\x^2+y^2+z^2=1.endarrayright.$$This system only has $2$ solutions: $(x,y,z,lambda)=pmleft(0,0,1,frac12right)$. So, you see what's the value of $z$ when $(x,y,z)=(0,0,pm1)$ and you deduce that the maximum and the minimum are attained at $(0,0,1)$ and at $(0,0,-1)$ respectively.
– José Carlos Santos
Jul 22 at 17:23

Suppose that your afte that maximum and the minimum of $z$ in the region $x^2+y^2+z^2=1$. Then you solve the system$$left{beginarrayl0=2lambda x\0=2lambda y\1=2lambda z\x^2+y^2+z^2=1.endarrayright.$$This system only has $2$ solutions: $(x,y,z,lambda)=pmleft(0,0,1,frac12right)$. So, you see what's the value of $z$ when $(x,y,z)=(0,0,pm1)$ and you deduce that the maximum and the minimum are attained at $(0,0,1)$ and at $(0,0,-1)$ respectively.
– José Carlos Santos
Jul 22 at 17:23

So in order to decide if it is a max or min I only have the 'deduction' method ?
– NPLS
Jul 22 at 17:34

So in order to decide if it is a max or min I only have the 'deduction' method ?
– NPLS
Jul 22 at 17:34

 | 
show 4 more comments

 

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