Mixing
(a bit difficult) finding maximum and minimum maybe using Lagrange's multiplier
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up vote
-1
down vote
favorite
thanks for reading this first :D
I'm trying to solve the problem
"Finding the maximum and minimum of
$x^2+2y$
under the conditions $x^3+3xy+y^3=5, ygeq0$"
I would super appriciate your help. Cheers.
calculus lagrange-multiplier
asked Jul 17 at 1:28
Sehun Kim
33
add a comment |Â
up vote
-1
down vote
favorite
thanks for reading this first :D
I'm trying to solve the problem
"Finding the maximum and minimum of
$x^2+2y$
under the conditions $x^3+3xy+y^3=5, ygeq0$"
I would super appriciate your help. Cheers.
calculus lagrange-multiplier
asked Jul 17 at 1:28
Sehun Kim
33
1
The first step is to check that a $min$ or $max$ exist. It is fairly straightforward to show that a $min$ exists, and a little work shows that the function is unbounded above.
– copper.hat
Jul 17 at 3:54
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
thanks for reading this first :D
I'm trying to solve the problem
"Finding the maximum and minimum of
$x^2+2y$
under the conditions $x^3+3xy+y^3=5, ygeq0$"
I would super appriciate your help. Cheers.
calculus lagrange-multiplier
asked Jul 17 at 1:28
Sehun Kim
33
thanks for reading this first :D
I'm trying to solve the problem
"Finding the maximum and minimum of
$x^2+2y$
under the conditions $x^3+3xy+y^3=5, ygeq0$"
I would super appriciate your help. Cheers.
calculus lagrange-multiplier
asked Jul 17 at 1:28
Sehun Kim
33
edited Jul 17 at 1:46
asked Jul 17 at 1:28
Sehun Kim
33
asked Jul 17 at 1:28
Sehun Kim
33
asked Jul 17 at 1:28
Sehun Kim
33
33
1
The first step is to check that a $min$ or $max$ exist. It is fairly straightforward to show that a $min$ exists, and a little work shows that the function is unbounded above.
– copper.hat
Jul 17 at 3:54
add a comment |Â
1
The first step is to check that a $min$ or $max$ exist. It is fairly straightforward to show that a $min$ exists, and a little work shows that the function is unbounded above.
– copper.hat
Jul 17 at 3:54
1
1
The first step is to check that a $min$ or $max$ exist. It is fairly straightforward to show that a $min$ exists, and a little work shows that the function is unbounded above.
– copper.hat
Jul 17 at 3:54
The first step is to check that a $min$ or $max$ exist. It is fairly straightforward to show that a $min$ exists, and a little work shows that the function is unbounded above.
– copper.hat
Jul 17 at 3:54
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
I'm giving you some hint to get proceed. Use Lagrange multiplier.
The two equations will be $ 2x=k(3x^2+3y) $ and $2= k(3x+3y^2)$. From 1st equation put $k=0$ , you can easily check this holds no good as it demands $2=0$
Next divide those equations. You will get conditions like $y=0$ and $xy=1$. From $y=0$, you will immediately get a value of $x$.
On the equation $x^3+3xy+y^3=5$, put $xy=1$. You get solutions like $x+y=2$. And I'm sure you can proceed thereafter.
answered Jul 17 at 2:36
Arnab Chowdhury
1489
I don't follow your hints. How can you just set $x=0$?
– copper.hat
Jul 17 at 4:15
I am sorry, I had mistakenly placed $x$ in place of $k$. Corrected now. And yes you have to check wheather maximum or minimum value exists.
– Arnab Chowdhury
Jul 17 at 4:22
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
I'm giving you some hint to get proceed. Use Lagrange multiplier.
The two equations will be $ 2x=k(3x^2+3y) $ and $2= k(3x+3y^2)$. From 1st equation put $k=0$ , you can easily check this holds no good as it demands $2=0$
Next divide those equations. You will get conditions like $y=0$ and $xy=1$. From $y=0$, you will immediately get a value of $x$.
On the equation $x^3+3xy+y^3=5$, put $xy=1$. You get solutions like $x+y=2$. And I'm sure you can proceed thereafter.
answered Jul 17 at 2:36
Arnab Chowdhury
1489
I don't follow your hints. How can you just set $x=0$?
