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Solve this Lagrange system with $f(x,y,z)=e^x$
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I have $f(x,y,z)=e^x$
and its constriction as follow: $g(x,y,z)=x^2+z^2-2=0$
$f_x = e^x$, $g_x = 2x$
$f_y = 0$, $g_y = 0$
$f_z = 0$, $g_z = 2z$
The system is:
$e^x +lambda 2x = 0$
$0 = 0$
$lambda 2z = 0$
$x^2+z^2-2 = 0$
My thought process is $rightarrow$ I see that the third equation is TRUE for $lambda=0$ or $z=0$, I then put $z=0$ in the fourth equation giving: $x^2 = 2$ so $x=pmsqrt2$
from here I tend to say that I found a point in $(x,y,z)=(pmsqrt2,0,0)$, I'd infer that ($(sqrt2,0,0)$) is a maximum and ($(-sqrt2,0,0)$) is a minimum. I know that I did something wrong (moreover I didnt even touched the first equation)
- So can you help me understand how to solve this system and giving me your thought process(not just a straight resolution!)?
- in this case the second equation is $0=0$ , what does it mean ? does It give me some useful informations?
calculus multivariable-calculus lagrange-multiplier
edited Jul 23 at 9:28
mechanodroid
22.2k52041
asked Jul 22 at 22:55
NPLS
1819
add a comment |Â
up vote
1
down vote
favorite
I have $f(x,y,z)=e^x$
and its constriction as follow: $g(x,y,z)=x^2+z^2-2=0$
$f_x = e^x$, $g_x = 2x$
$f_y = 0$, $g_y = 0$
$f_z = 0$, $g_z = 2z$
The system is:
$e^x +lambda 2x = 0$
$0 = 0$
$lambda 2z = 0$
$x^2+z^2-2 = 0$
My thought process is $rightarrow$ I see that the third equation is TRUE for $lambda=0$ or $z=0$, I then put $z=0$ in the fourth equation giving: $x^2 = 2$ so $x=pmsqrt2$
from here I tend to say that I found a point in $(x,y,z)=(pmsqrt2,0,0)$, I'd infer that ($(sqrt2,0,0)$) is a maximum and ($(-sqrt2,0,0)$) is a minimum. I know that I did something wrong (moreover I didnt even touched the first equation)
- So can you help me understand how to solve this system and giving me your thought process(not just a straight resolution!)?
- in this case the second equation is $0=0$ , what does it mean ? does It give me some useful informations?
calculus multivariable-calculus lagrange-multiplier
edited Jul 23 at 9:28
mechanodroid
22.2k52041
asked Jul 22 at 22:55
NPLS
1819
Hence there is an exponential function within the first equation it cannot be equal zero unless $lambda$ has a weird form like $lambda~=~c-frace^x2x$. So I am not sure what to conclude form this.
– mrtaurho
Jul 22 at 23:02
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have $f(x,y,z)=e^x$
and its constriction as follow: $g(x,y,z)=x^2+z^2-2=0$
$f_x = e^x$, $g_x = 2x$
$f_y = 0$, $g_y = 0$
$f_z = 0$, $g_z = 2z$
The system is:
$e^x +lambda 2x = 0$
$0 = 0$
$lambda 2z = 0$
$x^2+z^2-2 = 0$
My thought process is $rightarrow$ I see that the third equation is TRUE for $lambda=0$ or $z=0$, I then put $z=0$ in the fourth equation giving: $x^2 = 2$ so $x=pmsqrt2$
from here I tend to say that I found a point in $(x,y,z)=(pmsqrt2,0,0)$, I'd infer that ($(sqrt2,0,0)$) is a maximum and ($(-sqrt2,0,0)$) is a minimum. I know that I did something wrong (moreover I didnt even touched the first equation)
- So can you help me understand how to solve this system and giving me your thought process(not just a straight resolution!)?
- in this case the second equation is $0=0$ , what does it mean ? does It give me some useful informations?
calculus multivariable-calculus lagrange-multiplier
edited Jul 23 at 9:28
mechanodroid
22.2k52041
asked Jul 22 at 22:55
NPLS
1819
I have $f(x,y,z)=e^x$
and its constriction as follow: $g(x,y,z)=x^2+z^2-2=0$
$f_x = e^x$, $g_x = 2x$
$f_y = 0$, $g_y = 0$
$f_z = 0$, $g_z = 2z$
The system is:
$e^x +lambda 2x = 0$
$0 = 0$
$lambda 2z = 0$
$x^2+z^2-2 = 0$
My thought process is $rightarrow$ I see that the third equation is TRUE for $lambda=0$ or $z=0$, I then put $z=0$ in the fourth equation giving: $x^2 = 2$ so $x=pmsqrt2$
from here I tend to say that I found a point in $(x,y,z)=(pmsqrt2,0,0)$, I'd infer that ($(sqrt2,0,0)$) is a maximum and ($(-sqrt2,0,0)$) is a minimum. I know that I did something wrong (moreover I didnt even touched the first equation)
- So can you help me understand how to solve this system and giving me your thought process(not just a straight resolution!)?
