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Lagrange Multipliers $ln(3+2xy)$ function
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Can I solve this exercise using Lagrange multipliers? Find maximum and minimum value of the function $f(x,y) = ln(3+2xy)$ on the circle $x^2 + y^2 = 1$. I get a set of equalities that I can't solve.
lagrange-multiplier maxima-minima
edited 22 hours ago
José Carlos Santos
111k1695171
asked 22 hours ago
StudentMaths
266
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Can I solve this exercise using Lagrange multipliers? Find maximum and minimum value of the function $f(x,y) = ln(3+2xy)$ on the circle $x^2 + y^2 = 1$. I get a set of equalities that I can't solve.
lagrange-multiplier maxima-minima
edited 22 hours ago
José Carlos Santos
111k1695171
asked 22 hours ago
StudentMaths
266
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Can I solve this exercise using Lagrange multipliers? Find maximum and minimum value of the function $f(x,y) = ln(3+2xy)$ on the circle $x^2 + y^2 = 1$. I get a set of equalities that I can't solve.
lagrange-multiplier maxima-minima
edited 22 hours ago
José Carlos Santos
111k1695171
asked 22 hours ago
StudentMaths
266
Can I solve this exercise using Lagrange multipliers? Find maximum and minimum value of the function $f(x,y) = ln(3+2xy)$ on the circle $x^2 + y^2 = 1$. I get a set of equalities that I can't solve.
lagrange-multiplier maxima-minima
edited 22 hours ago
José Carlos Santos
111k1695171
asked 22 hours ago
StudentMaths
266
edited 22 hours ago
José Carlos Santos
111k1695171
edited 22 hours ago
José Carlos Santos
111k1695171
edited 22 hours ago
José Carlos Santos
111k1695171
111k1695171
asked 22 hours ago
StudentMaths
266
asked 22 hours ago
StudentMaths
266
asked 22 hours ago
StudentMaths
266
266
add a comment |Â
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
1
down vote
accepted
You will get the System
$$frac2y2xy+3+2lambda x=0$$
$$frac2x2xy+3+2lambda y=0$$
$$x^2+y^2=1$$
and then write
$$2xlambda=-frac2y2xy+3$$
$$2ylambda=-frac2x2xy+3$$
dividing both equations you will get
$$fracxy=fracyx$$
can you proceed now?
this leads to $$x^2-y^2=(x-y)(x+y)=0$$
answered 22 hours ago
Dr. Sonnhard Graubner
66.6k32659
1
And what is the rest, Sonnhard?
– Michael Rozenberg
22 hours ago
Thank you! Yes I get that system too, but how do I solve this to get the values for x and y? Then I can plug them into the constraint :)
– StudentMaths
22 hours ago
A part of the rest can you read now!
– Dr. Sonnhard Graubner
22 hours ago
Why by this way we'll get the absolute minimum and the absolute maximum?
– Michael Rozenberg
22 hours ago
1
Shouldn't it be $$x=pmfrac1sqrt2$$ etc?
– Dr. Sonnhard Graubner
21 hours ago
 |Â
show 3 more comments
up vote
2
down vote
Since $amapstoln(3+2a)$ is strictly increasing, your problem is equivalent to the problem of finding the maximum and the minimum of $xy$ on that circle. Note that the circle is compact and that threfore, since we are dealing with a continuous functions, it has to have a maximum and a minimum. Applying the method of Lagrange multipliers to this function leads to the system of equations$$left{beginarrayly=2lambda x\x=2lambda y\x^2+y^2=1.endarrayright.$$You get from the first two equations that $x=4lambda^2x$ and this implies that $lambda=pmfrac12$ or that $x=0$. But you can't have $x=0$, because the first equation would then imply that $y=0$, which is impossible, by the third equation.
If $lambda=frac12$, the solutions of the system are $pmleft(frac1sqrt2,frac1sqrt2right)$ and if $lambda=-frac12$ the solutions are $left(pmfrac1sqrt2,mpfrac1sqrt2right)$. At the first two of these points, the values of $xy$ is $frac12$, whereas at the other two the value of $xy$ is $-frac12$. So, the maximum of $xy$ is $frac12$ and it is attained at the first to points, whereas the minimum is $-frac12$, which is attained at the other two.
