How to find absolute mininum/maximum of a function on a set?
$begingroup$
I have this function :
$f(x,y)=x²+y²-2y-x$
I want to find the absolute min/max of the function on the set given in the figure below (where L3 is a piece of a circle with center (0, 0).
I've already describes lines 1 and 2 and found their critical points. Now working with L3 i see on the answer sheet that it is described as
$L3=((x,y): x²+y²=3², −3 ≤ x ≤ 0)$
May I ask you where they are getting the 3² ? I know it may be a dumb question but i just recently started studying the concept of absolute min and max, moreover with coordinates that make it quite harder. Thanks!
functions trigonometry analytic-geometry absolute-value
edited Dec 15 '18 at 15:46
Key Flex
8,32261233
asked Dec 15 '18 at 15:39
BM97BM97
758
$endgroup$
add a comment |
$begingroup$
I have this function :
$f(x,y)=x²+y²-2y-x$
I want to find the absolute min/max of the function on the set given in the figure below (where L3 is a piece of a circle with center (0, 0).
I've already describes lines 1 and 2 and found their critical points. Now working with L3 i see on the answer sheet that it is described as
$L3=((x,y): x²+y²=3², −3 ≤ x ≤ 0)$
May I ask you where they are getting the 3² ? I know it may be a dumb question but i just recently started studying the concept of absolute min and max, moreover with coordinates that make it quite harder. Thanks!
functions trigonometry analytic-geometry absolute-value
edited Dec 15 '18 at 15:46
Key Flex
8,32261233
asked Dec 15 '18 at 15:39
BM97BM97
758
$endgroup$
add a comment |
$begingroup$
I have this function :
$f(x,y)=x²+y²-2y-x$
I want to find the absolute min/max of the function on the set given in the figure below (where L3 is a piece of a circle with center (0, 0).
I've already describes lines 1 and 2 and found their critical points. Now working with L3 i see on the answer sheet that it is described as
$L3=((x,y): x²+y²=3², −3 ≤ x ≤ 0)$
May I ask you where they are getting the 3² ? I know it may be a dumb question but i just recently started studying the concept of absolute min and max, moreover with coordinates that make it quite harder. Thanks!
functions trigonometry analytic-geometry absolute-value
edited Dec 15 '18 at 15:46
Key Flex
8,32261233
asked Dec 15 '18 at 15:39
BM97BM97
758
$endgroup$
I have this function :
$f(x,y)=x²+y²-2y-x$
I want to find the absolute min/max of the function on the set given in the figure below (where L3 is a piece of a circle with center (0, 0).
I've already describes lines 1 and 2 and found their critical points. Now working with L3 i see on the answer sheet that it is described as
$L3=((x,y): x²+y²=3², −3 ≤ x ≤ 0)$
May I ask you where they are getting the 3² ? I know it may be a dumb question but i just recently started studying the concept of absolute min and max, moreover with coordinates that make it quite harder. Thanks!
functions trigonometry analytic-geometry absolute-value
functions trigonometry analytic-geometry absolute-value
edited Dec 15 '18 at 15:46
Key Flex
8,32261233
asked Dec 15 '18 at 15:39
BM97BM97
758
edited Dec 15 '18 at 15:46
Key Flex
8,32261233
asked Dec 15 '18 at 15:39
BM97BM97
758
edited Dec 15 '18 at 15:46
Key Flex
8,32261233
edited Dec 15 '18 at 15:46
Key Flex
8,32261233
edited Dec 15 '18 at 15:46
Key Flex
8,32261233
8,32261233
asked Dec 15 '18 at 15:39
BM97BM97
758
asked Dec 15 '18 at 15:39
BM97BM97
758
asked Dec 15 '18 at 15:39
BM97BM97
758
758
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The formula of the circle at origin with radius $r$ is $x^2+y^2=r^2$, here the radius is $3$.
As for the origin problem, you might like to visualize $f$ as
$$x^2+y^2-2y-x=(x-frac12)^2-frac14+(y-1)^2-1$$
answered Dec 15 '18 at 15:43
Siong Thye GohSiong Thye Goh
102k1467119
$endgroup$
$begingroup$
Thanks! Grealty appreciate your answer.. One last thing... how do i know the radius is 3?
