Minimize $x^2+6y^2+4z^2$ subject to $x+2y+z-4=0$ and $2x^2+y^2=16$
up vote
2
down vote
favorite
Minimize $x^2+6y^2+4z^2$ subject to $x+2y+z-4=0$ and $2x^2+y^2=16$
My try:
By Lagrange Multiplier method we have
$$L(x,y,z,lambda, mu)=(x^2+6y^2+4z^2)+lambda(x+2y+z-4)+mu(2x^2+y^2-16)$$
For $$L_x=0$$ we get
$$2x+lambda+4mu x=0 tag{1}$$
For $$L_y=0$$ we get
$$12y+2lambda+2mu y=0 tag{2}$$
For $$L_z=0$$ we get
$$8z+lambda=0 tag{3}$$
From $(1)$ and $(2)$ we get
$$x=frac{4 lambda}{1-2mu}$$
$$y=frac{8 lambda}{6-mu}$$
Substituting $x$ , $y$ and $z$ above in constrainst we get
$$2 frac{lambda^2}{(1-2mu)^2}+4 frac{lambda^2}{(6-mu)^2}=1 tag{4}$$
$$frac{4 lambda}{1-2mu}+frac{16 lambda}{6-mu}+frac{lambda}{8}=4 tag{5}$$
But its tedious to solve above equations for $lambda$ and $mu$
Any other approach?
optimization systems-of-equations lagrange-multiplier maxima-minima
asked Nov 22 at 13:37
Ekaveera Kumar Sharma
5,53511328
add a comment |
up vote
2
down vote
favorite
Minimize $x^2+6y^2+4z^2$ subject to $x+2y+z-4=0$ and $2x^2+y^2=16$
My try:
By Lagrange Multiplier method we have
$$L(x,y,z,lambda, mu)=(x^2+6y^2+4z^2)+lambda(x+2y+z-4)+mu(2x^2+y^2-16)$$
For $$L_x=0$$ we get
$$2x+lambda+4mu x=0 tag{1}$$
For $$L_y=0$$ we get
$$12y+2lambda+2mu y=0 tag{2}$$
For $$L_z=0$$ we get
$$8z+lambda=0 tag{3}$$
From $(1)$ and $(2)$ we get
$$x=frac{4 lambda}{1-2mu}$$
$$y=frac{8 lambda}{6-mu}$$
Substituting $x$ , $y$ and $z$ above in constrainst we get
$$2 frac{lambda^2}{(1-2mu)^2}+4 frac{lambda^2}{(6-mu)^2}=1 tag{4}$$
$$frac{4 lambda}{1-2mu}+frac{16 lambda}{6-mu}+frac{lambda}{8}=4 tag{5}$$
But its tedious to solve above equations for $lambda$ and $mu$
Any other approach?
optimization systems-of-equations lagrange-multiplier maxima-minima
asked Nov 22 at 13:37
Ekaveera Kumar Sharma
5,53511328
WLOG $sqrt2x=4cos t,y=4sin t, z=?$
– lab bhattacharjee
Nov 22 at 13:43
Good idea sir. we have $z=4-2sqrt{2} cos t-8 sin t$. To minimize $f(t)=8cos^2 t+96 sin^2 t+4left(4-2sqrt{2} cos t-8 sin t right)^2$. But this also seems lengthy
– Ekaveera Kumar Sharma
Nov 22 at 13:58
Try to solve equations $(4)$ and $(5)$ by setting $text{sin}(t) = frac{sqrt{2}lambda}{1-2mu}$ and $text{cos}(t) = frac{2lambda}{6-mu}$. Hence, use trigonometric identities to reduce the problem to a quadratic equation on $text{tan}(t)$.
– Alex Silva
Nov 22 at 15:29
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Minimize $x^2+6y^2+4z^2$ subject to $x+2y+z-4=0$ and $2x^2+y^2=16$
My try:
By Lagrange Multiplier method we have
$$L(x,y,z,lambda, mu)=(x^2+6y^2+4z^2)+lambda(x+2y+z-4)+mu(2x^2+y^2-16)$$
For $$L_x=0$$ we get
$$2x+lambda+4mu x=0 tag{1}$$
For $$L_y=0$$ we get
$$12y+2lambda+2mu y=0 tag{2}$$
For $$L_z=0$$ we get
$$8z+lambda=0 tag{3}$$
From $(1)$ and $(2)$ we get
$$x=frac{4 lambda}{1-2mu}$$
$$y=frac{8 lambda}{6-mu}$$
Substituting $x$ , $y$ and $z$ above in constrainst we get
$$2 frac{lambda^2}{(1-2mu)^2}+4 frac{lambda^2}{(6-mu)^2}=1 tag{4}$$
$$frac{4 lambda}{1-2mu}+frac{16 lambda}{6-mu}+frac{lambda}{8}=4 tag{5}$$
But its tedious to solve above equations for $lambda$ and $mu$
Any other approach?
