Mixing
how to use KKT conditions for an exponential function
$begingroup$
our teacher gave us a problem in the exam that I failed to answer it even after passing it, and I ask for an explanation from people here please...
this is the problem :
let K be a subset of $Bbb R^2$ such that : $$K={(x,y) in Bbb R^2, mathrm{e}^{-x} le y, quad y le 2-x,quad y le x}$$
and $$ f : Bbb R^2 to Bbb R $$
the function given by $$ f(x,y) = ycdotmathrm{e}^{-x} $$
1/ Show the existence of a solution (x*,y*) for the problem $$mathbf{min} underset{(x,y) in K} quad f(x,y) $$
I know there are many ways for doing that, like showing that f(x,y) is continuous and coercive and showing that K is closed, or showing that K is bounded and closed, means compact (I don't know if it is an enough condition for the existence), or showing that f(x,y) is strictly convex on the convex set K, please tell me which way it is the most efficient.
2/ justify and write the optimality conditions of KKT and solve the problem
and here the calamity, I find up to 9 cases that I couldn't at anyone find x or y --"
optimization karush-kuhn-tucker
asked Jan 23 at 21:20
Hansamu KaTuripu MyLudaHansamu KaTuripu MyLuda
31
$endgroup$
|
show 1 more comment
$begingroup$
our teacher gave us a problem in the exam that I failed to answer it even after passing it, and I ask for an explanation from people here please...
this is the problem :
let K be a subset of $Bbb R^2$ such that : $$K={(x,y) in Bbb R^2, mathrm{e}^{-x} le y, quad y le 2-x,quad y le x}$$
and $$ f : Bbb R^2 to Bbb R $$
the function given by $$ f(x,y) = ycdotmathrm{e}^{-x} $$
1/ Show the existence of a solution (x*,y*) for the problem $$mathbf{min} underset{(x,y) in K} quad f(x,y) $$
I know there are many ways for doing that, like showing that f(x,y) is continuous and coercive and showing that K is closed, or showing that K is bounded and closed, means compact (I don't know if it is an enough condition for the existence), or showing that f(x,y) is strictly convex on the convex set K, please tell me which way it is the most efficient.
2/ justify and write the optimality conditions of KKT and solve the problem
and here the calamity, I find up to 9 cases that I couldn't at anyone find x or y --"
optimization karush-kuhn-tucker
asked Jan 23 at 21:20
Hansamu KaTuripu MyLudaHansamu KaTuripu MyLuda
31
$endgroup$
$begingroup$
The Lagrangian is $$mathcal{L}(x,y,lambda,mu,nu)=ye^{-x}+lambda(e^{-x}-y)+mu(x+y-2)+nu(y-x).$$ The KKT conditions are $$begin{align*} -xye^{-x}-lambda e^{-x}+mu-nu=0\ e^{-x}-lambda+mu+nu=0\ e^{-x}leq y\ x+yleq 2\ yleq x\ lambda,mu,nugeq 0\ lambda(e^{-x}-y)=0\ mu(x+y-2)=0 \nu(y-x)=0 end{align*}$$ Now you just need to check each of the 8 combinations out of the complimentary slackness constraints (the last three).
$endgroup$
– nathan.j.mcdougall
Jan 23 at 23:53
$begingroup$
thank you so much, I really reached this step, but finding x and y, I always find a contradiction, I mean like landa is negative for example...
$endgroup$
– Hansamu KaTuripu MyLuda
Jan 24 at 0:24
$begingroup$
If you set $nu=mu=0$, and $e^{-x}=y$, then $(-1,e)$ is a solution with $lambda=egeq 0$. So although many of the cases are impossible, there is at least which is possible.
$endgroup$
– nathan.j.mcdougall
Jan 24 at 0:34
$begingroup$
thank you again, yea I tried them all, like the case that u just said, but it is impossible tho, because -1 is not in K and even e isn't in K...
$endgroup$
– Hansamu KaTuripu MyLuda
Jan 24 at 0:36
$begingroup$
Okay, you are right, my mistake. But still, there is $nu=0$, $y=e^{-x}$ and $x+y=2$ which gives $x=2+W(-e^{-2})$ and $y=expleft(-2-W(-e^{-2})right)$, and $$lambda=frac{1+x(2-x)}{y^{-1}-1}geq 0$$ $$mu=lambda-ygeq 0.$$ You can verify the inequalities by using the fact that $W$ is monotone and $W(-e^{-1})=-1$. Still, it seems strange that you'd be given this in an exam, I might have made a mistake.
