Mixing
Max and min of $f(x,y)=(3x^2+2y^2-x|x|+2)^{3 over 5}$ on $|x|+|y|le1$ [closed]
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D is a compact and f is continuous so for Weierstrass global max and min exist.Because f(x,y)=f(-x,-y) can I study for y>0?
real-analysis
asked Jan 31 at 22:35
user495707user495707
25
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closed as off-topic by RRL, metamorphy, José Carlos Santos, Adrian Keister, Aweygan Feb 1 at 15:57
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, metamorphy, José Carlos Santos, Adrian Keister, Aweygan
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
D is a compact and f is continuous so for Weierstrass global max and min exist.Because f(x,y)=f(-x,-y) can I study for y>0?
real-analysis
asked Jan 31 at 22:35
user495707user495707
25
$endgroup$
closed as off-topic by RRL, metamorphy, José Carlos Santos, Adrian Keister, Aweygan Feb 1 at 15:57
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, metamorphy, José Carlos Santos, Adrian Keister, Aweygan
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
This is not true $f(x,y)=f(-x,-y).$
$endgroup$
– user376343
Jan 31 at 22:40
$begingroup$
Did you intend to say $f(x,y)=f(x,-y)$? If so, then you only need consider $yge 0$
$endgroup$
– Henry
Jan 31 at 22:41
add a comment |
$begingroup$
D is a compact and f is continuous so for Weierstrass global max and min exist.Because f(x,y)=f(-x,-y) can I study for y>0?
real-analysis
asked Jan 31 at 22:35
user495707user495707
25
$endgroup$
D is a compact and f is continuous so for Weierstrass global max and min exist.Because f(x,y)=f(-x,-y) can I study for y>0?
real-analysis
real-analysis
asked Jan 31 at 22:35
user495707user495707
25
asked Jan 31 at 22:35
user495707user495707
25
asked Jan 31 at 22:35
user495707user495707
25
asked Jan 31 at 22:35
user495707user495707
25
asked Jan 31 at 22:35
user495707user495707
25
25
closed as off-topic by RRL, metamorphy, José Carlos Santos, Adrian Keister, Aweygan Feb 1 at 15:57
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, metamorphy, José Carlos Santos, Adrian Keister, Aweygan
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by RRL, metamorphy, José Carlos Santos, Adrian Keister, Aweygan Feb 1 at 15:57
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, metamorphy, José Carlos Santos, Adrian Keister, Aweygan
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
This is not true $f(x,y)=f(-x,-y).$
$endgroup$
– user376343
Jan 31 at 22:40
$begingroup$
Did you intend to say $f(x,y)=f(x,-y)$? If so, then you only need consider $yge 0$
$endgroup$
– Henry
Jan 31 at 22:41
add a comment |
1
$begingroup$
This is not true $f(x,y)=f(-x,-y).$
$endgroup$
– user376343
Jan 31 at 22:40
$begingroup$
Did you intend to say $f(x,y)=f(x,-y)$? If so, then you only need consider $yge 0$
$endgroup$
– Henry
Jan 31 at 22:41
1
1
$begingroup$
This is not true $f(x,y)=f(-x,-y).$
$endgroup$
– user376343
Jan 31 at 22:40
$begingroup$
This is not true $f(x,y)=f(-x,-y).$
$endgroup$
– user376343
Jan 31 at 22:40
$begingroup$
Did you intend to say $f(x,y)=f(x,-y)$? If so, then you only need consider $yge 0$
$endgroup$
– Henry
Jan 31 at 22:41
$begingroup$
Did you intend to say $f(x,y)=f(x,-y)$? If so, then you only need consider $yge 0$
$endgroup$
– Henry
Jan 31 at 22:41
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Some hints: it is easy to see that $3x^{2}+2y^{2}-x|x|+2 >0$. It is enough to find the max and min without the power and this simplifies a lot. Next, it is enough to find the max and min for $x >0$ and $x leq 0$ separately. This again simplifies the problem. Now you should find the problem much simpler.