optimization systems-of-equations lagrange-multiplier maxima-minima
asked Nov 22 at 13:37
Ekaveera Kumar Sharma
5,53511328
Minimize $x^2+6y^2+4z^2$ subject to $x+2y+z-4=0$ and $2x^2+y^2=16$
My try:
By Lagrange Multiplier method we have
$$L(x,y,z,lambda, mu)=(x^2+6y^2+4z^2)+lambda(x+2y+z-4)+mu(2x^2+y^2-16)$$
For $$L_x=0$$ we get
$$2x+lambda+4mu x=0 tag{1}$$
For $$L_y=0$$ we get
$$12y+2lambda+2mu y=0 tag{2}$$
For $$L_z=0$$ we get
$$8z+lambda=0 tag{3}$$
From $(1)$ and $(2)$ we get
$$x=frac{4 lambda}{1-2mu}$$
$$y=frac{8 lambda}{6-mu}$$
Substituting $x$ , $y$ and $z$ above in constrainst we get
$$2 frac{lambda^2}{(1-2mu)^2}+4 frac{lambda^2}{(6-mu)^2}=1 tag{4}$$
$$frac{4 lambda}{1-2mu}+frac{16 lambda}{6-mu}+frac{lambda}{8}=4 tag{5}$$
But its tedious to solve above equations for $lambda$ and $mu$
Any other approach?
optimization systems-of-equations lagrange-multiplier maxima-minima
optimization systems-of-equations lagrange-multiplier maxima-minima
asked Nov 22 at 13:37
Ekaveera Kumar Sharma
5,53511328
asked Nov 22 at 13:37
Ekaveera Kumar Sharma
5,53511328
asked Nov 22 at 13:37
Ekaveera Kumar Sharma
5,53511328
asked Nov 22 at 13:37
Ekaveera Kumar Sharma
5,53511328
asked Nov 22 at 13:37
Ekaveera Kumar Sharma
5,53511328
5,53511328
WLOG $sqrt2x=4cos t,y=4sin t, z=?$
– lab bhattacharjee
Nov 22 at 13:43
Good idea sir. we have $z=4-2sqrt{2} cos t-8 sin t$. To minimize $f(t)=8cos^2 t+96 sin^2 t+4left(4-2sqrt{2} cos t-8 sin t right)^2$. But this also seems lengthy
– Ekaveera Kumar Sharma
Nov 22 at 13:58
Try to solve equations $(4)$ and $(5)$ by setting $text{sin}(t) = frac{sqrt{2}lambda}{1-2mu}$ and $text{cos}(t) = frac{2lambda}{6-mu}$. Hence, use trigonometric identities to reduce the problem to a quadratic equation on $text{tan}(t)$.
– Alex Silva
Nov 22 at 15:29
add a comment |
WLOG $sqrt2x=4cos t,y=4sin t, z=?$
– lab bhattacharjee
Nov 22 at 13:43
Good idea sir. we have $z=4-2sqrt{2} cos t-8 sin t$. To minimize $f(t)=8cos^2 t+96 sin^2 t+4left(4-2sqrt{2} cos t-8 sin t right)^2$. But this also seems lengthy
– Ekaveera Kumar Sharma
Nov 22 at 13:58
Try to solve equations $(4)$ and $(5)$ by setting $text{sin}(t) = frac{sqrt{2}lambda}{1-2mu}$ and $text{cos}(t) = frac{2lambda}{6-mu}$. Hence, use trigonometric identities to reduce the problem to a quadratic equation on $text{tan}(t)$.
– Alex Silva
Nov 22 at 15:29
WLOG $sqrt2x=4cos t,y=4sin t, z=?$
– lab bhattacharjee
Nov 22 at 13:43
WLOG $sqrt2x=4cos t,y=4sin t, z=?$
– lab bhattacharjee
Nov 22 at 13:43
Good idea sir. we have $z=4-2sqrt{2} cos t-8 sin t$. To minimize $f(t)=8cos^2 t+96 sin^2 t+4left(4-2sqrt{2} cos t-8 sin t right)^2$. But this also seems lengthy
– Ekaveera Kumar Sharma
Nov 22 at 13:58
Good idea sir. we have $z=4-2sqrt{2} cos t-8 sin t$. To minimize $f(t)=8cos^2 t+96 sin^2 t+4left(4-2sqrt{2} cos t-8 sin t right)^2$. But this also seems lengthy
– Ekaveera Kumar Sharma
Nov 22 at 13:58
Try to solve equations $(4)$ and $(5)$ by setting $text{sin}(t) = frac{sqrt{2}lambda}{1-2mu}$ and $text{cos}(t) = frac{2lambda}{6-mu}$. Hence, use trigonometric identities to reduce the problem to a quadratic equation on $text{tan}(t)$.
