Mixing
Local maximum and level curves on a function
$begingroup$
This was an exam question I found and I need some help figuring it out
A village is founded in a mountain area of which the height (in meters) is calculated by
$$H(x,y)=500e^{-(x-2)^2-(y-x)^2}$$
a) Check if $H$ has a local maximum. Is this the global maximum? Where is this achieved and what is the height?
b) The centre of the village is found at point $(0,0)$. From there a road goes around the village at a constant height. On which point does it have the highest $y$-value and on which the lowest?
For a) I started with normal techniques to calculate the local max and found $(2,2)$ to be it. But i'm not quite sure how to get b).
multivariable-calculus
edited Jan 13 at 15:34
Parcly Taxel
41.8k1372101
asked Jan 13 at 15:32
TournaTourna
126
$endgroup$
add a comment |
$begingroup$
This was an exam question I found and I need some help figuring it out
A village is founded in a mountain area of which the height (in meters) is calculated by
$$H(x,y)=500e^{-(x-2)^2-(y-x)^2}$$
a) Check if $H$ has a local maximum. Is this the global maximum? Where is this achieved and what is the height?
b) The centre of the village is found at point $(0,0)$. From there a road goes around the village at a constant height. On which point does it have the highest $y$-value and on which the lowest?
For a) I started with normal techniques to calculate the local max and found $(2,2)$ to be it. But i'm not quite sure how to get b).
multivariable-calculus
edited Jan 13 at 15:34
Parcly Taxel
41.8k1372101
asked Jan 13 at 15:32
TournaTourna
126
$endgroup$
$begingroup$
There is a global maximum at $$(2,2)$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 13 at 15:39
$begingroup$
Hint: the height depends only on $(x-2)^2 + (y-x)^2$. At $(0,0)$ this is $4$.
$endgroup$
– Robert Israel
Jan 13 at 15:39
$begingroup$
@RobertIsrael Aha so I don't have to look at the rest anymore? And to answer b I would just have to calculate the maximum and minimum of that function?
$endgroup$
– Tourna
Jan 13 at 15:44
$begingroup$
The maximum and minimum of $y$ subject to $(x-2)^2 + (y-x)^2 = 4$.
$endgroup$
– Robert Israel
Jan 13 at 20:07
add a comment |
$begingroup$
This was an exam question I found and I need some help figuring it out
A village is founded in a mountain area of which the height (in meters) is calculated by
$$H(x,y)=500e^{-(x-2)^2-(y-x)^2}$$
a) Check if $H$ has a local maximum. Is this the global maximum? Where is this achieved and what is the height?
b) The centre of the village is found at point $(0,0)$. From there a road goes around the village at a constant height. On which point does it have the highest $y$-value and on which the lowest?
For a) I started with normal techniques to calculate the local max and found $(2,2)$ to be it. But i'm not quite sure how to get b).
multivariable-calculus
edited Jan 13 at 15:34
Parcly Taxel
41.8k1372101
asked Jan 13 at 15:32
TournaTourna
126
$endgroup$
This was an exam question I found and I need some help figuring it out
A village is founded in a mountain area of which the height (in meters) is calculated by
$$H(x,y)=500e^{-(x-2)^2-(y-x)^2}$$
a) Check if $H$ has a local maximum. Is this the global maximum? Where is this achieved and what is the height?
b) The centre of the village is found at point $(0,0)$. From there a road goes around the village at a constant height. On which point does it have the highest $y$-value and on which the lowest?
For a) I started with normal techniques to calculate the local max and found $(2,2)$ to be it. But i'm not quite sure how to get b).
multivariable-calculus
multivariable-calculus
edited Jan 13 at 15:34
Parcly Taxel
41.8k1372101
asked Jan 13 at 15:32
TournaTourna
126
edited Jan 13 at 15:34
Parcly Taxel
41.8k1372101
asked Jan 13 at 15:32
TournaTourna
126
edited Jan 13 at 15:34
Parcly Taxel
41.8k1372101
edited Jan 13 at 15:34
Parcly Taxel
41.8k1372101
edited Jan 13 at 15:34
Parcly Taxel
41.8k1372101
41.8k1372101
asked Jan 13 at 15:32
TournaTourna
126
asked Jan 13 at 15:32
TournaTourna
126
asked Jan 13 at 15:32
TournaTourna
126
126
$begingroup$
There is a global maximum at $$(2,2)$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 13 at 15:39
$begingroup$
Hint: the height depends only on $(x-2)^2 + (y-x)^2$. At $(0,0)$ this is $4$.
