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Is the local max point also a global max?

Unique critical point does not imply global maximum/global minimumFinding global max./min.Finding global minumum maximumGlobal Maximums and MinimumsGlobal max/min of surfacesFinding global maximizers and minimizersIf a function of two variables has a unique critical point, which is a local maximum, is it a global maximum?Finding if a crit point is a min/max/saddle after assembling the matrix.Debate on definition of Critical point and local / absolute extremasUsing the Hessian Matrix to classify points

0

1

$begingroup$

Consider the following function $f$ of three variables, defined on $mathbbR^3$: $$f(x,y,z) = 15x + xy - 4x^2 - 2y^2 - z^2 + 2yz + 7$$

Is each of the critical points a local maximum? Is each of them a global maximum? Explain.

I've found the critical points after taking the first-order derivatives $$f_x = 15 + y - 8x = 0,$$ $$f_y = x - 4y + 2z = 0,$$ $$fz=-2z + 2y = 0.$$

So there's only one critical point $left ( 2,1,1 right )$.

To figure out whether the critical point is a local maximum I've computed the partial derivatives and have obtained:

$$H = beginbmatrix
-f_xx &f_xy &f_xz \
f_yx&f_yy &f_yz \
f_zx &f_zy &zz
endbmatrix$$

$$ Rightarrow H = beginbmatrix
-8 &1 &0 \
1&-4 &2 \
0 &2 &-2
endbmatrix.$$

After using the Method of Leading principal minors, I've figured out that $H$ is negative definite so $left ( 2,1,1 right )$ is a local maximum.

However, I'm a bit confused about whether this local maximum point is also a global maximum point. I know that for a global maximum point, the first-order condition must be satisfied and $H$ (the matrix of partial derivatives) must be negative definite for all $x, y, z in mathbbR^3$.

So based on this, can I say that this local max point is also a global max because $H$ does not depend on the arguments?

calculus multivariable-calculus optimization

share|cite|improve this question

edited Oct 14 '18 at 0:01

Bernard

123k741116

asked Oct 13 '18 at 23:45

OGCOGC

1,43821229

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $f_z'$ should be $-2z+2y$, not $2z+2y$.
  $endgroup$
  – manooooh
  Oct 13 '18 at 23:53
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @manooooh Sorry, it's a typo. Have edited.
  $endgroup$
  – OGC
  Oct 13 '18 at 23:56
  

  
  
  
  

add a comment | 

0

1

$begingroup$

Consider the following function $f$ of three variables, defined on $mathbbR^3$: $$f(x,y,z) = 15x + xy - 4x^2 - 2y^2 - z^2 + 2yz + 7$$

Is each of the critical points a local maximum? Is each of them a global maximum? Explain.

I've found the critical points after taking the first-order derivatives $$f_x = 15 + y - 8x = 0,$$ $$f_y = x - 4y + 2z = 0,$$ $$fz=-2z + 2y = 0.$$

So there's only one critical point $left ( 2,1,1 right )$.

To figure out whether the critical point is a local maximum I've computed the partial derivatives and have obtained:

$$H = beginbmatrix
-f_xx &f_xy &f_xz \
f_yx&f_yy &f_yz \
f_zx &f_zy &zz
endbmatrix$$

$$ Rightarrow H = beginbmatrix
-8 &1 &0 \
1&-4 &2 \
0 &2 &-2
endbmatrix.$$

After using the Method of Leading principal minors, I've figured out that $H$ is negative definite so $left ( 2,1,1 right )$ is a local maximum.

However, I'm a bit confused about whether this local maximum point is also a global maximum point. I know that for a global maximum point, the first-order condition must be satisfied and $H$ (the matrix of partial derivatives) must be negative definite for all $x, y, z in mathbbR^3$.

So based on this, can I say that this local max point is also a global max because $H$ does not depend on the arguments?

calculus multivariable-calculus optimization

share|cite|improve this question

edited Oct 14 '18 at 0:01

Bernard

123k741116

asked Oct 13 '18 at 23:45

OGCOGC

1,43821229

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $f_z'$ should be $-2z+2y$, not $2z+2y$.
  $endgroup$
  – manooooh
  Oct 13 '18 at 23:53
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @manooooh Sorry, it's a typo. Have edited.
  $endgroup$
  – OGC
  Oct 13 '18 at 23:56
  

  
  
  
  

add a comment | 

$begingroup$

Consider the following function $f$ of three variables, defined on $mathbbR^3$: $$f(x,y,z) = 15x + xy - 4x^2 - 2y^2 - z^2 + 2yz + 7$$

Is each of the critical points a local maximum? Is each of them a global maximum? Explain.

I've found the critical points after taking the first-order derivatives $$f_x = 15 + y - 8x = 0,$$ $$f_y = x - 4y + 2z = 0,$$ $$fz=-2z + 2y = 0.$$

So there's only one critical point $left ( 2,1,1 right )$.