So, the maximum of the original function is $ln(4)$, whereas the minimum is $ln(2)$.
answered 22 hours ago
José Carlos Santos
111k1695171
Show please, why the minimum occurs for your points? Thank you.
– Michael Rozenberg
22 hours ago
@MichaelRozenberg I've edited my answer.
– José Carlos Santos
22 hours ago
Also, you need to say about continuous function on the compact. I think it's anti-maths to find maximum and minimum of $xy$ by the LM method.
– Michael Rozenberg
21 hours ago
1
@MichaelRozenberg You are right in both assertions. I will edit my answer once again because of your first remark. Concerning the second one, althought I fully agree with you, it's what the OP asked for. I would not have used Lagrange multipliers otherwise.
– José Carlos Santos
21 hours ago
1
If $x^2+y^2=1$, $x=costheta$ and $y=sintheta$, for some $thetain[0,2pi)$. Therefore$$xy=cos(theta)sin(theta)=frac12sin(2theta).$$So, the maximum and the minimum are $frac12$ and $-frac12$ respectively. Besides, the maximum is attained when $theta=fracpi4$ or $theta=-frac5pi4$ and the minimum is attained when $theta=frac3pi4$ or when $theta=frac7pi4$.
– José Carlos Santos
21 hours ago
 |Â
show 4 more comments
up vote
0
down vote
By AM-GM $$ln(3+2xy)leqln(3+2|xy|)leq ln(3+x^2+y^2)=2ln2.$$
THe equality occurs for $x=y=frac1sqrt2,$ which says that we got a maximal value.
Now, by AM-GM again
$$ln(3+2xy)geqln(3-2|xy|)geq ln(3-(x^2+y^2))=ln2.$$
The equality occurs for $x=frac1sqrt2$ and $y=-frac1sqrt2,$ which says that we got a minimal value.
answered 22 hours ago
Michael Rozenberg
86.9k1575178
1
This does not answer the question.
– uniquesolution
22 hours ago
@uniquesolution Why? See better the question.
– Michael Rozenberg
22 hours ago
1
The question asks "Can I solve this exercise using Lagrange multipliers"? What did you see?
– uniquesolution
22 hours ago
@MichaelRozenberg Can this be solved by using Lagrange? If so, how?
– StudentMaths
22 hours ago
@StudentMaths Yes I can, but I think it's an idiotism here. You can say the following words: "the Lagrange Multipliers Method is very useful method" and write my solution.
– Michael Rozenberg
22 hours ago
 |Â
show 7 more comments
up vote
0
down vote
You will get the System
$$frac2y2xy+3+2lambda x=0$$
$$frac2x2xy+3+2lambda y=0$$
$$x^2+y^2=1$$
you can write
$$2lambda x=-frac2y2xy+3$$
$$2lambda y=-frac2x2xy+3$$
dividing both equations you will ge
$$fracxy=fracyx$$
answered 22 hours ago
Dr. Sonnhard Graubner
66.6k32659
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You will get the System
$$frac2y2xy+3+2lambda x=0$$
$$frac2x2xy+3+2lambda y=0$$
$$x^2+y^2=1$$
and then write
$$2xlambda=-frac2y2xy+3$$
$$2ylambda=-frac2x2xy+3$$
dividing both equations you will get
$$fracxy=fracyx$$
can you proceed now?
this leads to $$x^2-y^2=(x-y)(x+y)=0$$
answered 22 hours ago
Dr. Sonnhard Graubner
66.6k32659
1
And what is the rest, Sonnhard?
– Michael Rozenberg
22 hours ago
Thank you! Yes I get that system too, but how do I solve this to get the values for x and y? Then I can plug them into the constraint :)
– StudentMaths
22 hours ago
A part of the rest can you read now!
– Dr. Sonnhard Graubner
22 hours ago
Why by this way we'll get the absolute minimum and the absolute maximum?
– Michael Rozenberg
22 hours ago
1
Shouldn't it be $$x=pmfrac1sqrt2$$ etc?