$endgroup$
– nathan.j.mcdougall
Jan 24 at 1:00
|
show 1 more comment
$begingroup$
our teacher gave us a problem in the exam that I failed to answer it even after passing it, and I ask for an explanation from people here please...
this is the problem :
let K be a subset of $Bbb R^2$ such that : $$K={(x,y) in Bbb R^2, mathrm{e}^{-x} le y, quad y le 2-x,quad y le x}$$
and $$ f : Bbb R^2 to Bbb R $$
the function given by $$ f(x,y) = ycdotmathrm{e}^{-x} $$
1/ Show the existence of a solution (x*,y*) for the problem $$mathbf{min} underset{(x,y) in K} quad f(x,y) $$
I know there are many ways for doing that, like showing that f(x,y) is continuous and coercive and showing that K is closed, or showing that K is bounded and closed, means compact (I don't know if it is an enough condition for the existence), or showing that f(x,y) is strictly convex on the convex set K, please tell me which way it is the most efficient.
2/ justify and write the optimality conditions of KKT and solve the problem
and here the calamity, I find up to 9 cases that I couldn't at anyone find x or y --"
optimization karush-kuhn-tucker
asked Jan 23 at 21:20
Hansamu KaTuripu MyLudaHansamu KaTuripu MyLuda
31
$endgroup$
our teacher gave us a problem in the exam that I failed to answer it even after passing it, and I ask for an explanation from people here please...
this is the problem :
let K be a subset of $Bbb R^2$ such that : $$K={(x,y) in Bbb R^2, mathrm{e}^{-x} le y, quad y le 2-x,quad y le x}$$
and $$ f : Bbb R^2 to Bbb R $$
the function given by $$ f(x,y) = ycdotmathrm{e}^{-x} $$
1/ Show the existence of a solution (x*,y*) for the problem $$mathbf{min} underset{(x,y) in K} quad f(x,y) $$
I know there are many ways for doing that, like showing that f(x,y) is continuous and coercive and showing that K is closed, or showing that K is bounded and closed, means compact (I don't know if it is an enough condition for the existence), or showing that f(x,y) is strictly convex on the convex set K, please tell me which way it is the most efficient.
2/ justify and write the optimality conditions of KKT and solve the problem
and here the calamity, I find up to 9 cases that I couldn't at anyone find x or y --"
optimization karush-kuhn-tucker
optimization karush-kuhn-tucker
asked Jan 23 at 21:20
Hansamu KaTuripu MyLudaHansamu KaTuripu MyLuda
31
asked Jan 23 at 21:20
Hansamu KaTuripu MyLudaHansamu KaTuripu MyLuda
31
asked Jan 23 at 21:20
Hansamu KaTuripu MyLudaHansamu KaTuripu MyLuda
31
asked Jan 23 at 21:20
Hansamu KaTuripu MyLudaHansamu KaTuripu MyLuda
31
asked Jan 23 at 21:20
Hansamu KaTuripu MyLudaHansamu KaTuripu MyLuda
31
31
$begingroup$
The Lagrangian is $$mathcal{L}(x,y,lambda,mu,nu)=ye^{-x}+lambda(e^{-x}-y)+mu(x+y-2)+nu(y-x).$$ The KKT conditions are $$begin{align*} -xye^{-x}-lambda e^{-x}+mu-nu=0\ e^{-x}-lambda+mu+nu=0\ e^{-x}leq y\ x+yleq 2\ yleq x\ lambda,mu,nugeq 0\ lambda(e^{-x}-y)=0\ mu(x+y-2)=0 \nu(y-x)=0 end{align*}$$ Now you just need to check each of the 8 combinations out of the complimentary slackness constraints (the last three).
$endgroup$
– nathan.j.mcdougall
Jan 23 at 23:53
$begingroup$
thank you so much, I really reached this step, but finding x and y, I always find a contradiction, I mean like landa is negative for example...
$endgroup$
– Hansamu KaTuripu MyLuda
Jan 24 at 0:24
$begingroup$
If you set $nu=mu=0$, and $e^{-x}=y$, then $(-1,e)$ is a solution with $lambda=egeq 0$. So although many of the cases are impossible, there is at least which is possible.
