Mixing
Find extrema (max,min) of a 3 variable function on a given domain
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$f(x,y,z) = (x^2+y^2)e^z$
with domain = $D=x^2+y^2=4 , (x-2)^2+y^2 +z^2<= 4$
I found that a possible min/max could be the point $(0,0)$ (grad(f)=0..)
- I can see that the boundary of $D$ is $z^2-4x+4=0$ which is the intersection between the cylinder $x^2+y^2=4$ and the sphere $(x-2)^2+y^2 +z^2<= 4$
- So I need to find the min/max on $D$ and on its boundary separately
- I checked the min/max using $Lagrange multiplier $ on the cylinder $x^2+y^2=4$, but here is the problem : The LaGrange system has 3 equation with 4 variables:
$f_x+gamma g_x=0$
$f_y+gamma g_y=0$
$x^2+y^2=4$
I dont know how to behave without a z in my constrain ->$x^2+y^2=4$
- I have the same problem with $z^2-4x+4=0$ if I do lagrange I dont know the $y$
Can you help me figure out the system clearly and find min/max constrained and not. Thx.
EDIT : I managed to find the max/min on the constrain -> I did the LaGrange system of 5 equations , so I used two constrains inside the system (not just one at a time). and I found $(2,0,+-2)$ which is correct. But I still can't figure out how to calculate max/min inside the domain $D$
calculus multivariable-calculus
edited Jul 21 at 13:40
Martin Argerami
116k1071164
asked Jul 21 at 12:11
NPLS
1819
add a comment |Â
up vote
2
down vote
favorite
$f(x,y,z) = (x^2+y^2)e^z$
with domain = $D=x^2+y^2=4 , (x-2)^2+y^2 +z^2<= 4$
I found that a possible min/max could be the point $(0,0)$ (grad(f)=0..)
- I can see that the boundary of $D$ is $z^2-4x+4=0$ which is the intersection between the cylinder $x^2+y^2=4$ and the sphere $(x-2)^2+y^2 +z^2<= 4$
- So I need to find the min/max on $D$ and on its boundary separately
- I checked the min/max using $Lagrange multiplier $ on the cylinder $x^2+y^2=4$, but here is the problem : The LaGrange system has 3 equation with 4 variables:
$f_x+gamma g_x=0$
$f_y+gamma g_y=0$
$x^2+y^2=4$
I dont know how to behave without a z in my constrain ->$x^2+y^2=4$
- I have the same problem with $z^2-4x+4=0$ if I do lagrange I dont know the $y$
Can you help me figure out the system clearly and find min/max constrained and not. Thx.
EDIT : I managed to find the max/min on the constrain -> I did the LaGrange system of 5 equations , so I used two constrains inside the system (not just one at a time). and I found $(2,0,+-2)$ which is correct. But I still can't figure out how to calculate max/min inside the domain $D$
calculus multivariable-calculus
edited Jul 21 at 13:40
Martin Argerami
116k1071164
asked Jul 21 at 12:11
NPLS
1819
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
$f(x,y,z) = (x^2+y^2)e^z$
with domain = $D=x^2+y^2=4 , (x-2)^2+y^2 +z^2<= 4$
I found that a possible min/max could be the point $(0,0)$ (grad(f)=0..)
- I can see that the boundary of $D$ is $z^2-4x+4=0$ which is the intersection between the cylinder $x^2+y^2=4$ and the sphere $(x-2)^2+y^2 +z^2<= 4$
- So I need to find the min/max on $D$ and on its boundary separately
- I checked the min/max using $Lagrange multiplier $ on the cylinder $x^2+y^2=4$, but here is the problem : The LaGrange system has 3 equation with 4 variables:
$f_x+gamma g_x=0$
$f_y+gamma g_y=0$
$x^2+y^2=4$
I dont know how to behave without a z in my constrain ->$x^2+y^2=4$
- I have the same problem with $z^2-4x+4=0$ if I do lagrange I dont know the $y$
Can you help me figure out the system clearly and find min/max constrained and not. Thx.
EDIT : I managed to find the max/min on the constrain -> I did the LaGrange system of 5 equations , so I used two constrains inside the system (not just one at a time). and I found $(2,0,+-2)$ which is correct. But I still can't figure out how to calculate max/min inside the domain $D$
calculus multivariable-calculus
edited Jul 21 at 13:40
Martin Argerami
116k1071164
asked Jul 21 at 12:11
NPLS
1819
$f(x,y,z) = (x^2+y^2)e^z$
with domain = $D=x^2+y^2=4 , (x-2)^2+y^2 +z^2<= 4$
I found that a possible min/max could be the point $(0,0)$ (grad(f)=0..)
