Find the maximum and minimum (there are four in total) that the function $f(x,y)=3xy$ achieves when $(x,y)$...
Find the maximum and minimum (there are four in total) that the function $f(x,y)=3xy$ achieves when $(x,y)$ travel the ellipse $x^2+y^2+xy=3$
I have thought about doing the following but I do not know if I am doing the right thing:
I'm using Lagrange multipliers and I start with $bigtriangledown f=lambdabigtriangledown g$, so from this I get $(3y,3x)=lambda(2x+y,2y+x)$ so $lambda=frac{3y}{2x+y}$ and $lambda=frac{3x}{2y+x}$ so $frac{3y}{2x+y}=frac{3x}{2y+x}$ so $x^2=y^2$ so $y=pm x$. So, I get to that $(0,0), (1,1), (-1,-1), (sqrt{3},-sqrt{3})$ and $(-sqrt{3},sqrt{3})$ are critical points, the problem is that I do not know how to determine if they are maximum or minimum, how can I do this? Thank you.
real-analysis multivariable-calculus optimization lagrange-multiplier maxima-minima
add a comment |
Find the maximum and minimum (there are four in total) that the function $f(x,y)=3xy$ achieves when $(x,y)$ travel the ellipse $x^2+y^2+xy=3$
I have thought about doing the following but I do not know if I am doing the right thing:
I'm using Lagrange multipliers and I start with $bigtriangledown f=lambdabigtriangledown g$, so from this I get $(3y,3x)=lambda(2x+y,2y+x)$ so $lambda=frac{3y}{2x+y}$ and $lambda=frac{3x}{2y+x}$ so $frac{3y}{2x+y}=frac{3x}{2y+x}$ so $x^2=y^2$ so $y=pm x$. So, I get to that $(0,0), (1,1), (-1,-1), (sqrt{3},-sqrt{3})$ and $(-sqrt{3},sqrt{3})$ are critical points, the problem is that I do not know how to determine if they are maximum or minimum, how can I do this? Thank you.
real-analysis multivariable-calculus optimization lagrange-multiplier maxima-minima
2
Just compute f at those 4 points. Btw, (0,0) is not on the ellipse.
– Martin R
Jan 4 at 4:10
@MartinR Is what I did fine?
– Nash
Jan 4 at 20:55
add a comment |
Find the maximum and minimum (there are four in total) that the function $f(x,y)=3xy$ achieves when $(x,y)$ travel the ellipse $x^2+y^2+xy=3$
I have thought about doing the following but I do not know if I am doing the right thing:
I'm using Lagrange multipliers and I start with $bigtriangledown f=lambdabigtriangledown g$, so from this I get $(3y,3x)=lambda(2x+y,2y+x)$ so $lambda=frac{3y}{2x+y}$ and $lambda=frac{3x}{2y+x}$ so $frac{3y}{2x+y}=frac{3x}{2y+x}$ so $x^2=y^2$ so $y=pm x$. So, I get to that $(0,0), (1,1), (-1,-1), (sqrt{3},-sqrt{3})$ and $(-sqrt{3},sqrt{3})$ are critical points, the problem is that I do not know how to determine if they are maximum or minimum, how can I do this? Thank you.
real-analysis multivariable-calculus optimization lagrange-multiplier maxima-minima
Find the maximum and minimum (there are four in total) that the function $f(x,y)=3xy$ achieves when $(x,y)$ travel the ellipse $x^2+y^2+xy=3$
I have thought about doing the following but I do not know if I am doing the right thing:
I'm using Lagrange multipliers and I start with $bigtriangledown f=lambdabigtriangledown g$, so from this I get $(3y,3x)=lambda(2x+y,2y+x)$ so $lambda=frac{3y}{2x+y}$ and $lambda=frac{3x}{2y+x}$ so $frac{3y}{2x+y}=frac{3x}{2y+x}$ so $x^2=y^2$ so $y=pm x$. So, I get to that $(0,0), (1,1), (-1,-1), (sqrt{3},-sqrt{3})$ and $(-sqrt{3},sqrt{3})$ are critical points, the problem is that I do not know how to determine if they are maximum or minimum, how can I do this? Thank you.
real-analysis multivariable-calculus optimization lagrange-multiplier maxima-minima
real-analysis multivariable-calculus optimization lagrange-multiplier maxima-minima
edited Jan 4 at 7:20
Michael Rozenberg
97.3k1589188
97.3k1589188
asked Jan 4 at 3:47
NashNash
48049
48049
2
Just compute f at those 4 points. Btw, (0,0) is not on the ellipse.
