Mixing
Prove Unique Lagrange Multipliers Equality Constraint
$begingroup$
I am working through some old test papers in preparation for exams an am trying to scout out potential sneaky questions that might be asked. I've stumbled across this one. Would you please verify or correct my solution. Thank you very much.
$textbf{Question}$:
Consider the problem of finding the extrema of $f(x,y)$ with constraints $h_1(x,y) = 0$ and $h_2(x,y)=0$. All three $f,h_1,h_2$ are $C^2$ functions (I imagine they are trying to say second-order conditions should old). Suppose $(x_0,y_0)$ satisfies $h_1(x_0,y_0)=h_2(x_0,y_0)=0$ and
$$ nabla f(x_0,y_0) + mu_1nabla h_1(x_0,y_0) + mu_2nabla h_2(x_0,y_0) =(0,0) $$
Is there a unique $mu_1,mu_2 in mathbb{R}$ that a unique solution for the above exists while satisfying for some $alpha in mathbb{R}$
$$ nabla h_1(x_0,y_0) = alpha nabla h_2(x_0,y_0) $$
$textbf{Solution}$
So what they gave us was the first-order neccessary conditions and essentially asks the Lagrangian to be zero. When this happens, $f$ is noted to be a multiple of the constraints. This would mean it is linearly dependent on it. I state this because I am trying to say that in finding a unique solution we need not bother ourselves with it. We shift our focus to
$$ mu_1nabla h_1(x_0,y_0) + mu_2nabla h_2(x_0,y_0) $$
Which we can rewrite as
$$ mu_1,alpha left[ begin{matrix} (h_2)_x\(h_2)_y end{matrix}right] + mu_2 left[ begin{matrix} (h_2)_x\(h_2)_y end{matrix}right] $$
Or for even further convenience
$$left[ begin{matrix} mu_1 alpha & mu_2 \ mu_1 alpha & mu_2 end{matrix} right] left[ begin{matrix} (h_1)_x \ (h_2)_y end{matrix} right] iff mathbf{Ax} $$
We already note that the rows are factors of one another such that they are not linearly independent. Thus a unique solution does not exist. To be fancy, we can also test for non-zero determinant.
$$ det(mathbf{A}) = alphamu_1mu_2 -alphamu_1mu_2= 0 $$
Thus we have linear dependance and non-unique solutions.
$textbf{Extra:}$ (dont't have to answer if you want to)
They go on to ask for linearly independent $h_1$ and $h_2$ which satisfy at $(x_0,y_0)$
$$ nabla h_1(x_0,y_0) = alpha nabla h_2(x_0,y_0) $$
I said that at (0,0) with $alpha = 2$ and constraints:
$$ h_1 = x^2 - y^2 $$
$$ h_2 = e^x - e^y$$
It would hold. Do you agree with this example.
Thank you for your time.
optimization nonlinear-optimization lagrange-multiplier constraints
asked Jan 19 at 16:07
YesYes
887
$endgroup$
add a comment |
$begingroup$
I am working through some old test papers in preparation for exams an am trying to scout out potential sneaky questions that might be asked. I've stumbled across this one. Would you please verify or correct my solution. Thank you very much.
