Mixing
Calculate max/min of a 3 variable function, restricted to g(x,y,z)=0
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
Calculate extrema of $f(x,y,z)=xe^yz$ on boundary $3x^2 +y^2 +z^2 =27$
- I did Lagrange multiplier (4 equations 4 variable) but I can't figure out how to solve that system.
$f_x + lambda g_x$,$f_y + lambda g_y,f_z + lambda g_z,g(x,y,z)=0$
$e^yz+lambda (6x)=0$
$xze^yz+lambda (2y)=0$
$xye^yz+lambda (2z)=0$
$3x^2 +y^2 +z^2 -27 = 0$
I tried a lot of combinations but I can't solve this. Can you help me ?
calculus multivariable-calculus lagrange-multiplier
asked Jul 21 at 16:45
NPLS
1819
add a comment |Â
up vote
1
down vote
favorite
Calculate extrema of $f(x,y,z)=xe^yz$ on boundary $3x^2 +y^2 +z^2 =27$
- I did Lagrange multiplier (4 equations 4 variable) but I can't figure out how to solve that system.
$f_x + lambda g_x$,$f_y + lambda g_y,f_z + lambda g_z,g(x,y,z)=0$
$e^yz+lambda (6x)=0$
$xze^yz+lambda (2y)=0$
$xye^yz+lambda (2z)=0$
$3x^2 +y^2 +z^2 -27 = 0$
I tried a lot of combinations but I can't solve this. Can you help me ?
calculus multivariable-calculus lagrange-multiplier
asked Jul 21 at 16:45
NPLS
1819
1
Although books rarely explain it this way, sometimes it's better to think about proportions here, eliminating $lambda$ and getting $$fracf_xg_x = fracf_yg_y = fracf_zg_z.$$ Of course, you have to be careful to consider the possibility that you've tried to divide by $0$.
– Ted Shifrin
Jul 21 at 17:03
I got to that point but I can't figure out how to proceed , like : 1/6x = xz/2y , how Do I need to reason in order to solve this proportion ?
– NPLS
Jul 21 at 17:07
1
So you end up with $$frac13x=fracxzy=fracxyz,$$ assuming none of $x,y,z$ is $0$. Start with the last equality, and you get $z^2=y^2$, so $z=pm y$. Therefore, this reduces to $frac13x=pm x$. Can you finish? With regard to the $0$ worries, $xne 0$, and $y=0$ if and only if $z=0$. So you should check points of the form $(x,0,0)$ that satisfy your constraint.
– Ted Shifrin
Jul 21 at 17:13
1
Thanks that's what I needed!
– NPLS
Jul 21 at 17:16
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Calculate extrema of $f(x,y,z)=xe^yz$ on boundary $3x^2 +y^2 +z^2 =27$
- I did Lagrange multiplier (4 equations 4 variable) but I can't figure out how to solve that system.
$f_x + lambda g_x$,$f_y + lambda g_y,f_z + lambda g_z,g(x,y,z)=0$
$e^yz+lambda (6x)=0$
$xze^yz+lambda (2y)=0$
$xye^yz+lambda (2z)=0$
$3x^2 +y^2 +z^2 -27 = 0$
I tried a lot of combinations but I can't solve this. Can you help me ?
calculus multivariable-calculus lagrange-multiplier
asked Jul 21 at 16:45
NPLS
1819
Calculate extrema of $f(x,y,z)=xe^yz$ on boundary $3x^2 +y^2 +z^2 =27$
- I did Lagrange multiplier (4 equations 4 variable) but I can't figure out how to solve that system.
$f_x + lambda g_x$,$f_y + lambda g_y,f_z + lambda g_z,g(x,y,z)=0$
$e^yz+lambda (6x)=0$
$xze^yz+lambda (2y)=0$
$xye^yz+lambda (2z)=0$
$3x^2 +y^2 +z^2 -27 = 0$
I tried a lot of combinations but I can't solve this. Can you help me ?
calculus multivariable-calculus lagrange-multiplier
asked Jul 21 at 16:45
NPLS
1819
asked Jul 21 at 16:45
NPLS
1819
asked Jul 21 at 16:45
NPLS
1819
asked Jul 21 at 16:45
NPLS
1819
1819
1
Although books rarely explain it this way, sometimes it's better to think about proportions here, eliminating $lambda$ and getting $$fracf_xg_x = fracf_yg_y = fracf_zg_z.$$ Of course, you have to be careful to consider the possibility that you've tried to divide by $0$.
– Ted Shifrin
Jul 21 at 17:03
I got to that point but I can't figure out how to proceed , like : 1/6x = xz/2y , how Do I need to reason in order to solve this proportion ?
– NPLS
Jul 21 at 17:07
1
So you end up with $$frac13x=fracxzy=fracxyz,$$ assuming none of $x,y,z$ is $0$. Start with the last equality, and you get $z^2=y^2$, so $z=pm y$. Therefore, this reduces to $frac13x=pm x$. Can you finish? With regard to the $0$ worries, $xne 0$, and $y=0$ if and only if $z=0$. So you should check points of the form $(x,0,0)$ that satisfy your constraint.