– Martin R
Jan 4 at 4:10
@MartinR Is what I did fine?
– Nash
Jan 4 at 20:55
add a comment |
2
Just compute f at those 4 points. Btw, (0,0) is not on the ellipse.
– Martin R
Jan 4 at 4:10
@MartinR Is what I did fine?
– Nash
Jan 4 at 20:55
2
2
Just compute f at those 4 points. Btw, (0,0) is not on the ellipse.
– Martin R
Jan 4 at 4:10
Just compute f at those 4 points. Btw, (0,0) is not on the ellipse.
– Martin R
Jan 4 at 4:10
@MartinR Is what I did fine?
– Nash
Jan 4 at 20:55
@MartinR Is what I did fine?
– Nash
Jan 4 at 20:55
add a comment |
7 Answers
7
active
oldest
votes
If $x,y$ have the same sign $f(x,y)>0$
and if they have opposite signs $f(x,y)<0$
With the 4 pairs you have, 2 share the same sign and and 2 do not. That should make it clear wich are the maxima and which are the minima.
But is calculus even necessary?
Make this substitution
$x = u+v\
y = u-v$
Then your objective and constraint become:
$f(u,v) = 3u^2 - 3v^2\
3u^2 + v^2 = 3$
Plugging the constraint into the objective.
$f(u,v) = 3 - 4v^2$
$f$ is maximized when $v = 0$
$u = pm 1\
x = pm 1\
y = pm 1$
$f$ is minimized when $u = 0$
$v = pm sqrt 3\
x = pm sqrt 3\
y = mp sqrt 3$
Is what I did fine?
– Nash
Jan 4 at 20:55
1
Your work on with the Lagrange multipliers was fine.
– Doug M
Jan 4 at 21:10
add a comment |
Let $3xy=k$.
Thus,
$$k(x^2+y^2+xy)=9xy$$ or
$$kx^2+(k-9)xy+ky^2=0$$ has solutions.
Let $kneq0$.
Thus, $$(k-9)^2-4k^2geq0$$ or
$$(k+9)(k-3)leq0$$ or
$$-9leq kleq3.$$
Now, we see that $k=0$ is not relevant and the extreme value of $k$ occurs for $x=-frac{k-9}{2k}y$,
which says that
$$max_{x^2+y^2+xy=3}3xy=3$$ and
$$min_{x^2+y^2+xy=3}3xy=-9$$
Is what I did fine?
– Nash
Jan 4 at 20:56
1
@Nash Without the point $(0,0)$ it looks fine.
– Michael Rozenberg
Jan 4 at 20:58
add a comment |
If Lagrange multipliers is not mandatory,
$$12=4x^2+4y^2+4xy=(2x+y)^2+(sqrt3y)^2$$
Let $sqrt3y=sqrt{12}cos tiff y=2cos t$
$2x+y=2sqrt3sin tiff 2x=2sqrt3sin t-2cos tiff x=sqrt3sin t-cos t=2sinleft(t-30^circright)$
$3xy=12cos tsinleft(t-30^circright)=6[sin(2t-30^circ)-sin30^circ]$
Now for real $x,y;t$ is real $$-1lesin(2t-30^circ)le1$$
Is what I did fine?
– Nash
Jan 4 at 20:56
add a comment |
Evaluate your function $f(x,y) = 3xy$ at each of these four points, $(0,0), (1,1), (-1,-1), (sqrt{3},-sqrt{3})$ and compare the results.You will find maximum is achieved at $(1,1)$ and $(-1,-1)$ and minimum is achieved at $ pm (sqrt 3, -sqrt 3)$
Is what I did fine?
– Nash
Jan 4 at 20:56
Yes your work is correct
– Mohammad Riazi-Kermani
2 days ago
add a comment |
As the equations involved are homogeneous, making $y = lambda x$ we have
$$
min(max)3x^2lambda mbox{s. t. } x^2(1+lambda^2+lambda) = 3
$$
which is equivalent to
$$
min_{lambda}(max_{lambda})f(lambda) = frac{9lambda}{lambda^2+lambda+1}
$$
so the stationary points are at
$$
f'(lambda) = -frac{9 left(lambda ^2-1right)}{left(lambda ^2+lambda +1right)^2} = 0Rightarrow lambda^* = {-1,1}
$$
then the solutions are at $y = pm x$ etc.
Is what I did fine?