$endgroup$
– nathan.j.mcdougall
Jan 24 at 0:34
$begingroup$
thank you again, yea I tried them all, like the case that u just said, but it is impossible tho, because -1 is not in K and even e isn't in K...
$endgroup$
– Hansamu KaTuripu MyLuda
Jan 24 at 0:36
$begingroup$
Okay, you are right, my mistake. But still, there is $nu=0$, $y=e^{-x}$ and $x+y=2$ which gives $x=2+W(-e^{-2})$ and $y=expleft(-2-W(-e^{-2})right)$, and $$lambda=frac{1+x(2-x)}{y^{-1}-1}geq 0$$ $$mu=lambda-ygeq 0.$$ You can verify the inequalities by using the fact that $W$ is monotone and $W(-e^{-1})=-1$. Still, it seems strange that you'd be given this in an exam, I might have made a mistake.
$endgroup$
– nathan.j.mcdougall
Jan 24 at 1:00
|
show 1 more comment
$begingroup$
The Lagrangian is $$mathcal{L}(x,y,lambda,mu,nu)=ye^{-x}+lambda(e^{-x}-y)+mu(x+y-2)+nu(y-x).$$ The KKT conditions are $$begin{align*} -xye^{-x}-lambda e^{-x}+mu-nu=0\ e^{-x}-lambda+mu+nu=0\ e^{-x}leq y\ x+yleq 2\ yleq x\ lambda,mu,nugeq 0\ lambda(e^{-x}-y)=0\ mu(x+y-2)=0 \nu(y-x)=0 end{align*}$$ Now you just need to check each of the 8 combinations out of the complimentary slackness constraints (the last three).
$endgroup$
– nathan.j.mcdougall
Jan 23 at 23:53
$begingroup$
thank you so much, I really reached this step, but finding x and y, I always find a contradiction, I mean like landa is negative for example...
$endgroup$
– Hansamu KaTuripu MyLuda
Jan 24 at 0:24
$begingroup$
If you set $nu=mu=0$, and $e^{-x}=y$, then $(-1,e)$ is a solution with $lambda=egeq 0$. So although many of the cases are impossible, there is at least which is possible.
$endgroup$
– nathan.j.mcdougall
Jan 24 at 0:34
$begingroup$
thank you again, yea I tried them all, like the case that u just said, but it is impossible tho, because -1 is not in K and even e isn't in K...
$endgroup$
– Hansamu KaTuripu MyLuda
Jan 24 at 0:36
$begingroup$
Okay, you are right, my mistake. But still, there is $nu=0$, $y=e^{-x}$ and $x+y=2$ which gives $x=2+W(-e^{-2})$ and $y=expleft(-2-W(-e^{-2})right)$, and $$lambda=frac{1+x(2-x)}{y^{-1}-1}geq 0$$ $$mu=lambda-ygeq 0.$$ You can verify the inequalities by using the fact that $W$ is monotone and $W(-e^{-1})=-1$. Still, it seems strange that you'd be given this in an exam, I might have made a mistake.
$endgroup$
– nathan.j.mcdougall
Jan 24 at 1:00
$begingroup$
The Lagrangian is $$mathcal{L}(x,y,lambda,mu,nu)=ye^{-x}+lambda(e^{-x}-y)+mu(x+y-2)+nu(y-x).$$ The KKT conditions are $$begin{align*} -xye^{-x}-lambda e^{-x}+mu-nu=0\ e^{-x}-lambda+mu+nu=0\ e^{-x}leq y\ x+yleq 2\ yleq x\ lambda,mu,nugeq 0\ lambda(e^{-x}-y)=0\ mu(x+y-2)=0 \nu(y-x)=0 end{align*}$$ Now you just need to check each of the 8 combinations out of the complimentary slackness constraints (the last three).
$endgroup$
– nathan.j.mcdougall
Jan 23 at 23:53
$begingroup$
The Lagrangian is $$mathcal{L}(x,y,lambda,mu,nu)=ye^{-x}+lambda(e^{-x}-y)+mu(x+y-2)+nu(y-x).$$ The KKT conditions are $$begin{align*} -xye^{-x}-lambda e^{-x}+mu-nu=0\ e^{-x}-lambda+mu+nu=0\ e^{-x}leq y\ x+yleq 2\ yleq x\ lambda,mu,nugeq 0\ lambda(e^{-x}-y)=0\ mu(x+y-2)=0 \nu(y-x)=0 end{align*}$$ Now you just need to check each of the 8 combinations out of the complimentary slackness constraints (the last three).