- I can see that the boundary of $D$ is $z^2-4x+4=0$ which is the intersection between the cylinder $x^2+y^2=4$ and the sphere $(x-2)^2+y^2 +z^2<= 4$
- So I need to find the min/max on $D$ and on its boundary separately
- I checked the min/max using $Lagrange multiplier $ on the cylinder $x^2+y^2=4$, but here is the problem : The LaGrange system has 3 equation with 4 variables:
$f_x+gamma g_x=0$
$f_y+gamma g_y=0$
$x^2+y^2=4$
I dont know how to behave without a z in my constrain ->$x^2+y^2=4$
- I have the same problem with $z^2-4x+4=0$ if I do lagrange I dont know the $y$
Can you help me figure out the system clearly and find min/max constrained and not. Thx.
EDIT : I managed to find the max/min on the constrain -> I did the LaGrange system of 5 equations , so I used two constrains inside the system (not just one at a time). and I found $(2,0,+-2)$ which is correct. But I still can't figure out how to calculate max/min inside the domain $D$
calculus multivariable-calculus
edited Jul 21 at 13:40
Martin Argerami
116k1071164
asked Jul 21 at 12:11
NPLS
1819
edited Jul 21 at 13:40
Martin Argerami
116k1071164
edited Jul 21 at 13:40
Martin Argerami
116k1071164
edited Jul 21 at 13:40
Martin Argerami
116k1071164
116k1071164
asked Jul 21 at 12:11
NPLS
1819
asked Jul 21 at 12:11
NPLS
1819
asked Jul 21 at 12:11
NPLS
1819
1819
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
In general, the method of Lagrange multipliers only applies to problems with equality constraints. Since this problem has inequality constraints, you would need to use the generalization of Lagrange multipliers, which are called the Karush-Kuhn-Tucker Conditions.
However, here's an alternative ad hoc way to solve this problem. Notice that the term $x^2+y^2$ appears in the objective function, but we have the constraint that $x^2+y^2=4$. Hence, we can reduce our problem to
$$
beginarrayrl
textoptimize & 4e^z \
texts.t. &(x-2)^2+y^2+z^2leqslant 4
endarray
$$
However, we can expand $(x-2)^2=x^2-4x+4$ and use the fact that $x^2+y^2=4$ to obtain
$$
beginarrayrl
textoptimize & 4e^z \
texts.t. &z^2leqslant 4(x-1) \
& -2leqslant xleqslant 2
endarray
$$
(the bounds on $x$ are required to satisfy the constraint $x^2+y^2=4$).
In this form, it's pretty easy to see what the minimum and maximum are in terms of $(x,z)$. For example, to maximize the function, we need $z$ to be as large as possible. Hence, we want $4(x-1)$ to be as large as possible, which happens when $x=2$. For this value of $x$, the largest possible value of $z$ is $2$. Hence the maximum occurs at $(x,y,z)=(2,0,2)$ (you can then solve for $y$ using the relation $x^2+y^2=4$).
answered Jul 21 at 12:57
David M.
1,314318
could you do a few more steps at the end (finding the point)? I'm not quite sure I understood
– NPLS
Jul 21 at 13:57
(I'm not sure if I can link links but whatever) tutorial.math.lamar.edu/Classes/CalcIII/… , if you see the 'example 5' in the link above , he just throw everything in the Lagrange fuction , is his method mathematically correct ?
– NPLS
Jul 21 at 14:06
@NPLS That example is correct because he’s solving a problem that only has equality constraints. He infers some inequality bounds, but they’re not necessary to state explicitly. See my edited answer
– David M.
Jul 21 at 18:17
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
In general, the method of Lagrange multipliers only applies to problems with equality constraints. Since this problem has inequality constraints, you would need to use the generalization of Lagrange multipliers, which are called the Karush-Kuhn-Tucker Conditions.
However, here's an alternative ad hoc way to solve this problem. Notice that the term $x^2+y^2$ appears in the objective function, but we have the constraint that $x^2+y^2=4$. Hence, we can reduce our problem to
$$
beginarrayrl
textoptimize & 4e^z \
texts.t. &(x-2)^2+y^2+z^2leqslant 4
endarray
$$
However, we can expand $(x-2)^2=x^2-4x+4$ and use the fact that $x^2+y^2=4$ to obtain
$$
beginarrayrl
textoptimize & 4e^z \
texts.t. &z^2leqslant 4(x-1) \
& -2leqslant xleqslant 2
endarray
$$
(the bounds on $x$ are required to satisfy the constraint $x^2+y^2=4$).