– Dr. Sonnhard Graubner
21 hours ago
 |Â
show 3 more comments
up vote
1
down vote
accepted
You will get the System
$$frac2y2xy+3+2lambda x=0$$
$$frac2x2xy+3+2lambda y=0$$
$$x^2+y^2=1$$
and then write
$$2xlambda=-frac2y2xy+3$$
$$2ylambda=-frac2x2xy+3$$
dividing both equations you will get
$$fracxy=fracyx$$
can you proceed now?
this leads to $$x^2-y^2=(x-y)(x+y)=0$$
answered 22 hours ago
Dr. Sonnhard Graubner
66.6k32659
1
And what is the rest, Sonnhard?
– Michael Rozenberg
22 hours ago
Thank you! Yes I get that system too, but how do I solve this to get the values for x and y? Then I can plug them into the constraint :)
– StudentMaths
22 hours ago
A part of the rest can you read now!
– Dr. Sonnhard Graubner
22 hours ago
Why by this way we'll get the absolute minimum and the absolute maximum?
– Michael Rozenberg
22 hours ago
1
Shouldn't it be $$x=pmfrac1sqrt2$$ etc?
– Dr. Sonnhard Graubner
21 hours ago
 |Â
show 3 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You will get the System
$$frac2y2xy+3+2lambda x=0$$
$$frac2x2xy+3+2lambda y=0$$
$$x^2+y^2=1$$
and then write
$$2xlambda=-frac2y2xy+3$$
$$2ylambda=-frac2x2xy+3$$
dividing both equations you will get
$$fracxy=fracyx$$
can you proceed now?
this leads to $$x^2-y^2=(x-y)(x+y)=0$$
answered 22 hours ago
Dr. Sonnhard Graubner
66.6k32659
You will get the System
$$frac2y2xy+3+2lambda x=0$$
$$frac2x2xy+3+2lambda y=0$$
$$x^2+y^2=1$$
and then write
$$2xlambda=-frac2y2xy+3$$
$$2ylambda=-frac2x2xy+3$$
dividing both equations you will get
$$fracxy=fracyx$$
can you proceed now?
this leads to $$x^2-y^2=(x-y)(x+y)=0$$
answered 22 hours ago
Dr. Sonnhard Graubner
66.6k32659
edited 21 hours ago
answered 22 hours ago
Dr. Sonnhard Graubner
66.6k32659
answered 22 hours ago
Dr. Sonnhard Graubner
66.6k32659
answered 22 hours ago
Dr. Sonnhard Graubner
66.6k32659
66.6k32659
1
And what is the rest, Sonnhard?
– Michael Rozenberg
22 hours ago
Thank you! Yes I get that system too, but how do I solve this to get the values for x and y? Then I can plug them into the constraint :)
– StudentMaths
22 hours ago
A part of the rest can you read now!
– Dr. Sonnhard Graubner
22 hours ago
Why by this way we'll get the absolute minimum and the absolute maximum?
– Michael Rozenberg
22 hours ago
1
Shouldn't it be $$x=pmfrac1sqrt2$$ etc?
– Dr. Sonnhard Graubner
21 hours ago
 |Â
show 3 more comments
1
And what is the rest, Sonnhard?
– Michael Rozenberg
22 hours ago
Thank you! Yes I get that system too, but how do I solve this to get the values for x and y? Then I can plug them into the constraint :)
– StudentMaths
22 hours ago
A part of the rest can you read now!
– Dr. Sonnhard Graubner
22 hours ago
Why by this way we'll get the absolute minimum and the absolute maximum?
– Michael Rozenberg
22 hours ago
1
Shouldn't it be $$x=pmfrac1sqrt2$$ etc?
– Dr. Sonnhard Graubner
21 hours ago
1
1
And what is the rest, Sonnhard?
– Michael Rozenberg
22 hours ago
And what is the rest, Sonnhard?
– Michael Rozenberg
22 hours ago
Thank you! Yes I get that system too, but how do I solve this to get the values for x and y? Then I can plug them into the constraint :)
– StudentMaths
22 hours ago
Thank you! Yes I get that system too, but how do I solve this to get the values for x and y? Then I can plug them into the constraint :)
– StudentMaths
22 hours ago
A part of the rest can you read now!
– Dr. Sonnhard Graubner
22 hours ago
A part of the rest can you read now!
– Dr. Sonnhard Graubner
22 hours ago
Why by this way we'll get the absolute minimum and the absolute maximum?