$endgroup$
– nathan.j.mcdougall
Jan 23 at 23:53
$begingroup$
thank you so much, I really reached this step, but finding x and y, I always find a contradiction, I mean like landa is negative for example...
$endgroup$
– Hansamu KaTuripu MyLuda
Jan 24 at 0:24
$begingroup$
thank you so much, I really reached this step, but finding x and y, I always find a contradiction, I mean like landa is negative for example...
$endgroup$
– Hansamu KaTuripu MyLuda
Jan 24 at 0:24
$begingroup$
If you set $nu=mu=0$, and $e^{-x}=y$, then $(-1,e)$ is a solution with $lambda=egeq 0$. So although many of the cases are impossible, there is at least which is possible.
$endgroup$
– nathan.j.mcdougall
Jan 24 at 0:34
$begingroup$
If you set $nu=mu=0$, and $e^{-x}=y$, then $(-1,e)$ is a solution with $lambda=egeq 0$. So although many of the cases are impossible, there is at least which is possible.
$endgroup$
– nathan.j.mcdougall
Jan 24 at 0:34
$begingroup$
thank you again, yea I tried them all, like the case that u just said, but it is impossible tho, because -1 is not in K and even e isn't in K...
$endgroup$
– Hansamu KaTuripu MyLuda
Jan 24 at 0:36
$begingroup$
thank you again, yea I tried them all, like the case that u just said, but it is impossible tho, because -1 is not in K and even e isn't in K...
$endgroup$
– Hansamu KaTuripu MyLuda
Jan 24 at 0:36
$begingroup$
Okay, you are right, my mistake. But still, there is $nu=0$, $y=e^{-x}$ and $x+y=2$ which gives $x=2+W(-e^{-2})$ and $y=expleft(-2-W(-e^{-2})right)$, and $$lambda=frac{1+x(2-x)}{y^{-1}-1}geq 0$$ $$mu=lambda-ygeq 0.$$ You can verify the inequalities by using the fact that $W$ is monotone and $W(-e^{-1})=-1$. Still, it seems strange that you'd be given this in an exam, I might have made a mistake.
$endgroup$
– nathan.j.mcdougall
Jan 24 at 1:00
$begingroup$
Okay, you are right, my mistake. But still, there is $nu=0$, $y=e^{-x}$ and $x+y=2$ which gives $x=2+W(-e^{-2})$ and $y=expleft(-2-W(-e^{-2})right)$, and $$lambda=frac{1+x(2-x)}{y^{-1}-1}geq 0$$ $$mu=lambda-ygeq 0.$$ You can verify the inequalities by using the fact that $W$ is monotone and $W(-e^{-1})=-1$. Still, it seems strange that you'd be given this in an exam, I might have made a mistake.
$endgroup$
– nathan.j.mcdougall
Jan 24 at 1:00
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
If $(x,y) in K$ then $0<e^{-x} leq y leq 2-x$ which gives $x leq 2$. Also $0<y leq xleq 2$. Hence $K$ is bounded. It is also closed, so is $K$ is compact. Since $f$ is continuous it attains its minimum at some point. To minimize $f$ we can minimize over $y$ first and then minimize over $x$. Hence we have to minimize $e^{-2x}$ subject to the conditions $e^{-x} leq min {x,2-x}$. It is easy to see that there is a unique $t$ such that $e^{t}(2-t)=1$ and $f$ is minimized when $x=t, y=e^{-t}$.
answered Jan 23 at 23:31
Kavi Rama MurthyKavi Rama Murthy
67.4k53067
$endgroup$
$begingroup$
oooh, thank you so much, I really needed for this, what about the second question please?