– Ted Shifrin
Jul 21 at 17:13
1
Thanks that's what I needed!
– NPLS
Jul 21 at 17:16
add a comment |Â
1
Although books rarely explain it this way, sometimes it's better to think about proportions here, eliminating $lambda$ and getting $$fracf_xg_x = fracf_yg_y = fracf_zg_z.$$ Of course, you have to be careful to consider the possibility that you've tried to divide by $0$.
– Ted Shifrin
Jul 21 at 17:03
I got to that point but I can't figure out how to proceed , like : 1/6x = xz/2y , how Do I need to reason in order to solve this proportion ?
– NPLS
Jul 21 at 17:07
1
So you end up with $$frac13x=fracxzy=fracxyz,$$ assuming none of $x,y,z$ is $0$. Start with the last equality, and you get $z^2=y^2$, so $z=pm y$. Therefore, this reduces to $frac13x=pm x$. Can you finish? With regard to the $0$ worries, $xne 0$, and $y=0$ if and only if $z=0$. So you should check points of the form $(x,0,0)$ that satisfy your constraint.
– Ted Shifrin
Jul 21 at 17:13
1
Thanks that's what I needed!
– NPLS
Jul 21 at 17:16
1
1
Although books rarely explain it this way, sometimes it's better to think about proportions here, eliminating $lambda$ and getting $$fracf_xg_x = fracf_yg_y = fracf_zg_z.$$ Of course, you have to be careful to consider the possibility that you've tried to divide by $0$.
– Ted Shifrin
Jul 21 at 17:03
Although books rarely explain it this way, sometimes it's better to think about proportions here, eliminating $lambda$ and getting $$fracf_xg_x = fracf_yg_y = fracf_zg_z.$$ Of course, you have to be careful to consider the possibility that you've tried to divide by $0$.
– Ted Shifrin
Jul 21 at 17:03
I got to that point but I can't figure out how to proceed , like : 1/6x = xz/2y , how Do I need to reason in order to solve this proportion ?
– NPLS
Jul 21 at 17:07
I got to that point but I can't figure out how to proceed , like : 1/6x = xz/2y , how Do I need to reason in order to solve this proportion ?
– NPLS
Jul 21 at 17:07
1
1
So you end up with $$frac13x=fracxzy=fracxyz,$$ assuming none of $x,y,z$ is $0$. Start with the last equality, and you get $z^2=y^2$, so $z=pm y$. Therefore, this reduces to $frac13x=pm x$. Can you finish? With regard to the $0$ worries, $xne 0$, and $y=0$ if and only if $z=0$. So you should check points of the form $(x,0,0)$ that satisfy your constraint.
– Ted Shifrin
Jul 21 at 17:13
So you end up with $$frac13x=fracxzy=fracxyz,$$ assuming none of $x,y,z$ is $0$. Start with the last equality, and you get $z^2=y^2$, so $z=pm y$. Therefore, this reduces to $frac13x=pm x$. Can you finish? With regard to the $0$ worries, $xne 0$, and $y=0$ if and only if $z=0$. So you should check points of the form $(x,0,0)$ that satisfy your constraint.
– Ted Shifrin
Jul 21 at 17:13
1
1
Thanks that's what I needed!
– NPLS
Jul 21 at 17:16
Thanks that's what I needed!
– NPLS
Jul 21 at 17:16
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
Hint.
Make $mu = fraclambdae^yz$ and then solve
$$
1+mu 6 x = 0\
x z + mu 2 y = 0\
xy + mu 2z = 0\
3x^2+y^2+z^2-27=0
$$
giving
$$
left[
beginarrayccccc
x & y & z & mu & f \
-3 & 0 & 0 & frac118 & 0 \
3 & 0 & 0 & -frac118 & 0 \
-frac1sqrt3 & -sqrt13 & -sqrt13 & frac12 sqrt3 & -frac13 esqrt3 \
-frac1sqrt3 & sqrt13 & sqrt13 & frac12 sqrt3 & -frac13 esqrt3 \
frac1sqrt3 & -sqrt13 & -sqrt13 & -frac12 sqrt3 & frac13 esqrt3 \
frac1sqrt3 & sqrt13 & sqrt13 & -frac12 sqrt3 & frac13 esqrt3 \
endarray
right]
$$
answered Jul 21 at 16:56
Cesareo
5,7252412
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Hint.
