Determine critical points of 2 variable functions without 2nd derivative test
I'm new to two variable calculus and having trouble classifying critical points for some functions which the second derivative test isn't really applicable.
E.g.
$$f(x,y) = x^2+y^2+x^2y+4 $$
$f_x=2x+2xy=0$
$f_y=2y+x^2=0$
Critical Point $(0,0)$
But I don't know how to determine the nature of this critical point.
calculus
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I'm new to two variable calculus and having trouble classifying critical points for some functions which the second derivative test isn't really applicable.
E.g.
$$f(x,y) = x^2+y^2+x^2y+4 $$
$f_x=2x+2xy=0$
$f_y=2y+x^2=0$
Critical Point $(0,0)$
But I don't know how to determine the nature of this critical point.
calculus
add a comment |
I'm new to two variable calculus and having trouble classifying critical points for some functions which the second derivative test isn't really applicable.
E.g.
$$f(x,y) = x^2+y^2+x^2y+4 $$
$f_x=2x+2xy=0$
$f_y=2y+x^2=0$
Critical Point $(0,0)$
But I don't know how to determine the nature of this critical point.
calculus
I'm new to two variable calculus and having trouble classifying critical points for some functions which the second derivative test isn't really applicable.
E.g.
$$f(x,y) = x^2+y^2+x^2y+4 $$
$f_x=2x+2xy=0$
$f_y=2y+x^2=0$
Critical Point $(0,0)$
But I don't know how to determine the nature of this critical point.
calculus
calculus
asked Nov 22 at 6:56
p.chives
1
1
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2 Answers
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Express your objective function as:
$$f(x,y)=x^2(1+y)+y^2+4.$$
Note that in the close neighborhood of $(0,0)$, the objective function is greater than or equal to $4$, hence $f(0,0)=4$ is the local minimum.
add a comment |
Two more critical points are $(sqrt 2,-1)$ and $(-sqrt 2,-1)$. For finding the nature we use the hessian matrix as following $$H=begin{bmatrix}2+2y&2x\2x& 2yend{bmatrix}$$The eigenvalues of $H$ for $(pm sqrt 2,-1)$ are $$lambda (lambda+2)-8=0\to lambda_1=2,lambda_2=-4$$therefore the points $(pm sqrt 2,-1)$ are saddle points. For $(0,0)$ we have $Hsucceq 0$. Hence $(0,0)$ is a local minimum.
Thank you. I fixed it....
– Mostafa Ayaz
Nov 22 at 7:19
add a comment |
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2 Answers
2
active
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2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
Express your objective function as:
$$f(x,y)=x^2(1+y)+y^2+4.$$
Note that in the close neighborhood of $(0,0)$, the objective function is greater than or equal to $4$, hence $f(0,0)=4$ is the local minimum.
add a comment |
Express your objective function as:
$$f(x,y)=x^2(1+y)+y^2+4.$$
Note that in the close neighborhood of $(0,0)$, the objective function is greater than or equal to $4$, hence $f(0,0)=4$ is the local minimum.
add a comment |
Express your objective function as:
$$f(x,y)=x^2(1+y)+y^2+4.$$
Note that in the close neighborhood of $(0,0)$, the objective function is greater than or equal to $4$, hence $f(0,0)=4$ is the local minimum.
Express your objective function as:
$$f(x,y)=x^2(1+y)+y^2+4.$$
Note that in the close neighborhood of $(0,0)$, the objective function is greater than or equal to $4$, hence $f(0,0)=4$ is the local minimum.
answered Nov 22 at 7:24
farruhota
18.9k2736
18.9k2736
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Two more critical points are $(sqrt 2,-1)$ and $(-sqrt 2,-1)$. For finding the nature we use the hessian matrix as following $$H=begin{bmatrix}2+2y&2x\2x& 2yend{bmatrix}$$The eigenvalues of $H$ for $(pm sqrt 2,-1)$ are $$lambda (lambda+2)-8=0\to lambda_1=2,lambda_2=-4$$therefore the points $(pm sqrt 2,-1)$ are saddle points. For $(0,0)$ we have $Hsucceq 0$. Hence $(0,0)$ is a local minimum.
Thank you. I fixed it....
– Mostafa Ayaz
Nov 22 at 7:19
add a comment |
Two more critical points are $(sqrt 2,-1)$ and $(-sqrt 2,-1)$. For finding the nature we use the hessian matrix as following $$H=begin{bmatrix}2+2y&2x\2x& 2yend{bmatrix}$$The eigenvalues of $H$ for $(pm sqrt 2,-1)$ are $$lambda (lambda+2)-8=0\to lambda_1=2,lambda_2=-4$$therefore the points $(pm sqrt 2,-1)$ are saddle points. For $(0,0)$ we have $Hsucceq 0$. Hence $(0,0)$ is a local minimum.
Thank you. I fixed it....
– Mostafa Ayaz
Nov 22 at 7:19
add a comment |
Two more critical points are $(sqrt 2,-1)$ and $(-sqrt 2,-1)$. For finding the nature we use the hessian matrix as following $$H=begin{bmatrix}2+2y&2x\2x& 2yend{bmatrix}$$The eigenvalues of $H$ for $(pm sqrt 2,-1)$ are $$lambda (lambda+2)-8=0\to lambda_1=2,lambda_2=-4$$therefore the points $(pm sqrt 2,-1)$ are saddle points. For $(0,0)$ we have $Hsucceq 0$. Hence $(0,0)$ is a local minimum.
Two more critical points are $(sqrt 2,-1)$ and $(-sqrt 2,-1)$. For finding the nature we use the hessian matrix as following $$H=begin{bmatrix}2+2y&2x\2x& 2yend{bmatrix}$$The eigenvalues of $H$ for $(pm sqrt 2,-1)$ are $$lambda (lambda+2)-8=0\to lambda_1=2,lambda_2=-4$$therefore the points $(pm sqrt 2,-1)$ are saddle points. For $(0,0)$ we have $Hsucceq 0$. Hence $(0,0)$ is a local minimum.
edited Nov 22 at 7:18
answered Nov 22 at 7:16
Mostafa Ayaz
13.7k3836
13.7k3836
Thank you. I fixed it....
– Mostafa Ayaz
Nov 22 at 7:19
add a comment |
Thank you. I fixed it....
– Mostafa Ayaz
Nov 22 at 7:19
Thank you. I fixed it....
– Mostafa Ayaz
Nov 22 at 7:19
Thank you. I fixed it....
– Mostafa Ayaz
Nov 22 at 7:19
add a comment |
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