– Nash
Jan 4 at 20:56
add a comment |
1) $(x+y)^2- xy= 3;$
$xy = (x+y)^2 -3;$
$f(x,y)=3xy = 3(x+y)^2-9.$
$f_{min}=-9$, at $x=-y$;
$-3x^2= -9; x=pm √3; y=mp √3$;
2) $(x-y)^2+3xy= 3;$
$f(x,y)= 3xy= 3-(x-y)^2$;
$f_{max}= 3$, at $x=y$.
$x^2=1$; $x =pm 1$, $y=pm 1$.
add a comment |
You can determine the "type" of the critical point using the partial derivatives. Suppose $(x,y)$ is a critical point that you are considering. Then,
- If $f_{xx} f_{yy}- (f_{xx})^2 > 0$ and $f_{xx} > 0$ when evaluated at $(x,y)$, then there is a relative minimum at $(x,y)$.
- If $f_{xx} f_{yy}- (f_{xx})^2 > 0$ and $f_{xx} < 0$, you have a relative minimum.
- If $f_{xx} f_{yy}- (f_{xx})^2 < 0$ then you've got a saddle point.
- If $f_{xx} f_{yy}- (f_{xx})^2 = 0$ its indeterminate.
Edit: Just looked at the function you've provided. Clearly, it's going to indeterminate since the second derivative for $x$ or $y$ is $0$. You'll have to just plug in the values of the critical points you've found and determine it based on that.
New contributor
1
This only provides local maxima or minima of the restricted function. For global one uses compactness and this criterion is then useless.
– Will M.
Jan 4 at 4:36
1
Examining the Jacobian determinant of the objective function in a constrained optimization problem doesn’t usually tell you anything useful. This problem is a case in point: the Jacobian determinant is identically equal to $-9$, so per these criteria the only critical points are saddle points. This is true of $f$ on $mathbb R^2$, but certainly not of its restriction to the ellipse.
– amd
Jan 4 at 4:47
add a comment |
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7 Answers
7
active
oldest
votes
7 Answers
7
active
oldest
votes
active
oldest
votes
active
oldest
votes
If $x,y$ have the same sign $f(x,y)>0$
and if they have opposite signs $f(x,y)<0$
With the 4 pairs you have, 2 share the same sign and and 2 do not. That should make it clear wich are the maxima and which are the minima.
But is calculus even necessary?
Make this substitution
$x = u+v\
y = u-v$
Then your objective and constraint become:
$f(u,v) = 3u^2 - 3v^2\
3u^2 + v^2 = 3$
Plugging the constraint into the objective.
$f(u,v) = 3 - 4v^2$
$f$ is maximized when $v = 0$
$u = pm 1\
x = pm 1\
y = pm 1$
$f$ is minimized when $u = 0$
$v = pm sqrt 3\
x = pm sqrt 3\
y = mp sqrt 3$
Is what I did fine?
– Nash
Jan 4 at 20:55
1
Your work on with the Lagrange multipliers was fine.
– Doug M
Jan 4 at 21:10
add a comment |
If $x,y$ have the same sign $f(x,y)>0$
and if they have opposite signs $f(x,y)<0$
With the 4 pairs you have, 2 share the same sign and and 2 do not. That should make it clear wich are the maxima and which are the minima.
But is calculus even necessary?
Make this substitution
$x = u+v\
y = u-v$
Then your objective and constraint become:
$f(u,v) = 3u^2 - 3v^2\
3u^2 + v^2 = 3$
Plugging the constraint into the objective.
$f(u,v) = 3 - 4v^2$
$f$ is maximized when $v = 0$
$u = pm 1\
x = pm 1\
y = pm 1$
$f$ is minimized when $u = 0$
$v = pm sqrt 3\
x = pm sqrt 3\
y = mp sqrt 3$
Is what I did fine?
– Nash
Jan 4 at 20:55
1
Your work on with the Lagrange multipliers was fine.
– Doug M
Jan 4 at 21:10
add a comment |
If $x,y$ have the same sign $f(x,y)>0$
and if they have opposite signs $f(x,y)<0$
With the 4 pairs you have, 2 share the same sign and and 2 do not. That should make it clear wich are the maxima and which are the minima.
But is calculus even necessary?
Make this substitution
$x = u+v\
y = u-v$
Then your objective and constraint become:
$f(u,v) = 3u^2 - 3v^2\
3u^2 + v^2 = 3$
Plugging the constraint into the objective.