$endgroup$
– BM97
Dec 15 '18 at 15:48
$begingroup$
$L_3$ is a piece of circle centered at origin, and it passes through $(0,3)$. The distance between $(0,3)$ and $(0,0)$ is $3$.
$endgroup$
– Siong Thye Goh
Dec 15 '18 at 15:51
$begingroup$
Why are you taking into consideration only point (0,3) ? doesn't it also pass through (-3,0)?
$endgroup$
– BM97
Dec 15 '18 at 15:57
$begingroup$
Sure, you can consider that point too, you will find that the distance to origin (the center of the circle) is constant, $3$.
$endgroup$
– Siong Thye Goh
Dec 15 '18 at 15:58
$begingroup$
Thank you !!!!!!!!!!!
$endgroup$
– BM97
Dec 15 '18 at 15:59
|
show 2 more comments
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The formula of the circle at origin with radius $r$ is $x^2+y^2=r^2$, here the radius is $3$.
As for the origin problem, you might like to visualize $f$ as
$$x^2+y^2-2y-x=(x-frac12)^2-frac14+(y-1)^2-1$$
answered Dec 15 '18 at 15:43
Siong Thye GohSiong Thye Goh
102k1467119
$endgroup$
$begingroup$
Thanks! Grealty appreciate your answer.. One last thing... how do i know the radius is 3?
$endgroup$
– BM97
Dec 15 '18 at 15:48
$begingroup$
$L_3$ is a piece of circle centered at origin, and it passes through $(0,3)$. The distance between $(0,3)$ and $(0,0)$ is $3$.
$endgroup$
– Siong Thye Goh
Dec 15 '18 at 15:51
$begingroup$
Why are you taking into consideration only point (0,3) ? doesn't it also pass through (-3,0)?
$endgroup$
– BM97
Dec 15 '18 at 15:57
$begingroup$
Sure, you can consider that point too, you will find that the distance to origin (the center of the circle) is constant, $3$.
$endgroup$
– Siong Thye Goh
Dec 15 '18 at 15:58
$begingroup$
Thank you !!!!!!!!!!!
$endgroup$
– BM97
Dec 15 '18 at 15:59
|
show 2 more comments
$begingroup$
The formula of the circle at origin with radius $r$ is $x^2+y^2=r^2$, here the radius is $3$.
As for the origin problem, you might like to visualize $f$ as
$$x^2+y^2-2y-x=(x-frac12)^2-frac14+(y-1)^2-1$$
answered Dec 15 '18 at 15:43
Siong Thye GohSiong Thye Goh
102k1467119
$endgroup$
$begingroup$
Thanks! Grealty appreciate your answer.. One last thing... how do i know the radius is 3?
$endgroup$
– BM97
Dec 15 '18 at 15:48
$begingroup$
$L_3$ is a piece of circle centered at origin, and it passes through $(0,3)$. The distance between $(0,3)$ and $(0,0)$ is $3$.
$endgroup$
– Siong Thye Goh
Dec 15 '18 at 15:51
$begingroup$
Why are you taking into consideration only point (0,3) ? doesn't it also pass through (-3,0)?
$endgroup$
– BM97
Dec 15 '18 at 15:57
$begingroup$
Sure, you can consider that point too, you will find that the distance to origin (the center of the circle) is constant, $3$.
$endgroup$
– Siong Thye Goh
Dec 15 '18 at 15:58
$begingroup$
Thank you !!!!!!!!!!!
$endgroup$
– BM97
Dec 15 '18 at 15:59
|
show 2 more comments
$begingroup$
The formula of the circle at origin with radius $r$ is $x^2+y^2=r^2$, here the radius is $3$.
As for the origin problem, you might like to visualize $f$ as
$$x^2+y^2-2y-x=(x-frac12)^2-frac14+(y-1)^2-1$$
answered Dec 15 '18 at 15:43
Siong Thye GohSiong Thye Goh
102k1467119
$endgroup$
The formula of the circle at origin with radius $r$ is $x^2+y^2=r^2$, here the radius is $3$.