– Michael Rozenberg
22 hours ago
Why by this way we'll get the absolute minimum and the absolute maximum?
– Michael Rozenberg
22 hours ago
1
1
Shouldn't it be $$x=pmfrac1sqrt2$$ etc?
– Dr. Sonnhard Graubner
21 hours ago
Shouldn't it be $$x=pmfrac1sqrt2$$ etc?
– Dr. Sonnhard Graubner
21 hours ago
 |Â
show 3 more comments
up vote
2
down vote
Since $amapstoln(3+2a)$ is strictly increasing, your problem is equivalent to the problem of finding the maximum and the minimum of $xy$ on that circle. Note that the circle is compact and that threfore, since we are dealing with a continuous functions, it has to have a maximum and a minimum. Applying the method of Lagrange multipliers to this function leads to the system of equations$$left{beginarrayly=2lambda x\x=2lambda y\x^2+y^2=1.endarrayright.$$You get from the first two equations that $x=4lambda^2x$ and this implies that $lambda=pmfrac12$ or that $x=0$. But you can't have $x=0$, because the first equation would then imply that $y=0$, which is impossible, by the third equation.
If $lambda=frac12$, the solutions of the system are $pmleft(frac1sqrt2,frac1sqrt2right)$ and if $lambda=-frac12$ the solutions are $left(pmfrac1sqrt2,mpfrac1sqrt2right)$. At the first two of these points, the values of $xy$ is $frac12$, whereas at the other two the value of $xy$ is $-frac12$. So, the maximum of $xy$ is $frac12$ and it is attained at the first to points, whereas the minimum is $-frac12$, which is attained at the other two.
So, the maximum of the original function is $ln(4)$, whereas the minimum is $ln(2)$.
answered 22 hours ago
José Carlos Santos
111k1695171
Show please, why the minimum occurs for your points? Thank you.
– Michael Rozenberg
22 hours ago
@MichaelRozenberg I've edited my answer.
– José Carlos Santos
22 hours ago
Also, you need to say about continuous function on the compact. I think it's anti-maths to find maximum and minimum of $xy$ by the LM method.
– Michael Rozenberg
21 hours ago
1
@MichaelRozenberg You are right in both assertions. I will edit my answer once again because of your first remark. Concerning the second one, althought I fully agree with you, it's what the OP asked for. I would not have used Lagrange multipliers otherwise.
– José Carlos Santos
21 hours ago
1
If $x^2+y^2=1$, $x=costheta$ and $y=sintheta$, for some $thetain[0,2pi)$. Therefore$$xy=cos(theta)sin(theta)=frac12sin(2theta).$$So, the maximum and the minimum are $frac12$ and $-frac12$ respectively. Besides, the maximum is attained when $theta=fracpi4$ or $theta=-frac5pi4$ and the minimum is attained when $theta=frac3pi4$ or when $theta=frac7pi4$.
– José Carlos Santos
21 hours ago
 |Â
show 4 more comments
up vote
2
down vote
Since $amapstoln(3+2a)$ is strictly increasing, your problem is equivalent to the problem of finding the maximum and the minimum of $xy$ on that circle. Note that the circle is compact and that threfore, since we are dealing with a continuous functions, it has to have a maximum and a minimum. Applying the method of Lagrange multipliers to this function leads to the system of equations$$left{beginarrayly=2lambda x\x=2lambda y\x^2+y^2=1.endarrayright.$$You get from the first two equations that $x=4lambda^2x$ and this implies that $lambda=pmfrac12$ or that $x=0$. But you can't have $x=0$, because the first equation would then imply that $y=0$, which is impossible, by the third equation.
If $lambda=frac12$, the solutions of the system are $pmleft(frac1sqrt2,frac1sqrt2right)$ and if $lambda=-frac12$ the solutions are $left(pmfrac1sqrt2,mpfrac1sqrt2right)$. At the first two of these points, the values of $xy$ is $frac12$, whereas at the other two the value of $xy$ is $-frac12$. So, the maximum of $xy$ is $frac12$ and it is attained at the first to points, whereas the minimum is $-frac12$, which is attained at the other two.
So, the maximum of the original function is $ln(4)$, whereas the minimum is $ln(2)$.
answered 22 hours ago
José Carlos Santos
111k1695171
Show please, why the minimum occurs for your points? Thank you.
