Mixing
Second derivative Test becomes zero
$begingroup$
Consider the domain $D = { (𝑥,𝑦) ∈ ℝ^2:𝑥 ≤ 𝑦 }$ and the function $ℎ: 𝐷 → ℝ$ defined by $ℎ((𝑥,𝑦)) = (𝑥 −2)^4 +(𝑦−1)^4$, $(𝑥,𝑦) ∈ 𝐷$.
Find the minimum value of $h$ in the domain $D$:
a) $1/2$
b) $1/4$
c) $1/8$
d) $1/16$
My work:
$f'(x) = 4(x-2)^3$
$f'(y) = 4(y-1)^3$
$f''(x) = 12(x-2)^2$
f_{xy} = $4(y-1)^3$ (partial derivative of $x,y$)
Critical points are $x = 2$ and $y = 1$,
Performing second derivative test with $x = 2$ and $y = 1$ I get zero ,
How do I proceed and where have I gone wrong? kindly help .
The answer is Option C: 1/8
self-learning maxima-minima
edited Jan 14 at 7:47
egreg
182k1485203
asked Jan 14 at 7:37
Dravid HemanthDravid Hemanth
56
$endgroup$
add a comment |
$begingroup$
Consider the domain $D = { (𝑥,𝑦) ∈ ℝ^2:𝑥 ≤ 𝑦 }$ and the function $ℎ: 𝐷 → ℝ$ defined by $ℎ((𝑥,𝑦)) = (𝑥 −2)^4 +(𝑦−1)^4$, $(𝑥,𝑦) ∈ 𝐷$.
Find the minimum value of $h$ in the domain $D$:
a) $1/2$
b) $1/4$
c) $1/8$
d) $1/16$
My work:
$f'(x) = 4(x-2)^3$
$f'(y) = 4(y-1)^3$
$f''(x) = 12(x-2)^2$
f_{xy} = $4(y-1)^3$ (partial derivative of $x,y$)
Critical points are $x = 2$ and $y = 1$,
Performing second derivative test with $x = 2$ and $y = 1$ I get zero ,
How do I proceed and where have I gone wrong? kindly help .
The answer is Option C: 1/8
self-learning maxima-minima
edited Jan 14 at 7:47
egreg
182k1485203
asked Jan 14 at 7:37
Dravid HemanthDravid Hemanth
56
$endgroup$
$begingroup$
The domain does not include $(x,y)=(2,1)$.
$endgroup$
– Kavi Rama Murthy
Jan 14 at 7:39
$begingroup$
How do I proceed then ? Can you help ? @KaviRamaMurthy
$endgroup$
– Dravid Hemanth
Jan 14 at 7:40
add a comment |
$begingroup$
Consider the domain $D = { (𝑥,𝑦) ∈ ℝ^2:𝑥 ≤ 𝑦 }$ and the function $ℎ: 𝐷 → ℝ$ defined by $ℎ((𝑥,𝑦)) = (𝑥 −2)^4 +(𝑦−1)^4$, $(𝑥,𝑦) ∈ 𝐷$.
Find the minimum value of $h$ in the domain $D$:
a) $1/2$
b) $1/4$
c) $1/8$
d) $1/16$
My work:
$f'(x) = 4(x-2)^3$
$f'(y) = 4(y-1)^3$
$f''(x) = 12(x-2)^2$
f_{xy} = $4(y-1)^3$ (partial derivative of $x,y$)
Critical points are $x = 2$ and $y = 1$,
Performing second derivative test with $x = 2$ and $y = 1$ I get zero ,
How do I proceed and where have I gone wrong? kindly help .
The answer is Option C: 1/8
self-learning maxima-minima
edited Jan 14 at 7:47
egreg
182k1485203
asked Jan 14 at 7:37
Dravid HemanthDravid Hemanth
56
$endgroup$
Consider the domain $D = { (𝑥,𝑦) ∈ ℝ^2:𝑥 ≤ 𝑦 }$ and the function $ℎ: 𝐷 → ℝ$ defined by $ℎ((𝑥,𝑦)) = (𝑥 −2)^4 +(𝑦−1)^4$, $(𝑥,𝑦) ∈ 𝐷$.
