Why it is necessary to have $yle 0$ in the given problem?
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Find the maximum and minimum values of $f(x,y)=xy$ subject to the constraint $x^2−y=12$. Assume that $y≤0$ for this problem. Why is this assumption needed?
NOTE:
I just want to know the necessity of the last part .why it should be $yle0$.please be elaborate and yes,graphical analysis will be appreciated . Thank you.
multivariable-calculus lagrange-multiplier
asked Nov 17 at 4:20
Rakibul Islam Prince
818211
add a comment |
up vote
1
down vote
favorite
Find the maximum and minimum values of $f(x,y)=xy$ subject to the constraint $x^2−y=12$. Assume that $y≤0$ for this problem. Why is this assumption needed?
NOTE:
I just want to know the necessity of the last part .why it should be $yle0$.please be elaborate and yes,graphical analysis will be appreciated . Thank you.
multivariable-calculus lagrange-multiplier
asked Nov 17 at 4:20
Rakibul Islam Prince
818211
@Decaf-Math I don't see the relevance of this comment.
– suchan
Nov 17 at 4:28
sorry?I didn't get your question.
– Rakibul Islam Prince
Nov 17 at 4:30
There was no question. Someone had left a misleading comment but it is deleted now.
– suchan
Nov 17 at 4:38
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Find the maximum and minimum values of $f(x,y)=xy$ subject to the constraint $x^2−y=12$. Assume that $y≤0$ for this problem. Why is this assumption needed?
NOTE:
I just want to know the necessity of the last part .why it should be $yle0$.please be elaborate and yes,graphical analysis will be appreciated . Thank you.
multivariable-calculus lagrange-multiplier
asked Nov 17 at 4:20
Rakibul Islam Prince
818211
Find the maximum and minimum values of $f(x,y)=xy$ subject to the constraint $x^2−y=12$. Assume that $y≤0$ for this problem. Why is this assumption needed?
NOTE:
I just want to know the necessity of the last part .why it should be $yle0$.please be elaborate and yes,graphical analysis will be appreciated . Thank you.
multivariable-calculus lagrange-multiplier
multivariable-calculus lagrange-multiplier
asked Nov 17 at 4:20
Rakibul Islam Prince
818211
asked Nov 17 at 4:20
Rakibul Islam Prince
818211
asked Nov 17 at 4:20
Rakibul Islam Prince
818211
asked Nov 17 at 4:20
Rakibul Islam Prince
818211
asked Nov 17 at 4:20
Rakibul Islam Prince
818211
818211
@Decaf-Math I don't see the relevance of this comment.
– suchan
Nov 17 at 4:28
sorry?I didn't get your question.
– Rakibul Islam Prince
Nov 17 at 4:30
There was no question. Someone had left a misleading comment but it is deleted now.
– suchan
Nov 17 at 4:38
add a comment |
@Decaf-Math I don't see the relevance of this comment.
– suchan
Nov 17 at 4:28
sorry?I didn't get your question.
– Rakibul Islam Prince
Nov 17 at 4:30
There was no question. Someone had left a misleading comment but it is deleted now.
– suchan
Nov 17 at 4:38
@Decaf-Math I don't see the relevance of this comment.
– suchan
Nov 17 at 4:28
@Decaf-Math I don't see the relevance of this comment.
– suchan
Nov 17 at 4:28
sorry?I didn't get your question.
– Rakibul Islam Prince
Nov 17 at 4:30
sorry?I didn't get your question.
– Rakibul Islam Prince
Nov 17 at 4:30
There was no question. Someone had left a misleading comment but it is deleted now.
– suchan
Nov 17 at 4:38
There was no question. Someone had left a misleading comment but it is deleted now.
– suchan
Nov 17 at 4:38
add a comment |
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
Let's see what happens if this constraint is not asserted.
The problem then become to optimize $x(x^2-12)$ which is an cubic equation which has no maximum nor minimum, making the question too simple for you.
answered Nov 17 at 4:26
Siong Thye Goh
95.7k1462116
But,why then the part is given?
– Rakibul Islam Prince
Nov 17 at 4:33
so that it is more interesting to you? rather than immediately saying that there is no maximum and minimum.
– Siong Thye Goh
Nov 17 at 4:35
As was explained, if a constraint like $y le 0$ were not given, then the problem would not make any sense since $f(x,y)$ would have no global extreme values.
– suchan
Nov 17 at 4:36
If you reduce the problem to a question in $x$ again given the constraint, the question has a maximum and minimum. It make things more fun.
– Siong Thye Goh
Nov 17 at 4:38
sorry...still i can't feel it.would you please please add a graph.
– Rakibul Islam Prince
Nov 17 at 5:00
|
show 5 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Let's see what happens if this constraint is not asserted.
The problem then become to optimize $x(x^2-12)$ which is an cubic equation which has no maximum nor minimum, making the question too simple for you.
answered Nov 17 at 4:26
Siong Thye Goh
95.7k1462116
But,why then the part is given?
– Rakibul Islam Prince
Nov 17 at 4:33
so that it is more interesting to you? rather than immediately saying that there is no maximum and minimum.
– Siong Thye Goh
Nov 17 at 4:35
As was explained, if a constraint like $y le 0$ were not given, then the problem would not make any sense since $f(x,y)$ would have no global extreme values.