- in this case the second equation is $0=0$ , what does it mean ? does It give me some useful informations?
calculus multivariable-calculus lagrange-multiplier
edited Jul 23 at 9:28
mechanodroid
22.2k52041
asked Jul 22 at 22:55
NPLS
1819
edited Jul 23 at 9:28
mechanodroid
22.2k52041
edited Jul 23 at 9:28
mechanodroid
22.2k52041
edited Jul 23 at 9:28
mechanodroid
22.2k52041
22.2k52041
asked Jul 22 at 22:55
NPLS
1819
asked Jul 22 at 22:55
NPLS
1819
asked Jul 22 at 22:55
NPLS
1819
1819
Hence there is an exponential function within the first equation it cannot be equal zero unless $lambda$ has a weird form like $lambda~=~c-frace^x2x$. So I am not sure what to conclude form this.
– mrtaurho
Jul 22 at 23:02
add a comment |Â
Hence there is an exponential function within the first equation it cannot be equal zero unless $lambda$ has a weird form like $lambda~=~c-frace^x2x$. So I am not sure what to conclude form this.
– mrtaurho
Jul 22 at 23:02
Hence there is an exponential function within the first equation it cannot be equal zero unless $lambda$ has a weird form like $lambda~=~c-frace^x2x$. So I am not sure what to conclude form this.
– mrtaurho
Jul 22 at 23:02
Hence there is an exponential function within the first equation it cannot be equal zero unless $lambda$ has a weird form like $lambda~=~c-frace^x2x$. So I am not sure what to conclude form this.
– mrtaurho
Jul 22 at 23:02
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
From the third equation, what you get is that $lambda=0$ or that $z=0$. But $lambda$ cannot be $0$, because of the first equation. Therefore, $z=0$. And so, $x=pmsqrt2$. So, the solutions of the system are the points of the form $left(pmsqrt2,y,0right)$, with $yinmathbb R$. Among these points, $f$ attains its maximum ($e^sqrt2$) at the points of the form $left(sqrt2,y,0right)$ and the minimum $(e^-sqrt2$) at the points of the form $left(-sqrt2,y,0right)$.
answered Jul 22 at 23:09
José Carlos Santos
113k1698176
Thank you again :)
– NPLS
Jul 22 at 23:12
I'm glad I could help.
– José Carlos Santos
Jul 22 at 23:13
add a comment |Â
up vote
1
down vote
You also had to check whether $lambda = 0$ is a valid solution. It is not because plugging in $lambda = 0$ into the first equation gives $e^x = 0$ which has no solutions.
This is where you use the first equation. The second equation $0 = 0$ indeed doesn't give any useful information so you can ignore it.
Therefore, $x= pmsqrt2$ and $z = 0$ so the extremal points are $(pmsqrt2, y, 0)$ for $y in mathbbR$.
answered Jul 22 at 23:08
mechanodroid
22.2k52041
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
From the third equation, what you get is that $lambda=0$ or that $z=0$. But $lambda$ cannot be $0$, because of the first equation. Therefore, $z=0$. And so, $x=pmsqrt2$. So, the solutions of the system are the points of the form $left(pmsqrt2,y,0right)$, with $yinmathbb R$. Among these points, $f$ attains its maximum ($e^sqrt2$) at the points of the form $left(sqrt2,y,0right)$ and the minimum $(e^-sqrt2$) at the points of the form $left(-sqrt2,y,0right)$.
answered Jul 22 at 23:09
José Carlos Santos
113k1698176
Thank you again :)
– NPLS
Jul 22 at 23:12
I'm glad I could help.
– José Carlos Santos
Jul 22 at 23:13
add a comment |Â
up vote
1
down vote
accepted
From the third equation, what you get is that $lambda=0$ or that $z=0$. But $lambda$ cannot be $0$, because of the first equation. Therefore, $z=0$. And so, $x=pmsqrt2$. So, the solutions of the system are the points of the form $left(pmsqrt2,y,0right)$, with $yinmathbb R$. Among these points, $f$ attains its maximum ($e^sqrt2$) at the points of the form $left(sqrt2,y,0right)$ and the minimum $(e^-sqrt2$) at the points of the form $left(-sqrt2,y,0right)$.
answered Jul 22 at 23:09
José Carlos Santos
113k1698176
Thank you again :)
– NPLS
Jul 22 at 23:12
I'm glad I could help.