In this form, it's pretty easy to see what the minimum and maximum are in terms of $(x,z)$. For example, to maximize the function, we need $z$ to be as large as possible. Hence, we want $4(x-1)$ to be as large as possible, which happens when $x=2$. For this value of $x$, the largest possible value of $z$ is $2$. Hence the maximum occurs at $(x,y,z)=(2,0,2)$ (you can then solve for $y$ using the relation $x^2+y^2=4$).
answered Jul 21 at 12:57
David M.
1,314318
could you do a few more steps at the end (finding the point)? I'm not quite sure I understood
– NPLS
Jul 21 at 13:57
(I'm not sure if I can link links but whatever) tutorial.math.lamar.edu/Classes/CalcIII/… , if you see the 'example 5' in the link above , he just throw everything in the Lagrange fuction , is his method mathematically correct ?
– NPLS
Jul 21 at 14:06
@NPLS That example is correct because he’s solving a problem that only has equality constraints. He infers some inequality bounds, but they’re not necessary to state explicitly. See my edited answer
– David M.
Jul 21 at 18:17
add a comment |Â
up vote
2
down vote
accepted
In general, the method of Lagrange multipliers only applies to problems with equality constraints. Since this problem has inequality constraints, you would need to use the generalization of Lagrange multipliers, which are called the Karush-Kuhn-Tucker Conditions.
However, here's an alternative ad hoc way to solve this problem. Notice that the term $x^2+y^2$ appears in the objective function, but we have the constraint that $x^2+y^2=4$. Hence, we can reduce our problem to
$$
beginarrayrl
textoptimize & 4e^z \
texts.t. &(x-2)^2+y^2+z^2leqslant 4
endarray
$$
However, we can expand $(x-2)^2=x^2-4x+4$ and use the fact that $x^2+y^2=4$ to obtain
$$
beginarrayrl
textoptimize & 4e^z \
texts.t. &z^2leqslant 4(x-1) \
& -2leqslant xleqslant 2
endarray
$$
(the bounds on $x$ are required to satisfy the constraint $x^2+y^2=4$).
In this form, it's pretty easy to see what the minimum and maximum are in terms of $(x,z)$. For example, to maximize the function, we need $z$ to be as large as possible. Hence, we want $4(x-1)$ to be as large as possible, which happens when $x=2$. For this value of $x$, the largest possible value of $z$ is $2$. Hence the maximum occurs at $(x,y,z)=(2,0,2)$ (you can then solve for $y$ using the relation $x^2+y^2=4$).
answered Jul 21 at 12:57
David M.
1,314318
could you do a few more steps at the end (finding the point)? I'm not quite sure I understood
– NPLS
Jul 21 at 13:57
(I'm not sure if I can link links but whatever) tutorial.math.lamar.edu/Classes/CalcIII/… , if you see the 'example 5' in the link above , he just throw everything in the Lagrange fuction , is his method mathematically correct ?
– NPLS
Jul 21 at 14:06
@NPLS That example is correct because he’s solving a problem that only has equality constraints. He infers some inequality bounds, but they’re not necessary to state explicitly. See my edited answer
– David M.
Jul 21 at 18:17
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
In general, the method of Lagrange multipliers only applies to problems with equality constraints. Since this problem has inequality constraints, you would need to use the generalization of Lagrange multipliers, which are called the Karush-Kuhn-Tucker Conditions.
However, here's an alternative ad hoc way to solve this problem. Notice that the term $x^2+y^2$ appears in the objective function, but we have the constraint that $x^2+y^2=4$. Hence, we can reduce our problem to
$$
beginarrayrl
textoptimize & 4e^z \
texts.t. &(x-2)^2+y^2+z^2leqslant 4
endarray
$$
However, we can expand $(x-2)^2=x^2-4x+4$ and use the fact that $x^2+y^2=4$ to obtain
$$
beginarrayrl
textoptimize & 4e^z \
texts.t. &z^2leqslant 4(x-1) \
& -2leqslant xleqslant 2
endarray
$$
(the bounds on $x$ are required to satisfy the constraint $x^2+y^2=4$).