$endgroup$
– Robert Israel
Jan 13 at 15:39
$begingroup$
@RobertIsrael Aha so I don't have to look at the rest anymore? And to answer b I would just have to calculate the maximum and minimum of that function?
$endgroup$
– Tourna
Jan 13 at 15:44
$begingroup$
The maximum and minimum of $y$ subject to $(x-2)^2 + (y-x)^2 = 4$.
$endgroup$
– Robert Israel
Jan 13 at 20:07
add a comment |
$begingroup$
There is a global maximum at $$(2,2)$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 13 at 15:39
$begingroup$
Hint: the height depends only on $(x-2)^2 + (y-x)^2$. At $(0,0)$ this is $4$.
$endgroup$
– Robert Israel
Jan 13 at 15:39
$begingroup$
@RobertIsrael Aha so I don't have to look at the rest anymore? And to answer b I would just have to calculate the maximum and minimum of that function?
$endgroup$
– Tourna
Jan 13 at 15:44
$begingroup$
The maximum and minimum of $y$ subject to $(x-2)^2 + (y-x)^2 = 4$.
$endgroup$
– Robert Israel
Jan 13 at 20:07
$begingroup$
There is a global maximum at $$(2,2)$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 13 at 15:39
$begingroup$
There is a global maximum at $$(2,2)$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 13 at 15:39
$begingroup$
Hint: the height depends only on $(x-2)^2 + (y-x)^2$. At $(0,0)$ this is $4$.
$endgroup$
– Robert Israel
Jan 13 at 15:39
$begingroup$
Hint: the height depends only on $(x-2)^2 + (y-x)^2$. At $(0,0)$ this is $4$.
$endgroup$
– Robert Israel
Jan 13 at 15:39
$begingroup$
@RobertIsrael Aha so I don't have to look at the rest anymore? And to answer b I would just have to calculate the maximum and minimum of that function?
$endgroup$
– Tourna
Jan 13 at 15:44
$begingroup$
@RobertIsrael Aha so I don't have to look at the rest anymore? And to answer b I would just have to calculate the maximum and minimum of that function?
$endgroup$
– Tourna
Jan 13 at 15:44
$begingroup$
The maximum and minimum of $y$ subject to $(x-2)^2 + (y-x)^2 = 4$.
$endgroup$
– Robert Israel
Jan 13 at 20:07
$begingroup$
The maximum and minimum of $y$ subject to $(x-2)^2 + (y-x)^2 = 4$.
$endgroup$
– Robert Israel
Jan 13 at 20:07
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint
The equation of whole the village would be$$e^{-4}=e^{-(x-2)^2-(x-y)^2}$$therefore $$(x-2)^2+(x-y)^2=4$$By defining $$x-2=2costheta\y-x=2sin theta$$we obtain $$x=2+2costheta\y=2+2sin theta+2costheta$$from which the maximum and minimum could be easily found.
answered Jan 13 at 16:21
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
$begingroup$
Thank you very much! So to find the highest y-value I just have to derive the y = 2 + 2sin + 2cos, find stationary points and do second derivate test?
$endgroup$
– Tourna
Jan 13 at 17:36
$begingroup$
You're welcome. That's right. After finding $theta$ substitute it in $x$ to find the full coordinates....
$endgroup$
– Mostafa Ayaz
Jan 13 at 17:41
add a comment |
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$begingroup$
Hint
The equation of whole the village would be$$e^{-4}=e^{-(x-2)^2-(x-y)^2}$$therefore $$(x-2)^2+(x-y)^2=4$$By defining $$x-2=2costheta\y-x=2sin theta$$we obtain $$x=2+2costheta\y=2+2sin theta+2costheta$$from which the maximum and minimum could be easily found.
answered Jan 13 at 16:21
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
$begingroup$
Thank you very much! So to find the highest y-value I just have to derive the y = 2 + 2sin + 2cos, find stationary points and do second derivate test?
$endgroup$
– Tourna
Jan 13 at 17:36
$begingroup$
You're welcome. That's right. After finding $theta$ substitute it in $x$ to find the full coordinates....