Find the minimum value of $h$ in the domain $D$:
a) $1/2$
b) $1/4$
c) $1/8$
d) $1/16$
My work:
$f'(x) = 4(x-2)^3$
$f'(y) = 4(y-1)^3$
$f''(x) = 12(x-2)^2$
f_{xy} = $4(y-1)^3$ (partial derivative of $x,y$)
Critical points are $x = 2$ and $y = 1$,
Performing second derivative test with $x = 2$ and $y = 1$ I get zero ,
How do I proceed and where have I gone wrong? kindly help .
The answer is Option C: 1/8
self-learning maxima-minima
self-learning maxima-minima
edited Jan 14 at 7:47
egreg
182k1485203
asked Jan 14 at 7:37
Dravid HemanthDravid Hemanth
56
edited Jan 14 at 7:47
egreg
182k1485203
asked Jan 14 at 7:37
Dravid HemanthDravid Hemanth
56
edited Jan 14 at 7:47
egreg
182k1485203
edited Jan 14 at 7:47
egreg
182k1485203
edited Jan 14 at 7:47
egreg
182k1485203
182k1485203
asked Jan 14 at 7:37
Dravid HemanthDravid Hemanth
56
asked Jan 14 at 7:37
Dravid HemanthDravid Hemanth
56
asked Jan 14 at 7:37
Dravid HemanthDravid Hemanth
56
56
$begingroup$
The domain does not include $(x,y)=(2,1)$.
$endgroup$
– Kavi Rama Murthy
Jan 14 at 7:39
$begingroup$
How do I proceed then ? Can you help ? @KaviRamaMurthy
$endgroup$
– Dravid Hemanth
Jan 14 at 7:40
add a comment |
$begingroup$
The domain does not include $(x,y)=(2,1)$.
$endgroup$
– Kavi Rama Murthy
Jan 14 at 7:39
$begingroup$
How do I proceed then ? Can you help ? @KaviRamaMurthy
$endgroup$
– Dravid Hemanth
Jan 14 at 7:40
$begingroup$
The domain does not include $(x,y)=(2,1)$.
$endgroup$
– Kavi Rama Murthy
Jan 14 at 7:39
$begingroup$
The domain does not include $(x,y)=(2,1)$.
$endgroup$
– Kavi Rama Murthy
Jan 14 at 7:39
$begingroup$
How do I proceed then ? Can you help ? @KaviRamaMurthy
$endgroup$
– Dravid Hemanth
Jan 14 at 7:40
$begingroup$
How do I proceed then ? Can you help ? @KaviRamaMurthy
$endgroup$
– Dravid Hemanth
Jan 14 at 7:40
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Since $(2,1)$ is not in the domain of $D$ of $h$, we have $operatorname{grad}h(x,y) ne (0,0)$ for all $(x,y) in D$. Thus investigate the function $h$ on $ partial D={(x,y): x=y}$.
Hence consider $f(x):=(x-2)^4+(x-1)^4$. Can you proceed?
edited Jan 14 at 7:52
egreg
182k1485203
answered Jan 14 at 7:51
FredFred
46.8k1848
$endgroup$
$begingroup$
Thank you for the answer ! should I try finding critical points to this by differentiating and equating to zero ?
$endgroup$
– Dravid Hemanth
Jan 14 at 8:14
$begingroup$
Yes, find $x_0$ with $f'(x_0)=0$.
$endgroup$
– Fred
Jan 14 at 8:18
$begingroup$
This is what I have done , I differentiated it and got $4(x-2)^3 + 4(x-1)^3$ ,After expanding and equating to zero I get $2x^3 -9x^2 +15x -9 = 0$ ,I am stuck here and not able to solve it further .
$endgroup$
– Dravid Hemanth
Jan 14 at 8:57
$begingroup$
I tried to solve but wasn't able to get an answer thats in the option
$endgroup$
– Dravid Hemanth
Jan 14 at 8:58
$begingroup$
From $4(x-2)^3 + 4(x-1)^3=0$ we get $(x-2)^3=-(x-1)^3$, hence $|x-2|=|x-1|$, thus $x=3/2$.
$endgroup$
– Fred
Jan 14 at 10:24
add a comment |
Your Answer
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
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active
oldest
votes
active
oldest
votes
$begingroup$
Since $(2,1)$ is not in the domain of $D$ of $h$, we have $operatorname{grad}h(x,y) ne (0,0)$ for all $(x,y) in D$. Thus investigate the function $h$ on $ partial D={(x,y): x=y}$.