Make $mu = fraclambdae^yz$ and then solve
$$
1+mu 6 x = 0\
x z + mu 2 y = 0\
xy + mu 2z = 0\
3x^2+y^2+z^2-27=0
$$
giving
$$
left[
beginarrayccccc
x & y & z & mu & f \
-3 & 0 & 0 & frac118 & 0 \
3 & 0 & 0 & -frac118 & 0 \
-frac1sqrt3 & -sqrt13 & -sqrt13 & frac12 sqrt3 & -frac13 esqrt3 \
-frac1sqrt3 & sqrt13 & sqrt13 & frac12 sqrt3 & -frac13 esqrt3 \
frac1sqrt3 & -sqrt13 & -sqrt13 & -frac12 sqrt3 & frac13 esqrt3 \
frac1sqrt3 & sqrt13 & sqrt13 & -frac12 sqrt3 & frac13 esqrt3 \
endarray
right]
$$
answered Jul 21 at 16:56
Cesareo
5,7252412
add a comment |Â
up vote
0
down vote
Hint.
Make $mu = fraclambdae^yz$ and then solve
$$
1+mu 6 x = 0\
x z + mu 2 y = 0\
xy + mu 2z = 0\
3x^2+y^2+z^2-27=0
$$
giving
$$
left[
beginarrayccccc
x & y & z & mu & f \
-3 & 0 & 0 & frac118 & 0 \
3 & 0 & 0 & -frac118 & 0 \
-frac1sqrt3 & -sqrt13 & -sqrt13 & frac12 sqrt3 & -frac13 esqrt3 \
-frac1sqrt3 & sqrt13 & sqrt13 & frac12 sqrt3 & -frac13 esqrt3 \
frac1sqrt3 & -sqrt13 & -sqrt13 & -frac12 sqrt3 & frac13 esqrt3 \
frac1sqrt3 & sqrt13 & sqrt13 & -frac12 sqrt3 & frac13 esqrt3 \
endarray
right]
$$
answered Jul 21 at 16:56
Cesareo
5,7252412
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Hint.
Make $mu = fraclambdae^yz$ and then solve
$$
1+mu 6 x = 0\
x z + mu 2 y = 0\
xy + mu 2z = 0\
3x^2+y^2+z^2-27=0
$$
giving
$$
left[
beginarrayccccc
x & y & z & mu & f \
-3 & 0 & 0 & frac118 & 0 \
3 & 0 & 0 & -frac118 & 0 \
-frac1sqrt3 & -sqrt13 & -sqrt13 & frac12 sqrt3 & -frac13 esqrt3 \
-frac1sqrt3 & sqrt13 & sqrt13 & frac12 sqrt3 & -frac13 esqrt3 \
frac1sqrt3 & -sqrt13 & -sqrt13 & -frac12 sqrt3 & frac13 esqrt3 \
frac1sqrt3 & sqrt13 & sqrt13 & -frac12 sqrt3 & frac13 esqrt3 \
endarray
right]
$$
answered Jul 21 at 16:56
Cesareo
5,7252412
Hint.
Make $mu = fraclambdae^yz$ and then solve
$$
1+mu 6 x = 0\
x z + mu 2 y = 0\
xy + mu 2z = 0\
3x^2+y^2+z^2-27=0
$$
giving
$$
left[
beginarrayccccc
x & y & z & mu & f \
-3 & 0 & 0 & frac118 & 0 \
3 & 0 & 0 & -frac118 & 0 \
-frac1sqrt3 & -sqrt13 & -sqrt13 & frac12 sqrt3 & -frac13 esqrt3 \
-frac1sqrt3 & sqrt13 & sqrt13 & frac12 sqrt3 & -frac13 esqrt3 \
frac1sqrt3 & -sqrt13 & -sqrt13 & -frac12 sqrt3 & frac13 esqrt3 \
frac1sqrt3 & sqrt13 & sqrt13 & -frac12 sqrt3 & frac13 esqrt3 \
endarray
right]
$$
answered Jul 21 at 16:56
Cesareo
5,7252412
edited Jul 21 at 17:05
answered Jul 21 at 16:56
Cesareo
5,7252412
answered Jul 21 at 16:56
Cesareo
5,7252412
answered Jul 21 at 16:56
Cesareo
5,7252412
5,7252412
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2858660%2fcalculate-max-min-of-a-3-variable-function-restricted-to-gx-y-z-0%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
Although books rarely explain it this way, sometimes it's better to think about proportions here, eliminating $lambda$ and getting $$fracf_xg_x = fracf_yg_y = fracf_zg_z.$$ Of course, you have to be careful to consider the possibility that you've tried to divide by $0$.
– Ted Shifrin
Jul 21 at 17:03
I got to that point but I can't figure out how to proceed , like : 1/6x = xz/2y , how Do I need to reason in order to solve this proportion ?
– NPLS
Jul 21 at 17:07
1
So you end up with $$frac13x=fracxzy=fracxyz,$$ assuming none of $x,y,z$ is $0$. Start with the last equality, and you get $z^2=y^2$, so $z=pm y$. Therefore, this reduces to $frac13x=pm x$. Can you finish? With regard to the $0$ worries, $xne 0$, and $y=0$ if and only if $z=0$. So you should check points of the form $(x,0,0)$ that satisfy your constraint.
– Ted Shifrin
Jul 21 at 17:13
1
Thanks that's what I needed!
– NPLS
Jul 21 at 17:16