– Michael Rozenberg
22 hours ago
@MichaelRozenberg I've edited my answer.
– José Carlos Santos
22 hours ago
Also, you need to say about continuous function on the compact. I think it's anti-maths to find maximum and minimum of $xy$ by the LM method.
– Michael Rozenberg
21 hours ago
1
@MichaelRozenberg You are right in both assertions. I will edit my answer once again because of your first remark. Concerning the second one, althought I fully agree with you, it's what the OP asked for. I would not have used Lagrange multipliers otherwise.
– José Carlos Santos
21 hours ago
1
If $x^2+y^2=1$, $x=costheta$ and $y=sintheta$, for some $thetain[0,2pi)$. Therefore$$xy=cos(theta)sin(theta)=frac12sin(2theta).$$So, the maximum and the minimum are $frac12$ and $-frac12$ respectively. Besides, the maximum is attained when $theta=fracpi4$ or $theta=-frac5pi4$ and the minimum is attained when $theta=frac3pi4$ or when $theta=frac7pi4$.
– José Carlos Santos
21 hours ago
 |Â
show 4 more comments
up vote
2
down vote
up vote
2
down vote
Since $amapstoln(3+2a)$ is strictly increasing, your problem is equivalent to the problem of finding the maximum and the minimum of $xy$ on that circle. Note that the circle is compact and that threfore, since we are dealing with a continuous functions, it has to have a maximum and a minimum. Applying the method of Lagrange multipliers to this function leads to the system of equations$$left{beginarrayly=2lambda x\x=2lambda y\x^2+y^2=1.endarrayright.$$You get from the first two equations that $x=4lambda^2x$ and this implies that $lambda=pmfrac12$ or that $x=0$. But you can't have $x=0$, because the first equation would then imply that $y=0$, which is impossible, by the third equation.
If $lambda=frac12$, the solutions of the system are $pmleft(frac1sqrt2,frac1sqrt2right)$ and if $lambda=-frac12$ the solutions are $left(pmfrac1sqrt2,mpfrac1sqrt2right)$. At the first two of these points, the values of $xy$ is $frac12$, whereas at the other two the value of $xy$ is $-frac12$. So, the maximum of $xy$ is $frac12$ and it is attained at the first to points, whereas the minimum is $-frac12$, which is attained at the other two.
So, the maximum of the original function is $ln(4)$, whereas the minimum is $ln(2)$.
answered 22 hours ago
José Carlos Santos
111k1695171
Since $amapstoln(3+2a)$ is strictly increasing, your problem is equivalent to the problem of finding the maximum and the minimum of $xy$ on that circle. Note that the circle is compact and that threfore, since we are dealing with a continuous functions, it has to have a maximum and a minimum. Applying the method of Lagrange multipliers to this function leads to the system of equations$$left{beginarrayly=2lambda x\x=2lambda y\x^2+y^2=1.endarrayright.$$You get from the first two equations that $x=4lambda^2x$ and this implies that $lambda=pmfrac12$ or that $x=0$. But you can't have $x=0$, because the first equation would then imply that $y=0$, which is impossible, by the third equation.
If $lambda=frac12$, the solutions of the system are $pmleft(frac1sqrt2,frac1sqrt2right)$ and if $lambda=-frac12$ the solutions are $left(pmfrac1sqrt2,mpfrac1sqrt2right)$. At the first two of these points, the values of $xy$ is $frac12$, whereas at the other two the value of $xy$ is $-frac12$. So, the maximum of $xy$ is $frac12$ and it is attained at the first to points, whereas the minimum is $-frac12$, which is attained at the other two.
So, the maximum of the original function is $ln(4)$, whereas the minimum is $ln(2)$.
answered 22 hours ago
José Carlos Santos
111k1695171
edited 21 hours ago
answered 22 hours ago
José Carlos Santos
111k1695171
answered 22 hours ago
José Carlos Santos
111k1695171
answered 22 hours ago
José Carlos Santos
111k1695171
111k1695171
Show please, why the minimum occurs for your points? Thank you.
– Michael Rozenberg
22 hours ago
@MichaelRozenberg I've edited my answer.