Hence consider $f(x):=(x-2)^4+(x-1)^4$. Can you proceed?
edited Jan 14 at 7:52
egreg
182k1485203
answered Jan 14 at 7:51
FredFred
46.8k1848
$endgroup$
$begingroup$
Thank you for the answer ! should I try finding critical points to this by differentiating and equating to zero ?
$endgroup$
– Dravid Hemanth
Jan 14 at 8:14
$begingroup$
Yes, find $x_0$ with $f'(x_0)=0$.
$endgroup$
– Fred
Jan 14 at 8:18
$begingroup$
This is what I have done , I differentiated it and got $4(x-2)^3 + 4(x-1)^3$ ,After expanding and equating to zero I get $2x^3 -9x^2 +15x -9 = 0$ ,I am stuck here and not able to solve it further .
$endgroup$
– Dravid Hemanth
Jan 14 at 8:57
$begingroup$
I tried to solve but wasn't able to get an answer thats in the option
$endgroup$
– Dravid Hemanth
Jan 14 at 8:58
$begingroup$
From $4(x-2)^3 + 4(x-1)^3=0$ we get $(x-2)^3=-(x-1)^3$, hence $|x-2|=|x-1|$, thus $x=3/2$.
$endgroup$
– Fred
Jan 14 at 10:24
add a comment |
$begingroup$
Since $(2,1)$ is not in the domain of $D$ of $h$, we have $operatorname{grad}h(x,y) ne (0,0)$ for all $(x,y) in D$. Thus investigate the function $h$ on $ partial D={(x,y): x=y}$.
Hence consider $f(x):=(x-2)^4+(x-1)^4$. Can you proceed?
edited Jan 14 at 7:52
egreg
182k1485203
answered Jan 14 at 7:51
FredFred
46.8k1848
$endgroup$
$begingroup$
Thank you for the answer ! should I try finding critical points to this by differentiating and equating to zero ?
$endgroup$
– Dravid Hemanth
Jan 14 at 8:14
$begingroup$
Yes, find $x_0$ with $f'(x_0)=0$.
$endgroup$
– Fred
Jan 14 at 8:18
$begingroup$
This is what I have done , I differentiated it and got $4(x-2)^3 + 4(x-1)^3$ ,After expanding and equating to zero I get $2x^3 -9x^2 +15x -9 = 0$ ,I am stuck here and not able to solve it further .
$endgroup$
– Dravid Hemanth
Jan 14 at 8:57
$begingroup$
I tried to solve but wasn't able to get an answer thats in the option
$endgroup$
– Dravid Hemanth
Jan 14 at 8:58
$begingroup$
From $4(x-2)^3 + 4(x-1)^3=0$ we get $(x-2)^3=-(x-1)^3$, hence $|x-2|=|x-1|$, thus $x=3/2$.
$endgroup$
– Fred
Jan 14 at 10:24
add a comment |
$begingroup$
Since $(2,1)$ is not in the domain of $D$ of $h$, we have $operatorname{grad}h(x,y) ne (0,0)$ for all $(x,y) in D$. Thus investigate the function $h$ on $ partial D={(x,y): x=y}$.
Hence consider $f(x):=(x-2)^4+(x-1)^4$. Can you proceed?
edited Jan 14 at 7:52
egreg
182k1485203
answered Jan 14 at 7:51
FredFred
46.8k1848
$endgroup$
Since $(2,1)$ is not in the domain of $D$ of $h$, we have $operatorname{grad}h(x,y) ne (0,0)$ for all $(x,y) in D$. Thus investigate the function $h$ on $ partial D={(x,y): x=y}$.
Hence consider $f(x):=(x-2)^4+(x-1)^4$. Can you proceed?
edited Jan 14 at 7:52
egreg
182k1485203
answered Jan 14 at 7:51
FredFred
46.8k1848
edited Jan 14 at 7:52
egreg
182k1485203
edited Jan 14 at 7:52
egreg
182k1485203
edited Jan 14 at 7:52
egreg
182k1485203
182k1485203
answered Jan 14 at 7:51
FredFred
46.8k1848
answered Jan 14 at 7:51
FredFred
46.8k1848
answered Jan 14 at 7:51
FredFred
46.8k1848
46.8k1848
$begingroup$
Thank you for the answer ! should I try finding critical points to this by differentiating and equating to zero ?
$endgroup$
– Dravid Hemanth
Jan 14 at 8:14
$begingroup$
Yes, find $x_0$ with $f'(x_0)=0$.