$f(u,v) = 3 - 4v^2$
$f$ is maximized when $v = 0$
$u = pm 1\
x = pm 1\
y = pm 1$
$f$ is minimized when $u = 0$
$v = pm sqrt 3\
x = pm sqrt 3\
y = mp sqrt 3$
If $x,y$ have the same sign $f(x,y)>0$
and if they have opposite signs $f(x,y)<0$
With the 4 pairs you have, 2 share the same sign and and 2 do not. That should make it clear wich are the maxima and which are the minima.
But is calculus even necessary?
Make this substitution
$x = u+v\
y = u-v$
Then your objective and constraint become:
$f(u,v) = 3u^2 - 3v^2\
3u^2 + v^2 = 3$
Plugging the constraint into the objective.
$f(u,v) = 3 - 4v^2$
$f$ is maximized when $v = 0$
$u = pm 1\
x = pm 1\
y = pm 1$
$f$ is minimized when $u = 0$
$v = pm sqrt 3\
x = pm sqrt 3\
y = mp sqrt 3$
edited Jan 4 at 7:24
answered Jan 4 at 4:25
Doug MDoug M
44.2k31854
44.2k31854
Is what I did fine?
– Nash
Jan 4 at 20:55
1
Your work on with the Lagrange multipliers was fine.
– Doug M
Jan 4 at 21:10
add a comment |
Is what I did fine?
– Nash
Jan 4 at 20:55
1
Your work on with the Lagrange multipliers was fine.
– Doug M
Jan 4 at 21:10
Is what I did fine?
– Nash
Jan 4 at 20:55
Is what I did fine?
– Nash
Jan 4 at 20:55
1
1
Your work on with the Lagrange multipliers was fine.
– Doug M
Jan 4 at 21:10
Your work on with the Lagrange multipliers was fine.
– Doug M
Jan 4 at 21:10
add a comment |
Let $3xy=k$.
Thus,
$$k(x^2+y^2+xy)=9xy$$ or
$$kx^2+(k-9)xy+ky^2=0$$ has solutions.
Let $kneq0$.
Thus, $$(k-9)^2-4k^2geq0$$ or
$$(k+9)(k-3)leq0$$ or
$$-9leq kleq3.$$
Now, we see that $k=0$ is not relevant and the extreme value of $k$ occurs for $x=-frac{k-9}{2k}y$,
which says that
$$max_{x^2+y^2+xy=3}3xy=3$$ and
$$min_{x^2+y^2+xy=3}3xy=-9$$
Is what I did fine?
– Nash
Jan 4 at 20:56
1
@Nash Without the point $(0,0)$ it looks fine.
– Michael Rozenberg
Jan 4 at 20:58
add a comment |
Let $3xy=k$.
Thus,
$$k(x^2+y^2+xy)=9xy$$ or
$$kx^2+(k-9)xy+ky^2=0$$ has solutions.
Let $kneq0$.
Thus, $$(k-9)^2-4k^2geq0$$ or
$$(k+9)(k-3)leq0$$ or
$$-9leq kleq3.$$
Now, we see that $k=0$ is not relevant and the extreme value of $k$ occurs for $x=-frac{k-9}{2k}y$,
which says that
$$max_{x^2+y^2+xy=3}3xy=3$$ and
$$min_{x^2+y^2+xy=3}3xy=-9$$
Is what I did fine?
– Nash
Jan 4 at 20:56
1
@Nash Without the point $(0,0)$ it looks fine.
– Michael Rozenberg
Jan 4 at 20:58
add a comment |
Let $3xy=k$.
Thus,
$$k(x^2+y^2+xy)=9xy$$ or
$$kx^2+(k-9)xy+ky^2=0$$ has solutions.
Let $kneq0$.
Thus, $$(k-9)^2-4k^2geq0$$ or
$$(k+9)(k-3)leq0$$ or
$$-9leq kleq3.$$
Now, we see that $k=0$ is not relevant and the extreme value of $k$ occurs for $x=-frac{k-9}{2k}y$,
which says that
$$max_{x^2+y^2+xy=3}3xy=3$$ and
$$min_{x^2+y^2+xy=3}3xy=-9$$
Let $3xy=k$.
Thus,
$$k(x^2+y^2+xy)=9xy$$ or
$$kx^2+(k-9)xy+ky^2=0$$ has solutions.
Let $kneq0$.
Thus, $$(k-9)^2-4k^2geq0$$ or
$$(k+9)(k-3)leq0$$ or
$$-9leq kleq3.$$
Now, we see that $k=0$ is not relevant and the extreme value of $k$ occurs for $x=-frac{k-9}{2k}y$,
which says that
$$max_{x^2+y^2+xy=3}3xy=3$$ and
$$min_{x^2+y^2+xy=3}3xy=-9$$
answered Jan 4 at 7:16
Michael RozenbergMichael Rozenberg
97.3k1589188
97.3k1589188
Is what I did fine?