– José Carlos Santos
22 hours ago
Also, you need to say about continuous function on the compact. I think it's anti-maths to find maximum and minimum of $xy$ by the LM method.
– Michael Rozenberg
21 hours ago
1
@MichaelRozenberg You are right in both assertions. I will edit my answer once again because of your first remark. Concerning the second one, althought I fully agree with you, it's what the OP asked for. I would not have used Lagrange multipliers otherwise.
– José Carlos Santos
21 hours ago
1
If $x^2+y^2=1$, $x=costheta$ and $y=sintheta$, for some $thetain[0,2pi)$. Therefore$$xy=cos(theta)sin(theta)=frac12sin(2theta).$$So, the maximum and the minimum are $frac12$ and $-frac12$ respectively. Besides, the maximum is attained when $theta=fracpi4$ or $theta=-frac5pi4$ and the minimum is attained when $theta=frac3pi4$ or when $theta=frac7pi4$.
– José Carlos Santos
21 hours ago
 |Â
show 4 more comments
Show please, why the minimum occurs for your points? Thank you.
– Michael Rozenberg
22 hours ago
@MichaelRozenberg I've edited my answer.
– José Carlos Santos
22 hours ago
Also, you need to say about continuous function on the compact. I think it's anti-maths to find maximum and minimum of $xy$ by the LM method.
– Michael Rozenberg
21 hours ago
1
@MichaelRozenberg You are right in both assertions. I will edit my answer once again because of your first remark. Concerning the second one, althought I fully agree with you, it's what the OP asked for. I would not have used Lagrange multipliers otherwise.
– José Carlos Santos
21 hours ago
1
If $x^2+y^2=1$, $x=costheta$ and $y=sintheta$, for some $thetain[0,2pi)$. Therefore$$xy=cos(theta)sin(theta)=frac12sin(2theta).$$So, the maximum and the minimum are $frac12$ and $-frac12$ respectively. Besides, the maximum is attained when $theta=fracpi4$ or $theta=-frac5pi4$ and the minimum is attained when $theta=frac3pi4$ or when $theta=frac7pi4$.
– José Carlos Santos
21 hours ago
Show please, why the minimum occurs for your points? Thank you.
– Michael Rozenberg
22 hours ago
Show please, why the minimum occurs for your points? Thank you.
– Michael Rozenberg
22 hours ago
@MichaelRozenberg I've edited my answer.
– José Carlos Santos
22 hours ago
@MichaelRozenberg I've edited my answer.
– José Carlos Santos
22 hours ago
Also, you need to say about continuous function on the compact. I think it's anti-maths to find maximum and minimum of $xy$ by the LM method.
– Michael Rozenberg
21 hours ago
Also, you need to say about continuous function on the compact. I think it's anti-maths to find maximum and minimum of $xy$ by the LM method.
– Michael Rozenberg
21 hours ago
1
1
@MichaelRozenberg You are right in both assertions. I will edit my answer once again because of your first remark. Concerning the second one, althought I fully agree with you, it's what the OP asked for. I would not have used Lagrange multipliers otherwise.
– José Carlos Santos
21 hours ago
@MichaelRozenberg You are right in both assertions. I will edit my answer once again because of your first remark. Concerning the second one, althought I fully agree with you, it's what the OP asked for. I would not have used Lagrange multipliers otherwise.
– José Carlos Santos
21 hours ago
1
1
If $x^2+y^2=1$, $x=costheta$ and $y=sintheta$, for some $thetain[0,2pi)$. Therefore$$xy=cos(theta)sin(theta)=frac12sin(2theta).$$So, the maximum and the minimum are $frac12$ and $-frac12$ respectively. Besides, the maximum is attained when $theta=fracpi4$ or $theta=-frac5pi4$ and the minimum is attained when $theta=frac3pi4$ or when $theta=frac7pi4$.
– José Carlos Santos
21 hours ago
If $x^2+y^2=1$, $x=costheta$ and $y=sintheta$, for some $thetain[0,2pi)$. Therefore$$xy=cos(theta)sin(theta)=frac12sin(2theta).$$So, the maximum and the minimum are $frac12$ and $-frac12$ respectively. Besides, the maximum is attained when $theta=fracpi4$ or $theta=-frac5pi4$ and the minimum is attained when $theta=frac3pi4$ or when $theta=frac7pi4$.