answered Jan 31 at 23:39
Kavi Rama MurthyKavi Rama Murthy
73.4k53170
$endgroup$
$begingroup$
So I found local min in $(0,0),(-1/3,2/3),(-1/3,-2/3),(1/2,1/2),(1/2,-1/2)$.Where is the max?(0,0) is the global min
$endgroup$
– user495707
Feb 1 at 8:30
$begingroup$
@user495707 No, need to find all local extrema. First consider $x>0$. Then $3x^{2}+2y^{2}-x^{2}+2=2x^{2}+2y^{2}+2$ can attain global maximum only when $x=1-|y|$ so you only have to maximize $2(1-|y|)^{2}+2y^{2}+2$ subject to $-1leq y leq 1$. The maximum value is $2$. Do a similar thing with $x<0$ and take the maximum of the two numbers you get.
$endgroup$
– Kavi Rama Murthy
Feb 1 at 8:41
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Some hints: it is easy to see that $3x^{2}+2y^{2}-x|x|+2 >0$. It is enough to find the max and min without the power and this simplifies a lot. Next, it is enough to find the max and min for $x >0$ and $x leq 0$ separately. This again simplifies the problem. Now you should find the problem much simpler.
answered Jan 31 at 23:39
Kavi Rama MurthyKavi Rama Murthy
73.4k53170
$endgroup$
$begingroup$
So I found local min in $(0,0),(-1/3,2/3),(-1/3,-2/3),(1/2,1/2),(1/2,-1/2)$.Where is the max?(0,0) is the global min
$endgroup$
– user495707
Feb 1 at 8:30
$begingroup$
@user495707 No, need to find all local extrema. First consider $x>0$. Then $3x^{2}+2y^{2}-x^{2}+2=2x^{2}+2y^{2}+2$ can attain global maximum only when $x=1-|y|$ so you only have to maximize $2(1-|y|)^{2}+2y^{2}+2$ subject to $-1leq y leq 1$. The maximum value is $2$. Do a similar thing with $x<0$ and take the maximum of the two numbers you get.
$endgroup$
– Kavi Rama Murthy
Feb 1 at 8:41
add a comment |
$begingroup$
Some hints: it is easy to see that $3x^{2}+2y^{2}-x|x|+2 >0$. It is enough to find the max and min without the power and this simplifies a lot. Next, it is enough to find the max and min for $x >0$ and $x leq 0$ separately. This again simplifies the problem. Now you should find the problem much simpler.
answered Jan 31 at 23:39
Kavi Rama MurthyKavi Rama Murthy
73.4k53170
$endgroup$
$begingroup$
So I found local min in $(0,0),(-1/3,2/3),(-1/3,-2/3),(1/2,1/2),(1/2,-1/2)$.Where is the max?(0,0) is the global min
$endgroup$
– user495707
Feb 1 at 8:30
$begingroup$
@user495707 No, need to find all local extrema. First consider $x>0$. Then $3x^{2}+2y^{2}-x^{2}+2=2x^{2}+2y^{2}+2$ can attain global maximum only when $x=1-|y|$ so you only have to maximize $2(1-|y|)^{2}+2y^{2}+2$ subject to $-1leq y leq 1$. The maximum value is $2$. Do a similar thing with $x<0$ and take the maximum of the two numbers you get.
$endgroup$
– Kavi Rama Murthy
Feb 1 at 8:41
add a comment |
$begingroup$
Some hints: it is easy to see that $3x^{2}+2y^{2}-x|x|+2 >0$. It is enough to find the max and min without the power and this simplifies a lot. Next, it is enough to find the max and min for $x >0$ and $x leq 0$ separately. This again simplifies the problem. Now you should find the problem much simpler.
answered Jan 31 at 23:39
Kavi Rama MurthyKavi Rama Murthy
73.4k53170
$endgroup$
Some hints: it is easy to see that $3x^{2}+2y^{2}-x|x|+2 >0$. It is enough to find the max and min without the power and this simplifies a lot. Next, it is enough to find the max and min for $x >0$ and $x leq 0$ separately. This again simplifies the problem. Now you should find the problem much simpler.