In this form, it's pretty easy to see what the minimum and maximum are in terms of $(x,z)$. For example, to maximize the function, we need $z$ to be as large as possible. Hence, we want $4(x-1)$ to be as large as possible, which happens when $x=2$. For this value of $x$, the largest possible value of $z$ is $2$. Hence the maximum occurs at $(x,y,z)=(2,0,2)$ (you can then solve for $y$ using the relation $x^2+y^2=4$).
answered Jul 21 at 12:57
David M.
1,314318
In general, the method of Lagrange multipliers only applies to problems with equality constraints. Since this problem has inequality constraints, you would need to use the generalization of Lagrange multipliers, which are called the Karush-Kuhn-Tucker Conditions.
However, here's an alternative ad hoc way to solve this problem. Notice that the term $x^2+y^2$ appears in the objective function, but we have the constraint that $x^2+y^2=4$. Hence, we can reduce our problem to
$$
beginarrayrl
textoptimize & 4e^z \
texts.t. &(x-2)^2+y^2+z^2leqslant 4
endarray
$$
However, we can expand $(x-2)^2=x^2-4x+4$ and use the fact that $x^2+y^2=4$ to obtain
$$
beginarrayrl
textoptimize & 4e^z \
texts.t. &z^2leqslant 4(x-1) \
& -2leqslant xleqslant 2
endarray
$$
(the bounds on $x$ are required to satisfy the constraint $x^2+y^2=4$).
In this form, it's pretty easy to see what the minimum and maximum are in terms of $(x,z)$. For example, to maximize the function, we need $z$ to be as large as possible. Hence, we want $4(x-1)$ to be as large as possible, which happens when $x=2$. For this value of $x$, the largest possible value of $z$ is $2$. Hence the maximum occurs at $(x,y,z)=(2,0,2)$ (you can then solve for $y$ using the relation $x^2+y^2=4$).
answered Jul 21 at 12:57
David M.
1,314318
edited Jul 21 at 18:21
answered Jul 21 at 12:57
David M.
1,314318
answered Jul 21 at 12:57
David M.
1,314318
answered Jul 21 at 12:57
David M.
1,314318
1,314318
could you do a few more steps at the end (finding the point)? I'm not quite sure I understood
– NPLS
Jul 21 at 13:57
(I'm not sure if I can link links but whatever) tutorial.math.lamar.edu/Classes/CalcIII/… , if you see the 'example 5' in the link above , he just throw everything in the Lagrange fuction , is his method mathematically correct ?
– NPLS
Jul 21 at 14:06
@NPLS That example is correct because he’s solving a problem that only has equality constraints. He infers some inequality bounds, but they’re not necessary to state explicitly. See my edited answer
– David M.
Jul 21 at 18:17
add a comment |Â
could you do a few more steps at the end (finding the point)? I'm not quite sure I understood
– NPLS
Jul 21 at 13:57
(I'm not sure if I can link links but whatever) tutorial.math.lamar.edu/Classes/CalcIII/… , if you see the 'example 5' in the link above , he just throw everything in the Lagrange fuction , is his method mathematically correct ?
– NPLS
Jul 21 at 14:06
@NPLS That example is correct because he’s solving a problem that only has equality constraints. He infers some inequality bounds, but they’re not necessary to state explicitly. See my edited answer
– David M.
Jul 21 at 18:17
could you do a few more steps at the end (finding the point)? I'm not quite sure I understood
– NPLS
Jul 21 at 13:57
could you do a few more steps at the end (finding the point)? I'm not quite sure I understood
– NPLS
Jul 21 at 13:57
(I'm not sure if I can link links but whatever) tutorial.math.lamar.edu/Classes/CalcIII/… , if you see the 'example 5' in the link above , he just throw everything in the Lagrange fuction , is his method mathematically correct ?
– NPLS
Jul 21 at 14:06
(I'm not sure if I can link links but whatever) tutorial.math.lamar.edu/Classes/CalcIII/… , if you see the 'example 5' in the link above , he just throw everything in the Lagrange fuction , is his method mathematically correct ?
– NPLS
Jul 21 at 14:06
@NPLS That example is correct because he’s solving a problem that only has equality constraints. He infers some inequality bounds, but they’re not necessary to state explicitly. See my edited answer
– David M.
Jul 21 at 18:17
@NPLS That example is correct because he’s solving a problem that only has equality constraints. He infers some inequality bounds, but they’re not necessary to state explicitly. See my edited answer
– David M.
Jul 21 at 18:17
add a comment |Â
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