$endgroup$
– Fred
Jan 14 at 8:18
$begingroup$
This is what I have done , I differentiated it and got $4(x-2)^3 + 4(x-1)^3$ ,After expanding and equating to zero I get $2x^3 -9x^2 +15x -9 = 0$ ,I am stuck here and not able to solve it further .
$endgroup$
– Dravid Hemanth
Jan 14 at 8:57
$begingroup$
I tried to solve but wasn't able to get an answer thats in the option
$endgroup$
– Dravid Hemanth
Jan 14 at 8:58
$begingroup$
From $4(x-2)^3 + 4(x-1)^3=0$ we get $(x-2)^3=-(x-1)^3$, hence $|x-2|=|x-1|$, thus $x=3/2$.
$endgroup$
– Fred
Jan 14 at 10:24
add a comment |
$begingroup$
Thank you for the answer ! should I try finding critical points to this by differentiating and equating to zero ?
$endgroup$
– Dravid Hemanth
Jan 14 at 8:14
$begingroup$
Yes, find $x_0$ with $f'(x_0)=0$.
$endgroup$
– Fred
Jan 14 at 8:18
$begingroup$
This is what I have done , I differentiated it and got $4(x-2)^3 + 4(x-1)^3$ ,After expanding and equating to zero I get $2x^3 -9x^2 +15x -9 = 0$ ,I am stuck here and not able to solve it further .
$endgroup$
– Dravid Hemanth
Jan 14 at 8:57
$begingroup$
I tried to solve but wasn't able to get an answer thats in the option
$endgroup$
– Dravid Hemanth
Jan 14 at 8:58
$begingroup$
From $4(x-2)^3 + 4(x-1)^3=0$ we get $(x-2)^3=-(x-1)^3$, hence $|x-2|=|x-1|$, thus $x=3/2$.
$endgroup$
– Fred
Jan 14 at 10:24
$begingroup$
Thank you for the answer ! should I try finding critical points to this by differentiating and equating to zero ?
$endgroup$
– Dravid Hemanth
Jan 14 at 8:14
$begingroup$
Thank you for the answer ! should I try finding critical points to this by differentiating and equating to zero ?
$endgroup$
– Dravid Hemanth
Jan 14 at 8:14
$begingroup$
Yes, find $x_0$ with $f'(x_0)=0$.
$endgroup$
– Fred
Jan 14 at 8:18
$begingroup$
Yes, find $x_0$ with $f'(x_0)=0$.
$endgroup$
– Fred
Jan 14 at 8:18
$begingroup$
This is what I have done , I differentiated it and got $4(x-2)^3 + 4(x-1)^3$ ,After expanding and equating to zero I get $2x^3 -9x^2 +15x -9 = 0$ ,I am stuck here and not able to solve it further .
$endgroup$
– Dravid Hemanth
Jan 14 at 8:57
$begingroup$
This is what I have done , I differentiated it and got $4(x-2)^3 + 4(x-1)^3$ ,After expanding and equating to zero I get $2x^3 -9x^2 +15x -9 = 0$ ,I am stuck here and not able to solve it further .
$endgroup$
– Dravid Hemanth
Jan 14 at 8:57
$begingroup$
I tried to solve but wasn't able to get an answer thats in the option
$endgroup$
– Dravid Hemanth
Jan 14 at 8:58
$begingroup$
I tried to solve but wasn't able to get an answer thats in the option
$endgroup$
– Dravid Hemanth
Jan 14 at 8:58
$begingroup$
From $4(x-2)^3 + 4(x-1)^3=0$ we get $(x-2)^3=-(x-1)^3$, hence $|x-2|=|x-1|$, thus $x=3/2$.
$endgroup$
– Fred
Jan 14 at 10:24
$begingroup$
From $4(x-2)^3 + 4(x-1)^3=0$ we get $(x-2)^3=-(x-1)^3$, hence $|x-2|=|x-1|$, thus $x=3/2$.
$endgroup$
– Fred
Jan 14 at 10:24
add a comment |
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$begingroup$
The domain does not include $(x,y)=(2,1)$.
$endgroup$
– Kavi Rama Murthy
Jan 14 at 7:39
$begingroup$
How do I proceed then ? Can you help ? @KaviRamaMurthy
$endgroup$
– Dravid Hemanth
Jan 14 at 7:40