$endgroup$
– Hansamu KaTuripu MyLuda
Jan 24 at 0:26
$begingroup$
@HansamuKaTuripuMyLuda I have attempted to solve part 2) without Lagrange Mulitpliers, etc. I hope I didn't make any mistake.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 0:48
$begingroup$
mmm,I thank you so much :"')
$endgroup$
– Hansamu KaTuripu MyLuda
Jan 24 at 1:10
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3085095%2fhow-to-use-kkt-conditions-for-an-exponential-function%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $(x,y) in K$ then $0<e^{-x} leq y leq 2-x$ which gives $x leq 2$. Also $0<y leq xleq 2$. Hence $K$ is bounded. It is also closed, so is $K$ is compact. Since $f$ is continuous it attains its minimum at some point. To minimize $f$ we can minimize over $y$ first and then minimize over $x$. Hence we have to minimize $e^{-2x}$ subject to the conditions $e^{-x} leq min {x,2-x}$. It is easy to see that there is a unique $t$ such that $e^{t}(2-t)=1$ and $f$ is minimized when $x=t, y=e^{-t}$.
answered Jan 23 at 23:31
Kavi Rama MurthyKavi Rama Murthy
67.4k53067
$endgroup$
$begingroup$
oooh, thank you so much, I really needed for this, what about the second question please?
$endgroup$
– Hansamu KaTuripu MyLuda
Jan 24 at 0:26
$begingroup$
@HansamuKaTuripuMyLuda I have attempted to solve part 2) without Lagrange Mulitpliers, etc. I hope I didn't make any mistake.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 0:48
$begingroup$
mmm,I thank you so much :"')
$endgroup$
– Hansamu KaTuripu MyLuda
Jan 24 at 1:10
add a comment |
$begingroup$
If $(x,y) in K$ then $0<e^{-x} leq y leq 2-x$ which gives $x leq 2$. Also $0<y leq xleq 2$. Hence $K$ is bounded. It is also closed, so is $K$ is compact. Since $f$ is continuous it attains its minimum at some point. To minimize $f$ we can minimize over $y$ first and then minimize over $x$. Hence we have to minimize $e^{-2x}$ subject to the conditions $e^{-x} leq min {x,2-x}$. It is easy to see that there is a unique $t$ such that $e^{t}(2-t)=1$ and $f$ is minimized when $x=t, y=e^{-t}$.
answered Jan 23 at 23:31
Kavi Rama MurthyKavi Rama Murthy
67.4k53067
$endgroup$
$begingroup$
oooh, thank you so much, I really needed for this, what about the second question please?
$endgroup$
– Hansamu KaTuripu MyLuda
Jan 24 at 0:26
$begingroup$
@HansamuKaTuripuMyLuda I have attempted to solve part 2) without Lagrange Mulitpliers, etc. I hope I didn't make any mistake.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 0:48
$begingroup$
mmm,I thank you so much :"')
$endgroup$
– Hansamu KaTuripu MyLuda
Jan 24 at 1:10
add a comment |
$begingroup$
If $(x,y) in K$ then $0<e^{-x} leq y leq 2-x$ which gives $x leq 2$. Also $0<y leq xleq 2$. Hence $K$ is bounded. It is also closed, so is $K$ is compact. Since $f$ is continuous it attains its minimum at some point. To minimize $f$ we can minimize over $y$ first and then minimize over $x$. Hence we have to minimize $e^{-2x}$ subject to the conditions $e^{-x} leq min {x,2-x}$. It is easy to see that there is a unique $t$ such that $e^{t}(2-t)=1$ and $f$ is minimized when $x=t, y=e^{-t}$.
answered Jan 23 at 23:31
Kavi Rama MurthyKavi Rama Murthy
67.4k53067
$endgroup$
If $(x,y) in K$ then $0<e^{-x} leq y leq 2-x$ which gives $x leq 2$. Also $0<y leq xleq 2$. Hence $K$ is bounded. It is also closed, so is $K$ is compact. Since $f$ is continuous it attains its minimum at some point. To minimize $f$ we can minimize over $y$ first and then minimize over $x$. Hence we have to minimize $e^{-2x}$ subject to the conditions $e^{-x} leq min {x,2-x}$. It is easy to see that there is a unique $t$ such that $e^{t}(2-t)=1$ and $f$ is minimized when $x=t, y=e^{-t}$.
answered Jan 23 at 23:31
Kavi Rama MurthyKavi Rama Murthy
67.4k53067
edited Jan 24 at 0:46
answered Jan 23 at 23:31
Kavi Rama MurthyKavi Rama Murthy
67.4k53067
answered Jan 23 at 23:31
Kavi Rama MurthyKavi Rama Murthy
67.4k53067
answered Jan 23 at 23:31
Kavi Rama MurthyKavi Rama Murthy
67.4k53067
67.4k53067
$begingroup$
oooh, thank you so much, I really needed for this, what about the second question please?