– José Carlos Santos
21 hours ago
 |Â
show 4 more comments
up vote
0
down vote
By AM-GM $$ln(3+2xy)leqln(3+2|xy|)leq ln(3+x^2+y^2)=2ln2.$$
THe equality occurs for $x=y=frac1sqrt2,$ which says that we got a maximal value.
Now, by AM-GM again
$$ln(3+2xy)geqln(3-2|xy|)geq ln(3-(x^2+y^2))=ln2.$$
The equality occurs for $x=frac1sqrt2$ and $y=-frac1sqrt2,$ which says that we got a minimal value.
answered 22 hours ago
Michael Rozenberg
86.9k1575178
1
This does not answer the question.
– uniquesolution
22 hours ago
@uniquesolution Why? See better the question.
– Michael Rozenberg
22 hours ago
1
The question asks "Can I solve this exercise using Lagrange multipliers"? What did you see?
– uniquesolution
22 hours ago
@MichaelRozenberg Can this be solved by using Lagrange? If so, how?
– StudentMaths
22 hours ago
@StudentMaths Yes I can, but I think it's an idiotism here. You can say the following words: "the Lagrange Multipliers Method is very useful method" and write my solution.
– Michael Rozenberg
22 hours ago
 |Â
show 7 more comments
up vote
0
down vote
By AM-GM $$ln(3+2xy)leqln(3+2|xy|)leq ln(3+x^2+y^2)=2ln2.$$
THe equality occurs for $x=y=frac1sqrt2,$ which says that we got a maximal value.
Now, by AM-GM again
$$ln(3+2xy)geqln(3-2|xy|)geq ln(3-(x^2+y^2))=ln2.$$
The equality occurs for $x=frac1sqrt2$ and $y=-frac1sqrt2,$ which says that we got a minimal value.
answered 22 hours ago
Michael Rozenberg
86.9k1575178
1
This does not answer the question.
– uniquesolution
22 hours ago
@uniquesolution Why? See better the question.
– Michael Rozenberg
22 hours ago
1
The question asks "Can I solve this exercise using Lagrange multipliers"? What did you see?
– uniquesolution
22 hours ago
@MichaelRozenberg Can this be solved by using Lagrange? If so, how?
– StudentMaths
22 hours ago
@StudentMaths Yes I can, but I think it's an idiotism here. You can say the following words: "the Lagrange Multipliers Method is very useful method" and write my solution.
– Michael Rozenberg
22 hours ago
 |Â
show 7 more comments
up vote
0
down vote
up vote
0
down vote
By AM-GM $$ln(3+2xy)leqln(3+2|xy|)leq ln(3+x^2+y^2)=2ln2.$$
THe equality occurs for $x=y=frac1sqrt2,$ which says that we got a maximal value.
Now, by AM-GM again
$$ln(3+2xy)geqln(3-2|xy|)geq ln(3-(x^2+y^2))=ln2.$$
The equality occurs for $x=frac1sqrt2$ and $y=-frac1sqrt2,$ which says that we got a minimal value.
answered 22 hours ago
Michael Rozenberg
86.9k1575178
By AM-GM $$ln(3+2xy)leqln(3+2|xy|)leq ln(3+x^2+y^2)=2ln2.$$
THe equality occurs for $x=y=frac1sqrt2,$ which says that we got a maximal value.
Now, by AM-GM again
$$ln(3+2xy)geqln(3-2|xy|)geq ln(3-(x^2+y^2))=ln2.$$
The equality occurs for $x=frac1sqrt2$ and $y=-frac1sqrt2,$ which says that we got a minimal value.
answered 22 hours ago
Michael Rozenberg
86.9k1575178
edited 22 hours ago
answered 22 hours ago
Michael Rozenberg
86.9k1575178
answered 22 hours ago
Michael Rozenberg
86.9k1575178
answered 22 hours ago
Michael Rozenberg
86.9k1575178
86.9k1575178
1
This does not answer the question.
– uniquesolution
22 hours ago
@uniquesolution Why? See better the question.
– Michael Rozenberg
22 hours ago
1
The question asks "Can I solve this exercise using Lagrange multipliers"? What did you see?