– Nash
Jan 4 at 20:56
1
@Nash Without the point $(0,0)$ it looks fine.
– Michael Rozenberg
Jan 4 at 20:58
add a comment |
Is what I did fine?
– Nash
Jan 4 at 20:56
1
@Nash Without the point $(0,0)$ it looks fine.
– Michael Rozenberg
Jan 4 at 20:58
Is what I did fine?
– Nash
Jan 4 at 20:56
Is what I did fine?
– Nash
Jan 4 at 20:56
1
1
@Nash Without the point $(0,0)$ it looks fine.
– Michael Rozenberg
Jan 4 at 20:58
@Nash Without the point $(0,0)$ it looks fine.
– Michael Rozenberg
Jan 4 at 20:58
add a comment |
If Lagrange multipliers is not mandatory,
$$12=4x^2+4y^2+4xy=(2x+y)^2+(sqrt3y)^2$$
Let $sqrt3y=sqrt{12}cos tiff y=2cos t$
$2x+y=2sqrt3sin tiff 2x=2sqrt3sin t-2cos tiff x=sqrt3sin t-cos t=2sinleft(t-30^circright)$
$3xy=12cos tsinleft(t-30^circright)=6[sin(2t-30^circ)-sin30^circ]$
Now for real $x,y;t$ is real $$-1lesin(2t-30^circ)le1$$
Is what I did fine?
– Nash
Jan 4 at 20:56
add a comment |
If Lagrange multipliers is not mandatory,
$$12=4x^2+4y^2+4xy=(2x+y)^2+(sqrt3y)^2$$
Let $sqrt3y=sqrt{12}cos tiff y=2cos t$
$2x+y=2sqrt3sin tiff 2x=2sqrt3sin t-2cos tiff x=sqrt3sin t-cos t=2sinleft(t-30^circright)$
$3xy=12cos tsinleft(t-30^circright)=6[sin(2t-30^circ)-sin30^circ]$
Now for real $x,y;t$ is real $$-1lesin(2t-30^circ)le1$$
Is what I did fine?
– Nash
Jan 4 at 20:56
add a comment |
If Lagrange multipliers is not mandatory,
$$12=4x^2+4y^2+4xy=(2x+y)^2+(sqrt3y)^2$$
Let $sqrt3y=sqrt{12}cos tiff y=2cos t$
$2x+y=2sqrt3sin tiff 2x=2sqrt3sin t-2cos tiff x=sqrt3sin t-cos t=2sinleft(t-30^circright)$
$3xy=12cos tsinleft(t-30^circright)=6[sin(2t-30^circ)-sin30^circ]$
Now for real $x,y;t$ is real $$-1lesin(2t-30^circ)le1$$
If Lagrange multipliers is not mandatory,
$$12=4x^2+4y^2+4xy=(2x+y)^2+(sqrt3y)^2$$
Let $sqrt3y=sqrt{12}cos tiff y=2cos t$
$2x+y=2sqrt3sin tiff 2x=2sqrt3sin t-2cos tiff x=sqrt3sin t-cos t=2sinleft(t-30^circright)$
$3xy=12cos tsinleft(t-30^circright)=6[sin(2t-30^circ)-sin30^circ]$
Now for real $x,y;t$ is real $$-1lesin(2t-30^circ)le1$$
answered Jan 4 at 11:47
lab bhattacharjeelab bhattacharjee
223k15156274
223k15156274
Is what I did fine?
– Nash
Jan 4 at 20:56
add a comment |
Is what I did fine?
– Nash
Jan 4 at 20:56
Is what I did fine?
– Nash
Jan 4 at 20:56
Is what I did fine?
– Nash
Jan 4 at 20:56
add a comment |
Evaluate your function $f(x,y) = 3xy$ at each of these four points, $(0,0), (1,1), (-1,-1), (sqrt{3},-sqrt{3})$ and compare the results.You will find maximum is achieved at $(1,1)$ and $(-1,-1)$ and minimum is achieved at $ pm (sqrt 3, -sqrt 3)$
Is what I did fine?
– Nash
Jan 4 at 20:56
Yes your work is correct
– Mohammad Riazi-Kermani
2 days ago
add a comment |
Evaluate your function $f(x,y) = 3xy$ at each of these four points, $(0,0), (1,1), (-1,-1), (sqrt{3},-sqrt{3})$ and compare the results.You will find maximum is achieved at $(1,1)$ and $(-1,-1)$ and minimum is achieved at $ pm (sqrt 3, -sqrt 3)$
Is what I did fine?