– Alex Silva
Nov 22 at 15:29
Try to solve equations $(4)$ and $(5)$ by setting $text{sin}(t) = frac{sqrt{2}lambda}{1-2mu}$ and $text{cos}(t) = frac{2lambda}{6-mu}$. Hence, use trigonometric identities to reduce the problem to a quadratic equation on $text{tan}(t)$.
– Alex Silva
Nov 22 at 15:29
add a comment |
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
Hint:
Let $u=x^2+6y^2+4(4-2y-x)^2$
$iff5x^2-16(2-y)x+22y^2-16y+16-u=0$
As $x$ is real, the discriminant will be $ge0$
$$implies256(y-2)^2ge20(22y^2-16y+16-u)$$
$$implies64(y-2)^2ge5(22y^2-16y+16-u)$$
$$implies5uge46y^2-176y+176=46left(y-dfrac{44}{23}right)^2+176-46left(dfrac{44}{23}right)^2$$
Now $y^2=16-2x^2le16implies-4le yle4iff-4-dfrac{44}{23}le y-dfrac{44}{23}le4-dfrac{44}{23}$
answered Nov 23 at 6:26
lab bhattacharjee
222k15155273
Very nice approach
– Ekaveera Kumar Sharma
Nov 23 at 7:10
add a comment |
up vote
1
down vote
Since you are assuming that $2x^2+y^2=16$, your problem is equivalent to the problem of minimizing $frac{11}2y^2+4z^2$. But then your first equation becomes $lambda+4mu x=0$, which is much simpler.
answered Nov 22 at 13:59
José Carlos Santos
146k22117217
Good ,so with your idea i got $x=frac{-lambda}{4 mu}$, $y=frac{-2lambda}{2mu+11}$, $z=frac{-lambda}{8}$. But i don't think its simple enough again?
– Ekaveera Kumar Sharma
Nov 22 at 14:29
add a comment |
up vote
0
down vote
With computer I get numeric solution
$$f_{min}=9.445967377367024$$
if
$$x=2.811441051354938\
y=0.4377195786259503\
z=0.3131197913931612$$
Exact values of $x,y,z$ is solutions of equations:
$$2033x^4-800x^3-13960x^2-1792x+6144=0,\
2033y^4-6016y^3-27920y^2+88064y-32768=0,\
2033z^4-19696z^3-84456z^2+176704z-46464=0.$$
We can solve these equations exact and get exact solutions.
However, there are too big expressions.
answered Nov 23 at 11:23
Aleksas Domarkas
8186
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Hint:
Let $u=x^2+6y^2+4(4-2y-x)^2$
$iff5x^2-16(2-y)x+22y^2-16y+16-u=0$
As $x$ is real, the discriminant will be $ge0$
$$implies256(y-2)^2ge20(22y^2-16y+16-u)$$
$$implies64(y-2)^2ge5(22y^2-16y+16-u)$$
$$implies5uge46y^2-176y+176=46left(y-dfrac{44}{23}right)^2+176-46left(dfrac{44}{23}right)^2$$
Now $y^2=16-2x^2le16implies-4le yle4iff-4-dfrac{44}{23}le y-dfrac{44}{23}le4-dfrac{44}{23}$
answered Nov 23 at 6:26
lab bhattacharjee
222k15155273
Very nice approach
– Ekaveera Kumar Sharma
Nov 23 at 7:10
add a comment |
up vote
1
down vote
accepted
Hint:
Let $u=x^2+6y^2+4(4-2y-x)^2$
$iff5x^2-16(2-y)x+22y^2-16y+16-u=0$
As $x$ is real, the discriminant will be $ge0$
$$implies256(y-2)^2ge20(22y^2-16y+16-u)$$
$$implies64(y-2)^2ge5(22y^2-16y+16-u)$$
$$implies5uge46y^2-176y+176=46left(y-dfrac{44}{23}right)^2+176-46left(dfrac{44}{23}right)^2$$
Now $y^2=16-2x^2le16implies-4le yle4iff-4-dfrac{44}{23}le y-dfrac{44}{23}le4-dfrac{44}{23}$
answered Nov 23 at 6:26
lab bhattacharjee
222k15155273
Very nice approach
– Ekaveera Kumar Sharma
Nov 23 at 7:10
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Hint:
Let $u=x^2+6y^2+4(4-2y-x)^2$
$iff5x^2-16(2-y)x+22y^2-16y+16-u=0$
As $x$ is real, the discriminant will be $ge0$