$textbf{Question}$:
Consider the problem of finding the extrema of $f(x,y)$ with constraints $h_1(x,y) = 0$ and $h_2(x,y)=0$. All three $f,h_1,h_2$ are $C^2$ functions (I imagine they are trying to say second-order conditions should old). Suppose $(x_0,y_0)$ satisfies $h_1(x_0,y_0)=h_2(x_0,y_0)=0$ and
$$ nabla f(x_0,y_0) + mu_1nabla h_1(x_0,y_0) + mu_2nabla h_2(x_0,y_0) =(0,0) $$
Is there a unique $mu_1,mu_2 in mathbb{R}$ that a unique solution for the above exists while satisfying for some $alpha in mathbb{R}$
$$ nabla h_1(x_0,y_0) = alpha nabla h_2(x_0,y_0) $$
$textbf{Solution}$
So what they gave us was the first-order neccessary conditions and essentially asks the Lagrangian to be zero. When this happens, $f$ is noted to be a multiple of the constraints. This would mean it is linearly dependent on it. I state this because I am trying to say that in finding a unique solution we need not bother ourselves with it. We shift our focus to
$$ mu_1nabla h_1(x_0,y_0) + mu_2nabla h_2(x_0,y_0) $$
Which we can rewrite as
$$ mu_1,alpha left[ begin{matrix} (h_2)_x\(h_2)_y end{matrix}right] + mu_2 left[ begin{matrix} (h_2)_x\(h_2)_y end{matrix}right] $$
Or for even further convenience
$$left[ begin{matrix} mu_1 alpha & mu_2 \ mu_1 alpha & mu_2 end{matrix} right] left[ begin{matrix} (h_1)_x \ (h_2)_y end{matrix} right] iff mathbf{Ax} $$
We already note that the rows are factors of one another such that they are not linearly independent. Thus a unique solution does not exist. To be fancy, we can also test for non-zero determinant.
$$ det(mathbf{A}) = alphamu_1mu_2 -alphamu_1mu_2= 0 $$
Thus we have linear dependance and non-unique solutions.
$textbf{Extra:}$ (dont't have to answer if you want to)
They go on to ask for linearly independent $h_1$ and $h_2$ which satisfy at $(x_0,y_0)$
$$ nabla h_1(x_0,y_0) = alpha nabla h_2(x_0,y_0) $$
I said that at (0,0) with $alpha = 2$ and constraints:
$$ h_1 = x^2 - y^2 $$
$$ h_2 = e^x - e^y$$
It would hold. Do you agree with this example.
Thank you for your time.
optimization nonlinear-optimization lagrange-multiplier constraints
asked Jan 19 at 16:07
YesYes
887
$endgroup$
add a comment |
$begingroup$
I am working through some old test papers in preparation for exams an am trying to scout out potential sneaky questions that might be asked. I've stumbled across this one. Would you please verify or correct my solution. Thank you very much.
$textbf{Question}$:
Consider the problem of finding the extrema of $f(x,y)$ with constraints $h_1(x,y) = 0$ and $h_2(x,y)=0$. All three $f,h_1,h_2$ are $C^2$ functions (I imagine they are trying to say second-order conditions should old). Suppose $(x_0,y_0)$ satisfies $h_1(x_0,y_0)=h_2(x_0,y_0)=0$ and
$$ nabla f(x_0,y_0) + mu_1nabla h_1(x_0,y_0) + mu_2nabla h_2(x_0,y_0) =(0,0) $$
Is there a unique $mu_1,mu_2 in mathbb{R}$ that a unique solution for the above exists while satisfying for some $alpha in mathbb{R}$
$$ nabla h_1(x_0,y_0) = alpha nabla h_2(x_0,y_0) $$
$textbf{Solution}$
So what they gave us was the first-order neccessary conditions and essentially asks the Lagrangian to be zero. When this happens, $f$ is noted to be a multiple of the constraints. This would mean it is linearly dependent on it. I state this because I am trying to say that in finding a unique solution we need not bother ourselves with it. We shift our focus to
$$ mu_1nabla h_1(x_0,y_0) + mu_2nabla h_2(x_0,y_0) $$
Which we can rewrite as
$$ mu_1,alpha left[ begin{matrix} (h_2)_x\(h_2)_y end{matrix}right] + mu_2 left[ begin{matrix} (h_2)_x\(h_2)_y end{matrix}right] $$
Or for even further convenience
$$left[ begin{matrix} mu_1 alpha & mu_2 \ mu_1 alpha & mu_2 end{matrix} right] left[ begin{matrix} (h_1)_x \ (h_2)_y end{matrix} right] iff mathbf{Ax} $$
We already note that the rows are factors of one another such that they are not linearly independent. Thus a unique solution does not exist. To be fancy, we can also test for non-zero determinant.
$$ det(mathbf{A}) = alphamu_1mu_2 -alphamu_1mu_2= 0 $$
Thus we have linear dependance and non-unique solutions.