– Nash
Jan 4 at 20:56
Yes your work is correct
– Mohammad Riazi-Kermani
2 days ago
add a comment |
Evaluate your function $f(x,y) = 3xy$ at each of these four points, $(0,0), (1,1), (-1,-1), (sqrt{3},-sqrt{3})$ and compare the results.You will find maximum is achieved at $(1,1)$ and $(-1,-1)$ and minimum is achieved at $ pm (sqrt 3, -sqrt 3)$
Evaluate your function $f(x,y) = 3xy$ at each of these four points, $(0,0), (1,1), (-1,-1), (sqrt{3},-sqrt{3})$ and compare the results.You will find maximum is achieved at $(1,1)$ and $(-1,-1)$ and minimum is achieved at $ pm (sqrt 3, -sqrt 3)$
answered Jan 4 at 4:17
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41k42059
41k42059
Is what I did fine?
– Nash
Jan 4 at 20:56
Yes your work is correct
– Mohammad Riazi-Kermani
2 days ago
add a comment |
Is what I did fine?
– Nash
Jan 4 at 20:56
Yes your work is correct
– Mohammad Riazi-Kermani
2 days ago
Is what I did fine?
– Nash
Jan 4 at 20:56
Is what I did fine?
– Nash
Jan 4 at 20:56
Yes your work is correct
– Mohammad Riazi-Kermani
2 days ago
Yes your work is correct
– Mohammad Riazi-Kermani
2 days ago
add a comment |
As the equations involved are homogeneous, making $y = lambda x$ we have
$$
min(max)3x^2lambda mbox{s. t. } x^2(1+lambda^2+lambda) = 3
$$
which is equivalent to
$$
min_{lambda}(max_{lambda})f(lambda) = frac{9lambda}{lambda^2+lambda+1}
$$
so the stationary points are at
$$
f'(lambda) = -frac{9 left(lambda ^2-1right)}{left(lambda ^2+lambda +1right)^2} = 0Rightarrow lambda^* = {-1,1}
$$
then the solutions are at $y = pm x$ etc.
Is what I did fine?
– Nash
Jan 4 at 20:56
add a comment |
As the equations involved are homogeneous, making $y = lambda x$ we have
$$
min(max)3x^2lambda mbox{s. t. } x^2(1+lambda^2+lambda) = 3
$$
which is equivalent to
$$
min_{lambda}(max_{lambda})f(lambda) = frac{9lambda}{lambda^2+lambda+1}
$$
so the stationary points are at
$$
f'(lambda) = -frac{9 left(lambda ^2-1right)}{left(lambda ^2+lambda +1right)^2} = 0Rightarrow lambda^* = {-1,1}
$$
then the solutions are at $y = pm x$ etc.
Is what I did fine?
– Nash
Jan 4 at 20:56
add a comment |
As the equations involved are homogeneous, making $y = lambda x$ we have
$$
min(max)3x^2lambda mbox{s. t. } x^2(1+lambda^2+lambda) = 3
$$
which is equivalent to
$$
min_{lambda}(max_{lambda})f(lambda) = frac{9lambda}{lambda^2+lambda+1}
$$
so the stationary points are at
$$
f'(lambda) = -frac{9 left(lambda ^2-1right)}{left(lambda ^2+lambda +1right)^2} = 0Rightarrow lambda^* = {-1,1}
$$
then the solutions are at $y = pm x$ etc.
As the equations involved are homogeneous, making $y = lambda x$ we have
$$
min(max)3x^2lambda mbox{s. t. } x^2(1+lambda^2+lambda) = 3
$$
which is equivalent to
$$
min_{lambda}(max_{lambda})f(lambda) = frac{9lambda}{lambda^2+lambda+1}
$$
so the stationary points are at
$$
f'(lambda) = -frac{9 left(lambda ^2-1right)}{left(lambda ^2+lambda +1right)^2} = 0Rightarrow lambda^* = {-1,1}
$$
then the solutions are at $y = pm x$ etc.
answered Jan 4 at 12:16
CesareoCesareo
8,3413516
8,3413516
Is what I did fine?
– Nash
Jan 4 at 20:56
add a comment |
Is what I did fine?
– Nash
Jan 4 at 20:56
Is what I did fine?
– Nash
Jan 4 at 20:56
Is what I did fine?