$endgroup$
– Hansamu KaTuripu MyLuda
Jan 24 at 0:26
$begingroup$
@HansamuKaTuripuMyLuda I have attempted to solve part 2) without Lagrange Mulitpliers, etc. I hope I didn't make any mistake.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 0:48
$begingroup$
mmm,I thank you so much :"')
$endgroup$
– Hansamu KaTuripu MyLuda
Jan 24 at 1:10
add a comment |
$begingroup$
oooh, thank you so much, I really needed for this, what about the second question please?
$endgroup$
– Hansamu KaTuripu MyLuda
Jan 24 at 0:26
$begingroup$
@HansamuKaTuripuMyLuda I have attempted to solve part 2) without Lagrange Mulitpliers, etc. I hope I didn't make any mistake.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 0:48
$begingroup$
mmm,I thank you so much :"')
$endgroup$
– Hansamu KaTuripu MyLuda
Jan 24 at 1:10
$begingroup$
oooh, thank you so much, I really needed for this, what about the second question please?
$endgroup$
– Hansamu KaTuripu MyLuda
Jan 24 at 0:26
$begingroup$
oooh, thank you so much, I really needed for this, what about the second question please?
$endgroup$
– Hansamu KaTuripu MyLuda
Jan 24 at 0:26
$begingroup$
@HansamuKaTuripuMyLuda I have attempted to solve part 2) without Lagrange Mulitpliers, etc. I hope I didn't make any mistake.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 0:48
$begingroup$
@HansamuKaTuripuMyLuda I have attempted to solve part 2) without Lagrange Mulitpliers, etc. I hope I didn't make any mistake.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 0:48
$begingroup$
mmm,I thank you so much :"')
$endgroup$
– Hansamu KaTuripu MyLuda
Jan 24 at 1:10
$begingroup$
mmm,I thank you so much :"')
$endgroup$
– Hansamu KaTuripu MyLuda
Jan 24 at 1:10
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3085095%2fhow-to-use-kkt-conditions-for-an-exponential-function%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
The Lagrangian is $$mathcal{L}(x,y,lambda,mu,nu)=ye^{-x}+lambda(e^{-x}-y)+mu(x+y-2)+nu(y-x).$$ The KKT conditions are $$begin{align*} -xye^{-x}-lambda e^{-x}+mu-nu=0\ e^{-x}-lambda+mu+nu=0\ e^{-x}leq y\ x+yleq 2\ yleq x\ lambda,mu,nugeq 0\ lambda(e^{-x}-y)=0\ mu(x+y-2)=0 \nu(y-x)=0 end{align*}$$ Now you just need to check each of the 8 combinations out of the complimentary slackness constraints (the last three).
$endgroup$
– nathan.j.mcdougall
Jan 23 at 23:53
$begingroup$
thank you so much, I really reached this step, but finding x and y, I always find a contradiction, I mean like landa is negative for example...
$endgroup$
– Hansamu KaTuripu MyLuda
Jan 24 at 0:24
$begingroup$
If you set $nu=mu=0$, and $e^{-x}=y$, then $(-1,e)$ is a solution with $lambda=egeq 0$. So although many of the cases are impossible, there is at least which is possible.
$endgroup$
– nathan.j.mcdougall
Jan 24 at 0:34
$begingroup$
thank you again, yea I tried them all, like the case that u just said, but it is impossible tho, because -1 is not in K and even e isn't in K...
$endgroup$
– Hansamu KaTuripu MyLuda
Jan 24 at 0:36
$begingroup$
Okay, you are right, my mistake. But still, there is $nu=0$, $y=e^{-x}$ and $x+y=2$ which gives $x=2+W(-e^{-2})$ and $y=expleft(-2-W(-e^{-2})right)$, and $$lambda=frac{1+x(2-x)}{y^{-1}-1}geq 0$$ $$mu=lambda-ygeq 0.$$ You can verify the inequalities by using the fact that $W$ is monotone and $W(-e^{-1})=-1$. Still, it seems strange that you'd be given this in an exam, I might have made a mistake.
$endgroup$
– nathan.j.mcdougall
Jan 24 at 1:00