– copper.hat
Jul 17 at 4:15
I am sorry, I had mistakenly placed $x$ in place of $k$. Corrected now. And yes you have to check wheather maximum or minimum value exists.
– Arnab Chowdhury
Jul 17 at 4:22
add a comment |Â
up vote
2
down vote
accepted
I'm giving you some hint to get proceed. Use Lagrange multiplier.
The two equations will be $ 2x=k(3x^2+3y) $ and $2= k(3x+3y^2)$. From 1st equation put $k=0$ , you can easily check this holds no good as it demands $2=0$
Next divide those equations. You will get conditions like $y=0$ and $xy=1$. From $y=0$, you will immediately get a value of $x$.
On the equation $x^3+3xy+y^3=5$, put $xy=1$. You get solutions like $x+y=2$. And I'm sure you can proceed thereafter.
answered Jul 17 at 2:36
Arnab Chowdhury
1489
I don't follow your hints. How can you just set $x=0$?
– copper.hat
Jul 17 at 4:15
I am sorry, I had mistakenly placed $x$ in place of $k$. Corrected now. And yes you have to check wheather maximum or minimum value exists.
– Arnab Chowdhury
Jul 17 at 4:22
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
I'm giving you some hint to get proceed. Use Lagrange multiplier.
The two equations will be $ 2x=k(3x^2+3y) $ and $2= k(3x+3y^2)$. From 1st equation put $k=0$ , you can easily check this holds no good as it demands $2=0$
Next divide those equations. You will get conditions like $y=0$ and $xy=1$. From $y=0$, you will immediately get a value of $x$.
On the equation $x^3+3xy+y^3=5$, put $xy=1$. You get solutions like $x+y=2$. And I'm sure you can proceed thereafter.
answered Jul 17 at 2:36
Arnab Chowdhury
1489
I'm giving you some hint to get proceed. Use Lagrange multiplier.
The two equations will be $ 2x=k(3x^2+3y) $ and $2= k(3x+3y^2)$. From 1st equation put $k=0$ , you can easily check this holds no good as it demands $2=0$
Next divide those equations. You will get conditions like $y=0$ and $xy=1$. From $y=0$, you will immediately get a value of $x$.
On the equation $x^3+3xy+y^3=5$, put $xy=1$. You get solutions like $x+y=2$. And I'm sure you can proceed thereafter.
answered Jul 17 at 2:36
Arnab Chowdhury
1489
edited Jul 17 at 8:00
answered Jul 17 at 2:36
Arnab Chowdhury
1489
answered Jul 17 at 2:36
Arnab Chowdhury
1489
answered Jul 17 at 2:36
Arnab Chowdhury
1489
1489
I don't follow your hints. How can you just set $x=0$?
– copper.hat
Jul 17 at 4:15
I am sorry, I had mistakenly placed $x$ in place of $k$. Corrected now. And yes you have to check wheather maximum or minimum value exists.
– Arnab Chowdhury
Jul 17 at 4:22
add a comment |Â
I don't follow your hints. How can you just set $x=0$?
– copper.hat
Jul 17 at 4:15
I am sorry, I had mistakenly placed $x$ in place of $k$. Corrected now. And yes you have to check wheather maximum or minimum value exists.
– Arnab Chowdhury
Jul 17 at 4:22
I don't follow your hints. How can you just set $x=0$?
– copper.hat
Jul 17 at 4:15
I don't follow your hints. How can you just set $x=0$?
– copper.hat
Jul 17 at 4:15
I am sorry, I had mistakenly placed $x$ in place of $k$. Corrected now. And yes you have to check wheather maximum or minimum value exists.
– Arnab Chowdhury
Jul 17 at 4:22
I am sorry, I had mistakenly placed $x$ in place of $k$. Corrected now. And yes you have to check wheather maximum or minimum value exists.
– Arnab Chowdhury
Jul 17 at 4:22
add a comment |Â
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1
The first step is to check that a $min$ or $max$ exist. It is fairly straightforward to show that a $min$ exists, and a little work shows that the function is unbounded above.
– copper.hat
Jul 17 at 3:54