To figure out whether the critical point is a local maximum I've computed the partial derivatives and have obtained:

$$H = beginbmatrix
-f_xx &f_xy &f_xz \
f_yx&f_yy &f_yz \
f_zx &f_zy &zz
endbmatrix$$

$$ Rightarrow H = beginbmatrix
-8 &1 &0 \
1&-4 &2 \
0 &2 &-2
endbmatrix.$$

After using the Method of Leading principal minors, I've figured out that $H$ is negative definite so $left ( 2,1,1 right )$ is a local maximum.

However, I'm a bit confused about whether this local maximum point is also a global maximum point. I know that for a global maximum point, the first-order condition must be satisfied and $H$ (the matrix of partial derivatives) must be negative definite for all $x, y, z in mathbbR^3$.

So based on this, can I say that this local max point is also a global max because $H$ does not depend on the arguments?

calculus multivariable-calculus optimization

share|cite|improve this question

edited Oct 14 '18 at 0:01

Bernard

123k741116

asked Oct 13 '18 at 23:45

OGCOGC

1,43821229

$endgroup$

share|cite|improve this question

edited Oct 14 '18 at 0:01

Bernard

123k741116

asked Oct 13 '18 at 23:45

OGCOGC

1,43821229

share|cite|improve this question

edited Oct 14 '18 at 0:01

Bernard

123k741116

asked Oct 13 '18 at 23:45

OGCOGC

1,43821229

edited Oct 14 '18 at 0:01

Bernard

123k741116

edited Oct 14 '18 at 0:01

Bernard

123k741116

asked Oct 13 '18 at 23:45

OGCOGC

1,43821229

asked Oct 13 '18 at 23:45

OGCOGC

1,43821229

1 Answer
1

active

oldest

votes

-2

$begingroup$

When there is only one critical point and the critical point is a local maximum, that critical point is global maximum as well.

share|cite|improve this answer

answered Oct 13 '18 at 23:56

Mohammad Riazi-KermaniMohammad Riazi-Kermani

41.6k42061

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  This is false. Consider $f(x,y) = -(x^2+y^2(1+x)^3)$. It has a unique critical point at $(0,0)$ which is a local maximum, but restricted to the line $y=x$, it's a polynomial of odd degree which must be unbounded.
  $endgroup$
  – KReiser
  Oct 14 '18 at 0:04
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It is true that in this case, the given local maximum is a global maximum, but your proof as currently presented requires some adjustment.
  $endgroup$
  – KReiser
  Oct 14 '18 at 0:16
  

  
  
  
  

add a comment | 

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-2

$begingroup$

When there is only one critical point and the critical point is a local maximum, that critical point is global maximum as well.

share|cite|improve this answer

answered Oct 13 '18 at 23:56

Mohammad Riazi-KermaniMohammad Riazi-Kermani

41.6k42061

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  This is false. Consider $f(x,y) = -(x^2+y^2(1+x)^3)$. It has a unique critical point at $(0,0)$ which is a local maximum, but restricted to the line $y=x$, it's a polynomial of odd degree which must be unbounded.
  $endgroup$
  – KReiser
  Oct 14 '18 at 0:04
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It is true that in this case, the given local maximum is a global maximum, but your proof as currently presented requires some adjustment.
  $endgroup$
  – KReiser
  Oct 14 '18 at 0:16
  

  
  
  
  

add a comment | 

-2

$begingroup$

When there is only one critical point and the critical point is a local maximum, that critical point is global maximum as well.

share|cite|improve this answer

answered Oct 13 '18 at 23:56

Mohammad Riazi-KermaniMohammad Riazi-Kermani

41.6k42061

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  This is false. Consider $f(x,y) = -(x^2+y^2(1+x)^3)$. It has a unique critical point at $(0,0)$ which is a local maximum, but restricted to the line $y=x$, it's a polynomial of odd degree which must be unbounded.
  $endgroup$
  – KReiser
  Oct 14 '18 at 0:04
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It is true that in this case, the given local maximum is a global maximum, but your proof as currently presented requires some adjustment.
  $endgroup$
  – KReiser
  Oct 14 '18 at 0:16
  

  
  
  
  

add a comment | 

$begingroup$

When there is only one critical point and the critical point is a local maximum, that critical point is global maximum as well.

share|cite|improve this answer

answered Oct 13 '18 at 23:56

Mohammad Riazi-KermaniMohammad Riazi-Kermani

41.6k42061

$endgroup$

share|cite|improve this answer

answered Oct 13 '18 at 23:56

Mohammad Riazi-KermaniMohammad Riazi-Kermani

41.6k42061

answered Oct 13 '18 at 23:56

Mohammad Riazi-KermaniMohammad Riazi-Kermani

41.6k42061

answered Oct 13 '18 at 23:56

Mohammad Riazi-KermaniMohammad Riazi-Kermani

41.6k42061