– Nash
Jan 4 at 20:56
add a comment |
1) $(x+y)^2- xy= 3;$
$xy = (x+y)^2 -3;$
$f(x,y)=3xy = 3(x+y)^2-9.$
$f_{min}=-9$, at $x=-y$;
$-3x^2= -9; x=pm √3; y=mp √3$;
2) $(x-y)^2+3xy= 3;$
$f(x,y)= 3xy= 3-(x-y)^2$;
$f_{max}= 3$, at $x=y$.
$x^2=1$; $x =pm 1$, $y=pm 1$.
add a comment |
1) $(x+y)^2- xy= 3;$
$xy = (x+y)^2 -3;$
$f(x,y)=3xy = 3(x+y)^2-9.$
$f_{min}=-9$, at $x=-y$;
$-3x^2= -9; x=pm √3; y=mp √3$;
2) $(x-y)^2+3xy= 3;$
$f(x,y)= 3xy= 3-(x-y)^2$;
$f_{max}= 3$, at $x=y$.
$x^2=1$; $x =pm 1$, $y=pm 1$.
add a comment |
1) $(x+y)^2- xy= 3;$
$xy = (x+y)^2 -3;$
$f(x,y)=3xy = 3(x+y)^2-9.$
$f_{min}=-9$, at $x=-y$;
$-3x^2= -9; x=pm √3; y=mp √3$;
2) $(x-y)^2+3xy= 3;$
$f(x,y)= 3xy= 3-(x-y)^2$;
$f_{max}= 3$, at $x=y$.
$x^2=1$; $x =pm 1$, $y=pm 1$.
1) $(x+y)^2- xy= 3;$
$xy = (x+y)^2 -3;$
$f(x,y)=3xy = 3(x+y)^2-9.$
$f_{min}=-9$, at $x=-y$;
$-3x^2= -9; x=pm √3; y=mp √3$;
2) $(x-y)^2+3xy= 3;$
$f(x,y)= 3xy= 3-(x-y)^2$;
$f_{max}= 3$, at $x=y$.
$x^2=1$; $x =pm 1$, $y=pm 1$.
edited Jan 4 at 21:12
answered Jan 4 at 7:22
Peter SzilasPeter Szilas
10.8k2720
10.8k2720
add a comment |
add a comment |
You can determine the "type" of the critical point using the partial derivatives. Suppose $(x,y)$ is a critical point that you are considering. Then,
- If $f_{xx} f_{yy}- (f_{xx})^2 > 0$ and $f_{xx} > 0$ when evaluated at $(x,y)$, then there is a relative minimum at $(x,y)$.
- If $f_{xx} f_{yy}- (f_{xx})^2 > 0$ and $f_{xx} < 0$, you have a relative minimum.
- If $f_{xx} f_{yy}- (f_{xx})^2 < 0$ then you've got a saddle point.
- If $f_{xx} f_{yy}- (f_{xx})^2 = 0$ its indeterminate.
Edit: Just looked at the function you've provided. Clearly, it's going to indeterminate since the second derivative for $x$ or $y$ is $0$. You'll have to just plug in the values of the critical points you've found and determine it based on that.
New contributor
1
This only provides local maxima or minima of the restricted function. For global one uses compactness and this criterion is then useless.
– Will M.
Jan 4 at 4:36
1
Examining the Jacobian determinant of the objective function in a constrained optimization problem doesn’t usually tell you anything useful. This problem is a case in point: the Jacobian determinant is identically equal to $-9$, so per these criteria the only critical points are saddle points. This is true of $f$ on $mathbb R^2$, but certainly not of its restriction to the ellipse.
– amd
Jan 4 at 4:47
add a comment |
You can determine the "type" of the critical point using the partial derivatives. Suppose $(x,y)$ is a critical point that you are considering. Then,
- If $f_{xx} f_{yy}- (f_{xx})^2 > 0$ and $f_{xx} > 0$ when evaluated at $(x,y)$, then there is a relative minimum at $(x,y)$.
- If $f_{xx} f_{yy}- (f_{xx})^2 > 0$ and $f_{xx} < 0$, you have a relative minimum.
- If $f_{xx} f_{yy}- (f_{xx})^2 < 0$ then you've got a saddle point.
- If $f_{xx} f_{yy}- (f_{xx})^2 = 0$ its indeterminate.
Edit: Just looked at the function you've provided. Clearly, it's going to indeterminate since the second derivative for $x$ or $y$ is $0$. You'll have to just plug in the values of the critical points you've found and determine it based on that.
New contributor
1
This only provides local maxima or minima of the restricted function. For global one uses compactness and this criterion is then useless.