$textbf{Extra:}$ (dont't have to answer if you want to)
They go on to ask for linearly independent $h_1$ and $h_2$ which satisfy at $(x_0,y_0)$
$$ nabla h_1(x_0,y_0) = alpha nabla h_2(x_0,y_0) $$
I said that at (0,0) with $alpha = 2$ and constraints:
$$ h_1 = x^2 - y^2 $$
$$ h_2 = e^x - e^y$$
It would hold. Do you agree with this example.
Thank you for your time.
optimization nonlinear-optimization lagrange-multiplier constraints
asked Jan 19 at 16:07
YesYes
887
$endgroup$
I am working through some old test papers in preparation for exams an am trying to scout out potential sneaky questions that might be asked. I've stumbled across this one. Would you please verify or correct my solution. Thank you very much.
$textbf{Question}$:
Consider the problem of finding the extrema of $f(x,y)$ with constraints $h_1(x,y) = 0$ and $h_2(x,y)=0$. All three $f,h_1,h_2$ are $C^2$ functions (I imagine they are trying to say second-order conditions should old). Suppose $(x_0,y_0)$ satisfies $h_1(x_0,y_0)=h_2(x_0,y_0)=0$ and
$$ nabla f(x_0,y_0) + mu_1nabla h_1(x_0,y_0) + mu_2nabla h_2(x_0,y_0) =(0,0) $$
Is there a unique $mu_1,mu_2 in mathbb{R}$ that a unique solution for the above exists while satisfying for some $alpha in mathbb{R}$
$$ nabla h_1(x_0,y_0) = alpha nabla h_2(x_0,y_0) $$
$textbf{Solution}$
So what they gave us was the first-order neccessary conditions and essentially asks the Lagrangian to be zero. When this happens, $f$ is noted to be a multiple of the constraints. This would mean it is linearly dependent on it. I state this because I am trying to say that in finding a unique solution we need not bother ourselves with it. We shift our focus to
$$ mu_1nabla h_1(x_0,y_0) + mu_2nabla h_2(x_0,y_0) $$
Which we can rewrite as
$$ mu_1,alpha left[ begin{matrix} (h_2)_x\(h_2)_y end{matrix}right] + mu_2 left[ begin{matrix} (h_2)_x\(h_2)_y end{matrix}right] $$
Or for even further convenience
$$left[ begin{matrix} mu_1 alpha & mu_2 \ mu_1 alpha & mu_2 end{matrix} right] left[ begin{matrix} (h_1)_x \ (h_2)_y end{matrix} right] iff mathbf{Ax} $$
We already note that the rows are factors of one another such that they are not linearly independent. Thus a unique solution does not exist. To be fancy, we can also test for non-zero determinant.
$$ det(mathbf{A}) = alphamu_1mu_2 -alphamu_1mu_2= 0 $$
Thus we have linear dependance and non-unique solutions.
$textbf{Extra:}$ (dont't have to answer if you want to)
They go on to ask for linearly independent $h_1$ and $h_2$ which satisfy at $(x_0,y_0)$
$$ nabla h_1(x_0,y_0) = alpha nabla h_2(x_0,y_0) $$
I said that at (0,0) with $alpha = 2$ and constraints:
$$ h_1 = x^2 - y^2 $$
$$ h_2 = e^x - e^y$$
It would hold. Do you agree with this example.
Thank you for your time.
optimization nonlinear-optimization lagrange-multiplier constraints
optimization nonlinear-optimization lagrange-multiplier constraints
asked Jan 19 at 16:07
YesYes
887
asked Jan 19 at 16:07
YesYes
887
asked Jan 19 at 16:07
YesYes
887
asked Jan 19 at 16:07
YesYes
887
asked Jan 19 at 16:07
YesYes
887
887
add a comment |
add a comment |
0
active
oldest
votes
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%2f3079505%2fprove-unique-lagrange-multipliers-equality-constraint%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3079505%2fprove-unique-lagrange-multipliers-equality-constraint%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