$$implies256(y-2)^2ge20(22y^2-16y+16-u)$$
$$implies64(y-2)^2ge5(22y^2-16y+16-u)$$
$$implies5uge46y^2-176y+176=46left(y-dfrac{44}{23}right)^2+176-46left(dfrac{44}{23}right)^2$$
Now $y^2=16-2x^2le16implies-4le yle4iff-4-dfrac{44}{23}le y-dfrac{44}{23}le4-dfrac{44}{23}$
answered Nov 23 at 6:26
lab bhattacharjee
222k15155273
Hint:
Let $u=x^2+6y^2+4(4-2y-x)^2$
$iff5x^2-16(2-y)x+22y^2-16y+16-u=0$
As $x$ is real, the discriminant will be $ge0$
$$implies256(y-2)^2ge20(22y^2-16y+16-u)$$
$$implies64(y-2)^2ge5(22y^2-16y+16-u)$$
$$implies5uge46y^2-176y+176=46left(y-dfrac{44}{23}right)^2+176-46left(dfrac{44}{23}right)^2$$
Now $y^2=16-2x^2le16implies-4le yle4iff-4-dfrac{44}{23}le y-dfrac{44}{23}le4-dfrac{44}{23}$
answered Nov 23 at 6:26
lab bhattacharjee
222k15155273
answered Nov 23 at 6:26
lab bhattacharjee
222k15155273
answered Nov 23 at 6:26
lab bhattacharjee
222k15155273
answered Nov 23 at 6:26
lab bhattacharjee
222k15155273
222k15155273
Very nice approach
– Ekaveera Kumar Sharma
Nov 23 at 7:10
add a comment |
Very nice approach
– Ekaveera Kumar Sharma
Nov 23 at 7:10
Very nice approach
– Ekaveera Kumar Sharma
Nov 23 at 7:10
Very nice approach
– Ekaveera Kumar Sharma
Nov 23 at 7:10
add a comment |
up vote
1
down vote
Since you are assuming that $2x^2+y^2=16$, your problem is equivalent to the problem of minimizing $frac{11}2y^2+4z^2$. But then your first equation becomes $lambda+4mu x=0$, which is much simpler.
answered Nov 22 at 13:59
José Carlos Santos
146k22117217
Good ,so with your idea i got $x=frac{-lambda}{4 mu}$, $y=frac{-2lambda}{2mu+11}$, $z=frac{-lambda}{8}$. But i don't think its simple enough again?
– Ekaveera Kumar Sharma
Nov 22 at 14:29
add a comment |
up vote
1
down vote
Since you are assuming that $2x^2+y^2=16$, your problem is equivalent to the problem of minimizing $frac{11}2y^2+4z^2$. But then your first equation becomes $lambda+4mu x=0$, which is much simpler.
answered Nov 22 at 13:59
José Carlos Santos
146k22117217
Good ,so with your idea i got $x=frac{-lambda}{4 mu}$, $y=frac{-2lambda}{2mu+11}$, $z=frac{-lambda}{8}$. But i don't think its simple enough again?
– Ekaveera Kumar Sharma
Nov 22 at 14:29
add a comment |
up vote
1
down vote
up vote
1
down vote
Since you are assuming that $2x^2+y^2=16$, your problem is equivalent to the problem of minimizing $frac{11}2y^2+4z^2$. But then your first equation becomes $lambda+4mu x=0$, which is much simpler.
answered Nov 22 at 13:59
José Carlos Santos
146k22117217
Since you are assuming that $2x^2+y^2=16$, your problem is equivalent to the problem of minimizing $frac{11}2y^2+4z^2$. But then your first equation becomes $lambda+4mu x=0$, which is much simpler.
answered Nov 22 at 13:59
José Carlos Santos
146k22117217
edited Nov 22 at 14:21
answered Nov 22 at 13:59
José Carlos Santos
146k22117217
answered Nov 22 at 13:59
José Carlos Santos
146k22117217
answered Nov 22 at 13:59
José Carlos Santos
146k22117217
146k22117217
Good ,so with your idea i got $x=frac{-lambda}{4 mu}$, $y=frac{-2lambda}{2mu+11}$, $z=frac{-lambda}{8}$. But i don't think its simple enough again?