answered Jan 31 at 23:39
Kavi Rama MurthyKavi Rama Murthy
73.4k53170
answered Jan 31 at 23:39
Kavi Rama MurthyKavi Rama Murthy
73.4k53170
answered Jan 31 at 23:39
Kavi Rama MurthyKavi Rama Murthy
73.4k53170
answered Jan 31 at 23:39
Kavi Rama MurthyKavi Rama Murthy
73.4k53170
73.4k53170
$begingroup$
So I found local min in $(0,0),(-1/3,2/3),(-1/3,-2/3),(1/2,1/2),(1/2,-1/2)$.Where is the max?(0,0) is the global min
$endgroup$
– user495707
Feb 1 at 8:30
$begingroup$
@user495707 No, need to find all local extrema. First consider $x>0$. Then $3x^{2}+2y^{2}-x^{2}+2=2x^{2}+2y^{2}+2$ can attain global maximum only when $x=1-|y|$ so you only have to maximize $2(1-|y|)^{2}+2y^{2}+2$ subject to $-1leq y leq 1$. The maximum value is $2$. Do a similar thing with $x<0$ and take the maximum of the two numbers you get.
$endgroup$
– Kavi Rama Murthy
Feb 1 at 8:41
add a comment |
$begingroup$
So I found local min in $(0,0),(-1/3,2/3),(-1/3,-2/3),(1/2,1/2),(1/2,-1/2)$.Where is the max?(0,0) is the global min
$endgroup$
– user495707
Feb 1 at 8:30
$begingroup$
@user495707 No, need to find all local extrema. First consider $x>0$. Then $3x^{2}+2y^{2}-x^{2}+2=2x^{2}+2y^{2}+2$ can attain global maximum only when $x=1-|y|$ so you only have to maximize $2(1-|y|)^{2}+2y^{2}+2$ subject to $-1leq y leq 1$. The maximum value is $2$. Do a similar thing with $x<0$ and take the maximum of the two numbers you get.
$endgroup$
– Kavi Rama Murthy
Feb 1 at 8:41
$begingroup$
So I found local min in $(0,0),(-1/3,2/3),(-1/3,-2/3),(1/2,1/2),(1/2,-1/2)$.Where is the max?(0,0) is the global min
$endgroup$
– user495707
Feb 1 at 8:30
$begingroup$
So I found local min in $(0,0),(-1/3,2/3),(-1/3,-2/3),(1/2,1/2),(1/2,-1/2)$.Where is the max?(0,0) is the global min
$endgroup$
– user495707
Feb 1 at 8:30
$begingroup$
@user495707 No, need to find all local extrema. First consider $x>0$. Then $3x^{2}+2y^{2}-x^{2}+2=2x^{2}+2y^{2}+2$ can attain global maximum only when $x=1-|y|$ so you only have to maximize $2(1-|y|)^{2}+2y^{2}+2$ subject to $-1leq y leq 1$. The maximum value is $2$. Do a similar thing with $x<0$ and take the maximum of the two numbers you get.
$endgroup$
– Kavi Rama Murthy
Feb 1 at 8:41
$begingroup$
@user495707 No, need to find all local extrema. First consider $x>0$. Then $3x^{2}+2y^{2}-x^{2}+2=2x^{2}+2y^{2}+2$ can attain global maximum only when $x=1-|y|$ so you only have to maximize $2(1-|y|)^{2}+2y^{2}+2$ subject to $-1leq y leq 1$. The maximum value is $2$. Do a similar thing with $x<0$ and take the maximum of the two numbers you get.
$endgroup$
– Kavi Rama Murthy
Feb 1 at 8:41
add a comment |
1
$begingroup$
This is not true $f(x,y)=f(-x,-y).$
$endgroup$
– user376343
Jan 31 at 22:40
$begingroup$
Did you intend to say $f(x,y)=f(x,-y)$? If so, then you only need consider $yge 0$
$endgroup$
– Henry
Jan 31 at 22:41