As for the origin problem, you might like to visualize $f$ as
$$x^2+y^2-2y-x=(x-frac12)^2-frac14+(y-1)^2-1$$
answered Dec 15 '18 at 15:43
Siong Thye GohSiong Thye Goh
102k1467119
answered Dec 15 '18 at 15:43
Siong Thye GohSiong Thye Goh
102k1467119
answered Dec 15 '18 at 15:43
Siong Thye GohSiong Thye Goh
102k1467119
answered Dec 15 '18 at 15:43
Siong Thye GohSiong Thye Goh
102k1467119
102k1467119
$begingroup$
Thanks! Grealty appreciate your answer.. One last thing... how do i know the radius is 3?
$endgroup$
– BM97
Dec 15 '18 at 15:48
$begingroup$
$L_3$ is a piece of circle centered at origin, and it passes through $(0,3)$. The distance between $(0,3)$ and $(0,0)$ is $3$.
$endgroup$
– Siong Thye Goh
Dec 15 '18 at 15:51
$begingroup$
Why are you taking into consideration only point (0,3) ? doesn't it also pass through (-3,0)?
$endgroup$
– BM97
Dec 15 '18 at 15:57
$begingroup$
Sure, you can consider that point too, you will find that the distance to origin (the center of the circle) is constant, $3$.
$endgroup$
– Siong Thye Goh
Dec 15 '18 at 15:58
$begingroup$
Thank you !!!!!!!!!!!
$endgroup$
– BM97
Dec 15 '18 at 15:59
|
show 2 more comments
$begingroup$
Thanks! Grealty appreciate your answer.. One last thing... how do i know the radius is 3?
$endgroup$
– BM97
Dec 15 '18 at 15:48
$begingroup$
$L_3$ is a piece of circle centered at origin, and it passes through $(0,3)$. The distance between $(0,3)$ and $(0,0)$ is $3$.
$endgroup$
– Siong Thye Goh
Dec 15 '18 at 15:51
$begingroup$
Why are you taking into consideration only point (0,3) ? doesn't it also pass through (-3,0)?
$endgroup$
– BM97
Dec 15 '18 at 15:57
$begingroup$
Sure, you can consider that point too, you will find that the distance to origin (the center of the circle) is constant, $3$.
$endgroup$
– Siong Thye Goh
Dec 15 '18 at 15:58
$begingroup$
Thank you !!!!!!!!!!!
$endgroup$
– BM97
Dec 15 '18 at 15:59
$begingroup$
Thanks! Grealty appreciate your answer.. One last thing... how do i know the radius is 3?
$endgroup$
– BM97
Dec 15 '18 at 15:48
$begingroup$
Thanks! Grealty appreciate your answer.. One last thing... how do i know the radius is 3?
$endgroup$
– BM97
Dec 15 '18 at 15:48
$begingroup$
$L_3$ is a piece of circle centered at origin, and it passes through $(0,3)$. The distance between $(0,3)$ and $(0,0)$ is $3$.
$endgroup$
– Siong Thye Goh
Dec 15 '18 at 15:51
$begingroup$
$L_3$ is a piece of circle centered at origin, and it passes through $(0,3)$. The distance between $(0,3)$ and $(0,0)$ is $3$.
$endgroup$
– Siong Thye Goh
Dec 15 '18 at 15:51
$begingroup$
Why are you taking into consideration only point (0,3) ? doesn't it also pass through (-3,0)?
$endgroup$
– BM97
Dec 15 '18 at 15:57
$begingroup$
Why are you taking into consideration only point (0,3) ? doesn't it also pass through (-3,0)?
$endgroup$
– BM97
Dec 15 '18 at 15:57
$begingroup$
Sure, you can consider that point too, you will find that the distance to origin (the center of the circle) is constant, $3$.
$endgroup$
– Siong Thye Goh
Dec 15 '18 at 15:58
$begingroup$
Sure, you can consider that point too, you will find that the distance to origin (the center of the circle) is constant, $3$.
$endgroup$
– Siong Thye Goh
Dec 15 '18 at 15:58
$begingroup$
Thank you !!!!!!!!!!!
$endgroup$
– BM97
Dec 15 '18 at 15:59
$begingroup$
Thank you !!!!!!!!!!!
$endgroup$
– BM97
Dec 15 '18 at 15:59
|
show 2 more comments
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