– suchan
Nov 17 at 4:36
If you reduce the problem to a question in $x$ again given the constraint, the question has a maximum and minimum. It make things more fun.
– Siong Thye Goh
Nov 17 at 4:38
sorry...still i can't feel it.would you please please add a graph.
– Rakibul Islam Prince
Nov 17 at 5:00
|
show 5 more comments
up vote
3
down vote
accepted
Let's see what happens if this constraint is not asserted.
The problem then become to optimize $x(x^2-12)$ which is an cubic equation which has no maximum nor minimum, making the question too simple for you.
answered Nov 17 at 4:26
Siong Thye Goh
95.7k1462116
But,why then the part is given?
– Rakibul Islam Prince
Nov 17 at 4:33
so that it is more interesting to you? rather than immediately saying that there is no maximum and minimum.
– Siong Thye Goh
Nov 17 at 4:35
As was explained, if a constraint like $y le 0$ were not given, then the problem would not make any sense since $f(x,y)$ would have no global extreme values.
– suchan
Nov 17 at 4:36
If you reduce the problem to a question in $x$ again given the constraint, the question has a maximum and minimum. It make things more fun.
– Siong Thye Goh
Nov 17 at 4:38
sorry...still i can't feel it.would you please please add a graph.
– Rakibul Islam Prince
Nov 17 at 5:00
|
show 5 more comments
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Let's see what happens if this constraint is not asserted.
The problem then become to optimize $x(x^2-12)$ which is an cubic equation which has no maximum nor minimum, making the question too simple for you.
answered Nov 17 at 4:26
Siong Thye Goh
95.7k1462116
Let's see what happens if this constraint is not asserted.
The problem then become to optimize $x(x^2-12)$ which is an cubic equation which has no maximum nor minimum, making the question too simple for you.
answered Nov 17 at 4:26
Siong Thye Goh
95.7k1462116
edited Nov 17 at 5:23
answered Nov 17 at 4:26
Siong Thye Goh
95.7k1462116
answered Nov 17 at 4:26
Siong Thye Goh
95.7k1462116
answered Nov 17 at 4:26
Siong Thye Goh
95.7k1462116
95.7k1462116
But,why then the part is given?
– Rakibul Islam Prince
Nov 17 at 4:33
so that it is more interesting to you? rather than immediately saying that there is no maximum and minimum.
– Siong Thye Goh
Nov 17 at 4:35
As was explained, if a constraint like $y le 0$ were not given, then the problem would not make any sense since $f(x,y)$ would have no global extreme values.
– suchan
Nov 17 at 4:36
If you reduce the problem to a question in $x$ again given the constraint, the question has a maximum and minimum. It make things more fun.
– Siong Thye Goh
Nov 17 at 4:38
sorry...still i can't feel it.would you please please add a graph.
– Rakibul Islam Prince
Nov 17 at 5:00
|
show 5 more comments
But,why then the part is given?
– Rakibul Islam Prince
Nov 17 at 4:33
so that it is more interesting to you? rather than immediately saying that there is no maximum and minimum.
– Siong Thye Goh
Nov 17 at 4:35
As was explained, if a constraint like $y le 0$ were not given, then the problem would not make any sense since $f(x,y)$ would have no global extreme values.
– suchan
Nov 17 at 4:36
If you reduce the problem to a question in $x$ again given the constraint, the question has a maximum and minimum. It make things more fun.
– Siong Thye Goh
Nov 17 at 4:38
sorry...still i can't feel it.would you please please add a graph.
– Rakibul Islam Prince
Nov 17 at 5:00
But,why then the part is given?
– Rakibul Islam Prince
Nov 17 at 4:33
But,why then the part is given?
– Rakibul Islam Prince
Nov 17 at 4:33
so that it is more interesting to you? rather than immediately saying that there is no maximum and minimum.
– Siong Thye Goh
Nov 17 at 4:35
so that it is more interesting to you? rather than immediately saying that there is no maximum and minimum.
– Siong Thye Goh
Nov 17 at 4:35
As was explained, if a constraint like $y le 0$ were not given, then the problem would not make any sense since $f(x,y)$ would have no global extreme values.
– suchan
Nov 17 at 4:36
As was explained, if a constraint like $y le 0$ were not given, then the problem would not make any sense since $f(x,y)$ would have no global extreme values.
– suchan
Nov 17 at 4:36
If you reduce the problem to a question in $x$ again given the constraint, the question has a maximum and minimum. It make things more fun.
– Siong Thye Goh
Nov 17 at 4:38
If you reduce the problem to a question in $x$ again given the constraint, the question has a maximum and minimum. It make things more fun.
– Siong Thye Goh
Nov 17 at 4:38
sorry...still i can't feel it.would you please please add a graph.
– Rakibul Islam Prince
Nov 17 at 5:00
sorry...still i can't feel it.would you please please add a graph.
– Rakibul Islam Prince
Nov 17 at 5:00
|
show 5 more comments
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@Decaf-Math I don't see the relevance of this comment.
– suchan
Nov 17 at 4:28
sorry?I didn't get your question.
– Rakibul Islam Prince
Nov 17 at 4:30
There was no question. Someone had left a misleading comment but it is deleted now.
– suchan
Nov 17 at 4:38