$endgroup$
– Mostafa Ayaz
Jan 13 at 17:41
add a comment |
$begingroup$
Hint
The equation of whole the village would be$$e^{-4}=e^{-(x-2)^2-(x-y)^2}$$therefore $$(x-2)^2+(x-y)^2=4$$By defining $$x-2=2costheta\y-x=2sin theta$$we obtain $$x=2+2costheta\y=2+2sin theta+2costheta$$from which the maximum and minimum could be easily found.
answered Jan 13 at 16:21
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
$begingroup$
Thank you very much! So to find the highest y-value I just have to derive the y = 2 + 2sin + 2cos, find stationary points and do second derivate test?
$endgroup$
– Tourna
Jan 13 at 17:36
$begingroup$
You're welcome. That's right. After finding $theta$ substitute it in $x$ to find the full coordinates....
$endgroup$
– Mostafa Ayaz
Jan 13 at 17:41
add a comment |
$begingroup$
Hint
The equation of whole the village would be$$e^{-4}=e^{-(x-2)^2-(x-y)^2}$$therefore $$(x-2)^2+(x-y)^2=4$$By defining $$x-2=2costheta\y-x=2sin theta$$we obtain $$x=2+2costheta\y=2+2sin theta+2costheta$$from which the maximum and minimum could be easily found.
answered Jan 13 at 16:21
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
Hint
The equation of whole the village would be$$e^{-4}=e^{-(x-2)^2-(x-y)^2}$$therefore $$(x-2)^2+(x-y)^2=4$$By defining $$x-2=2costheta\y-x=2sin theta$$we obtain $$x=2+2costheta\y=2+2sin theta+2costheta$$from which the maximum and minimum could be easily found.
answered Jan 13 at 16:21
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 13 at 16:21
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 13 at 16:21
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 13 at 16:21
Mostafa AyazMostafa Ayaz
15.6k3939
15.6k3939
$begingroup$
Thank you very much! So to find the highest y-value I just have to derive the y = 2 + 2sin + 2cos, find stationary points and do second derivate test?
$endgroup$
– Tourna
Jan 13 at 17:36
$begingroup$
You're welcome. That's right. After finding $theta$ substitute it in $x$ to find the full coordinates....
$endgroup$
– Mostafa Ayaz
Jan 13 at 17:41
add a comment |
$begingroup$
Thank you very much! So to find the highest y-value I just have to derive the y = 2 + 2sin + 2cos, find stationary points and do second derivate test?
$endgroup$
– Tourna
Jan 13 at 17:36
$begingroup$
You're welcome. That's right. After finding $theta$ substitute it in $x$ to find the full coordinates....
$endgroup$
– Mostafa Ayaz
Jan 13 at 17:41
$begingroup$
Thank you very much! So to find the highest y-value I just have to derive the y = 2 + 2sin + 2cos, find stationary points and do second derivate test?
$endgroup$
– Tourna
Jan 13 at 17:36
$begingroup$
Thank you very much! So to find the highest y-value I just have to derive the y = 2 + 2sin + 2cos, find stationary points and do second derivate test?
$endgroup$
– Tourna
Jan 13 at 17:36
$begingroup$
You're welcome. That's right. After finding $theta$ substitute it in $x$ to find the full coordinates....
$endgroup$
– Mostafa Ayaz
Jan 13 at 17:41
$begingroup$
You're welcome. That's right. After finding $theta$ substitute it in $x$ to find the full coordinates....
$endgroup$
– Mostafa Ayaz
Jan 13 at 17:41
add a comment |
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$begingroup$
There is a global maximum at $$(2,2)$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 13 at 15:39
$begingroup$
Hint: the height depends only on $(x-2)^2 + (y-x)^2$. At $(0,0)$ this is $4$.
$endgroup$
– Robert Israel
Jan 13 at 15:39
$begingroup$
@RobertIsrael Aha so I don't have to look at the rest anymore? And to answer b I would just have to calculate the maximum and minimum of that function?
$endgroup$
– Tourna
Jan 13 at 15:44
$begingroup$
The maximum and minimum of $y$ subject to $(x-2)^2 + (y-x)^2 = 4$.
$endgroup$
– Robert Israel
Jan 13 at 20:07