– José Carlos Santos
Jul 22 at 23:13
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
From the third equation, what you get is that $lambda=0$ or that $z=0$. But $lambda$ cannot be $0$, because of the first equation. Therefore, $z=0$. And so, $x=pmsqrt2$. So, the solutions of the system are the points of the form $left(pmsqrt2,y,0right)$, with $yinmathbb R$. Among these points, $f$ attains its maximum ($e^sqrt2$) at the points of the form $left(sqrt2,y,0right)$ and the minimum $(e^-sqrt2$) at the points of the form $left(-sqrt2,y,0right)$.
answered Jul 22 at 23:09
José Carlos Santos
113k1698176
From the third equation, what you get is that $lambda=0$ or that $z=0$. But $lambda$ cannot be $0$, because of the first equation. Therefore, $z=0$. And so, $x=pmsqrt2$. So, the solutions of the system are the points of the form $left(pmsqrt2,y,0right)$, with $yinmathbb R$. Among these points, $f$ attains its maximum ($e^sqrt2$) at the points of the form $left(sqrt2,y,0right)$ and the minimum $(e^-sqrt2$) at the points of the form $left(-sqrt2,y,0right)$.
answered Jul 22 at 23:09
José Carlos Santos
113k1698176
answered Jul 22 at 23:09
José Carlos Santos
113k1698176
answered Jul 22 at 23:09
José Carlos Santos
113k1698176
answered Jul 22 at 23:09
José Carlos Santos
113k1698176
113k1698176
Thank you again :)
– NPLS
Jul 22 at 23:12
I'm glad I could help.
– José Carlos Santos
Jul 22 at 23:13
add a comment |Â
Thank you again :)
– NPLS
Jul 22 at 23:12
I'm glad I could help.
– José Carlos Santos
Jul 22 at 23:13
Thank you again :)
– NPLS
Jul 22 at 23:12
Thank you again :)
– NPLS
Jul 22 at 23:12
I'm glad I could help.
– José Carlos Santos
Jul 22 at 23:13
I'm glad I could help.
– José Carlos Santos
Jul 22 at 23:13
add a comment |Â
up vote
1
down vote
You also had to check whether $lambda = 0$ is a valid solution. It is not because plugging in $lambda = 0$ into the first equation gives $e^x = 0$ which has no solutions.
This is where you use the first equation. The second equation $0 = 0$ indeed doesn't give any useful information so you can ignore it.
Therefore, $x= pmsqrt2$ and $z = 0$ so the extremal points are $(pmsqrt2, y, 0)$ for $y in mathbbR$.
answered Jul 22 at 23:08
mechanodroid
22.2k52041
add a comment |Â
up vote
1
down vote
You also had to check whether $lambda = 0$ is a valid solution. It is not because plugging in $lambda = 0$ into the first equation gives $e^x = 0$ which has no solutions.
This is where you use the first equation. The second equation $0 = 0$ indeed doesn't give any useful information so you can ignore it.
Therefore, $x= pmsqrt2$ and $z = 0$ so the extremal points are $(pmsqrt2, y, 0)$ for $y in mathbbR$.
answered Jul 22 at 23:08
mechanodroid
22.2k52041
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You also had to check whether $lambda = 0$ is a valid solution. It is not because plugging in $lambda = 0$ into the first equation gives $e^x = 0$ which has no solutions.
This is where you use the first equation. The second equation $0 = 0$ indeed doesn't give any useful information so you can ignore it.
Therefore, $x= pmsqrt2$ and $z = 0$ so the extremal points are $(pmsqrt2, y, 0)$ for $y in mathbbR$.
answered Jul 22 at 23:08
mechanodroid
22.2k52041
You also had to check whether $lambda = 0$ is a valid solution. It is not because plugging in $lambda = 0$ into the first equation gives $e^x = 0$ which has no solutions.
This is where you use the first equation. The second equation $0 = 0$ indeed doesn't give any useful information so you can ignore it.
Therefore, $x= pmsqrt2$ and $z = 0$ so the extremal points are $(pmsqrt2, y, 0)$ for $y in mathbbR$.
answered Jul 22 at 23:08
mechanodroid
22.2k52041
answered Jul 22 at 23:08
mechanodroid
22.2k52041
answered Jul 22 at 23:08
mechanodroid
22.2k52041
answered Jul 22 at 23:08
mechanodroid
22.2k52041
22.2k52041
add a comment |Â
add a comment |Â
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Hence there is an exponential function within the first equation it cannot be equal zero unless $lambda$ has a weird form like $lambda~=~c-frace^x2x$. So I am not sure what to conclude form this.
– mrtaurho
Jul 22 at 23:02