– Ekaveera Kumar Sharma
Nov 22 at 14:29
add a comment |
Good ,so with your idea i got $x=frac{-lambda}{4 mu}$, $y=frac{-2lambda}{2mu+11}$, $z=frac{-lambda}{8}$. But i don't think its simple enough again?
– Ekaveera Kumar Sharma
Nov 22 at 14:29
Good ,so with your idea i got $x=frac{-lambda}{4 mu}$, $y=frac{-2lambda}{2mu+11}$, $z=frac{-lambda}{8}$. But i don't think its simple enough again?
– Ekaveera Kumar Sharma
Nov 22 at 14:29
Good ,so with your idea i got $x=frac{-lambda}{4 mu}$, $y=frac{-2lambda}{2mu+11}$, $z=frac{-lambda}{8}$. But i don't think its simple enough again?
– Ekaveera Kumar Sharma
Nov 22 at 14:29
add a comment |
up vote
0
down vote
With computer I get numeric solution
$$f_{min}=9.445967377367024$$
if
$$x=2.811441051354938\
y=0.4377195786259503\
z=0.3131197913931612$$
Exact values of $x,y,z$ is solutions of equations:
$$2033x^4-800x^3-13960x^2-1792x+6144=0,\
2033y^4-6016y^3-27920y^2+88064y-32768=0,\
2033z^4-19696z^3-84456z^2+176704z-46464=0.$$
We can solve these equations exact and get exact solutions.
However, there are too big expressions.
answered Nov 23 at 11:23
Aleksas Domarkas
8186
add a comment |
up vote
0
down vote
With computer I get numeric solution
$$f_{min}=9.445967377367024$$
if
$$x=2.811441051354938\
y=0.4377195786259503\
z=0.3131197913931612$$
Exact values of $x,y,z$ is solutions of equations:
$$2033x^4-800x^3-13960x^2-1792x+6144=0,\
2033y^4-6016y^3-27920y^2+88064y-32768=0,\
2033z^4-19696z^3-84456z^2+176704z-46464=0.$$
We can solve these equations exact and get exact solutions.
However, there are too big expressions.
answered Nov 23 at 11:23
Aleksas Domarkas
8186
add a comment |
up vote
0
down vote
up vote
0
down vote
With computer I get numeric solution
$$f_{min}=9.445967377367024$$
if
$$x=2.811441051354938\
y=0.4377195786259503\
z=0.3131197913931612$$
Exact values of $x,y,z$ is solutions of equations:
$$2033x^4-800x^3-13960x^2-1792x+6144=0,\
2033y^4-6016y^3-27920y^2+88064y-32768=0,\
2033z^4-19696z^3-84456z^2+176704z-46464=0.$$
We can solve these equations exact and get exact solutions.
However, there are too big expressions.
answered Nov 23 at 11:23
Aleksas Domarkas
8186
With computer I get numeric solution
$$f_{min}=9.445967377367024$$
if
$$x=2.811441051354938\
y=0.4377195786259503\
z=0.3131197913931612$$
Exact values of $x,y,z$ is solutions of equations:
$$2033x^4-800x^3-13960x^2-1792x+6144=0,\
2033y^4-6016y^3-27920y^2+88064y-32768=0,\
2033z^4-19696z^3-84456z^2+176704z-46464=0.$$
We can solve these equations exact and get exact solutions.
However, there are too big expressions.
answered Nov 23 at 11:23
Aleksas Domarkas
8186
answered Nov 23 at 11:23
Aleksas Domarkas
8186
answered Nov 23 at 11:23
Aleksas Domarkas
8186
answered Nov 23 at 11:23
Aleksas Domarkas
8186
8186
add a comment |
add a comment |
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WLOG $sqrt2x=4cos t,y=4sin t, z=?$
– lab bhattacharjee
Nov 22 at 13:43
Good idea sir. we have $z=4-2sqrt{2} cos t-8 sin t$. To minimize $f(t)=8cos^2 t+96 sin^2 t+4left(4-2sqrt{2} cos t-8 sin t right)^2$. But this also seems lengthy
– Ekaveera Kumar Sharma
Nov 22 at 13:58
Try to solve equations $(4)$ and $(5)$ by setting $text{sin}(t) = frac{sqrt{2}lambda}{1-2mu}$ and $text{cos}(t) = frac{2lambda}{6-mu}$. Hence, use trigonometric identities to reduce the problem to a quadratic equation on $text{tan}(t)$.
– Alex Silva
Nov 22 at 15:29