– uniquesolution
22 hours ago
@MichaelRozenberg Can this be solved by using Lagrange? If so, how?
– StudentMaths
22 hours ago
@StudentMaths Yes I can, but I think it's an idiotism here. You can say the following words: "the Lagrange Multipliers Method is very useful method" and write my solution.
– Michael Rozenberg
22 hours ago
 |Â
show 7 more comments
1
This does not answer the question.
– uniquesolution
22 hours ago
@uniquesolution Why? See better the question.
– Michael Rozenberg
22 hours ago
1
The question asks "Can I solve this exercise using Lagrange multipliers"? What did you see?
– uniquesolution
22 hours ago
@MichaelRozenberg Can this be solved by using Lagrange? If so, how?
– StudentMaths
22 hours ago
@StudentMaths Yes I can, but I think it's an idiotism here. You can say the following words: "the Lagrange Multipliers Method is very useful method" and write my solution.
– Michael Rozenberg
22 hours ago
1
1
This does not answer the question.
– uniquesolution
22 hours ago
This does not answer the question.
– uniquesolution
22 hours ago
@uniquesolution Why? See better the question.
– Michael Rozenberg
22 hours ago
@uniquesolution Why? See better the question.
– Michael Rozenberg
22 hours ago
1
1
The question asks "Can I solve this exercise using Lagrange multipliers"? What did you see?
– uniquesolution
22 hours ago
The question asks "Can I solve this exercise using Lagrange multipliers"? What did you see?
– uniquesolution
22 hours ago
@MichaelRozenberg Can this be solved by using Lagrange? If so, how?
– StudentMaths
22 hours ago
@MichaelRozenberg Can this be solved by using Lagrange? If so, how?
– StudentMaths
22 hours ago
@StudentMaths Yes I can, but I think it's an idiotism here. You can say the following words: "the Lagrange Multipliers Method is very useful method" and write my solution.
– Michael Rozenberg
22 hours ago
@StudentMaths Yes I can, but I think it's an idiotism here. You can say the following words: "the Lagrange Multipliers Method is very useful method" and write my solution.
– Michael Rozenberg
22 hours ago
 |Â
show 7 more comments
up vote
0
down vote
You will get the System
$$frac2y2xy+3+2lambda x=0$$
$$frac2x2xy+3+2lambda y=0$$
$$x^2+y^2=1$$
you can write
$$2lambda x=-frac2y2xy+3$$
$$2lambda y=-frac2x2xy+3$$
dividing both equations you will ge
$$fracxy=fracyx$$
answered 22 hours ago
Dr. Sonnhard Graubner
66.6k32659
add a comment |Â
up vote
0
down vote
You will get the System
$$frac2y2xy+3+2lambda x=0$$
$$frac2x2xy+3+2lambda y=0$$
$$x^2+y^2=1$$
you can write
$$2lambda x=-frac2y2xy+3$$
$$2lambda y=-frac2x2xy+3$$
dividing both equations you will ge
$$fracxy=fracyx$$
answered 22 hours ago
Dr. Sonnhard Graubner
66.6k32659
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You will get the System
$$frac2y2xy+3+2lambda x=0$$
$$frac2x2xy+3+2lambda y=0$$
$$x^2+y^2=1$$
you can write
$$2lambda x=-frac2y2xy+3$$
$$2lambda y=-frac2x2xy+3$$
dividing both equations you will ge
$$fracxy=fracyx$$
answered 22 hours ago
Dr. Sonnhard Graubner
66.6k32659
You will get the System
$$frac2y2xy+3+2lambda x=0$$
$$frac2x2xy+3+2lambda y=0$$
$$x^2+y^2=1$$
you can write
$$2lambda x=-frac2y2xy+3$$
$$2lambda y=-frac2x2xy+3$$
dividing both equations you will ge
$$fracxy=fracyx$$
answered 22 hours ago
Dr. Sonnhard Graubner
66.6k32659
answered 22 hours ago
Dr. Sonnhard Graubner
66.6k32659
answered 22 hours ago
Dr. Sonnhard Graubner
66.6k32659
answered 22 hours ago
Dr. Sonnhard Graubner
66.6k32659
66.6k32659
add a comment |Â
add a comment |Â
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