– Will M.
Jan 4 at 4:36
1
Examining the Jacobian determinant of the objective function in a constrained optimization problem doesn’t usually tell you anything useful. This problem is a case in point: the Jacobian determinant is identically equal to $-9$, so per these criteria the only critical points are saddle points. This is true of $f$ on $mathbb R^2$, but certainly not of its restriction to the ellipse.
– amd
Jan 4 at 4:47
add a comment |
You can determine the "type" of the critical point using the partial derivatives. Suppose $(x,y)$ is a critical point that you are considering. Then,
- If $f_{xx} f_{yy}- (f_{xx})^2 > 0$ and $f_{xx} > 0$ when evaluated at $(x,y)$, then there is a relative minimum at $(x,y)$.
- If $f_{xx} f_{yy}- (f_{xx})^2 > 0$ and $f_{xx} < 0$, you have a relative minimum.
- If $f_{xx} f_{yy}- (f_{xx})^2 < 0$ then you've got a saddle point.
- If $f_{xx} f_{yy}- (f_{xx})^2 = 0$ its indeterminate.
Edit: Just looked at the function you've provided. Clearly, it's going to indeterminate since the second derivative for $x$ or $y$ is $0$. You'll have to just plug in the values of the critical points you've found and determine it based on that.
New contributor
You can determine the "type" of the critical point using the partial derivatives. Suppose $(x,y)$ is a critical point that you are considering. Then,
- If $f_{xx} f_{yy}- (f_{xx})^2 > 0$ and $f_{xx} > 0$ when evaluated at $(x,y)$, then there is a relative minimum at $(x,y)$.
- If $f_{xx} f_{yy}- (f_{xx})^2 > 0$ and $f_{xx} < 0$, you have a relative minimum.
- If $f_{xx} f_{yy}- (f_{xx})^2 < 0$ then you've got a saddle point.
- If $f_{xx} f_{yy}- (f_{xx})^2 = 0$ its indeterminate.
Edit: Just looked at the function you've provided. Clearly, it's going to indeterminate since the second derivative for $x$ or $y$ is $0$. You'll have to just plug in the values of the critical points you've found and determine it based on that.
New contributor
New contributor
answered Jan 4 at 4:13
kkckkc
1308
1308
New contributor
New contributor
1
This only provides local maxima or minima of the restricted function. For global one uses compactness and this criterion is then useless.
– Will M.
Jan 4 at 4:36
1
Examining the Jacobian determinant of the objective function in a constrained optimization problem doesn’t usually tell you anything useful. This problem is a case in point: the Jacobian determinant is identically equal to $-9$, so per these criteria the only critical points are saddle points. This is true of $f$ on $mathbb R^2$, but certainly not of its restriction to the ellipse.
– amd
Jan 4 at 4:47
add a comment |
1
This only provides local maxima or minima of the restricted function. For global one uses compactness and this criterion is then useless.
– Will M.
Jan 4 at 4:36
1
Examining the Jacobian determinant of the objective function in a constrained optimization problem doesn’t usually tell you anything useful. This problem is a case in point: the Jacobian determinant is identically equal to $-9$, so per these criteria the only critical points are saddle points. This is true of $f$ on $mathbb R^2$, but certainly not of its restriction to the ellipse.
– amd
Jan 4 at 4:47
1
1
This only provides local maxima or minima of the restricted function. For global one uses compactness and this criterion is then useless.
– Will M.
Jan 4 at 4:36
This only provides local maxima or minima of the restricted function. For global one uses compactness and this criterion is then useless.
– Will M.
Jan 4 at 4:36
1
1
Examining the Jacobian determinant of the objective function in a constrained optimization problem doesn’t usually tell you anything useful. This problem is a case in point: the Jacobian determinant is identically equal to $-9$, so per these criteria the only critical points are saddle points. This is true of $f$ on $mathbb R^2$, but certainly not of its restriction to the ellipse.
– amd
Jan 4 at 4:47
Examining the Jacobian determinant of the objective function in a constrained optimization problem doesn’t usually tell you anything useful. This problem is a case in point: the Jacobian determinant is identically equal to $-9$, so per these criteria the only critical points are saddle points. This is true of $f$ on $mathbb R^2$, but certainly not of its restriction to the ellipse.
– amd
Jan 4 at 4:47
add a comment |
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Just compute f at those 4 points. Btw, (0,0) is not on the ellipse.
– Martin R
Jan 4 at 4:10
@MartinR Is what I did fine?
– Nash
Jan 4 at 20:55