Maximizing perimeter to area ratio of a function
$begingroup$
I was trying to find the curve, $y = f(x)$, from $x = x_i$ to $x = x_f$, constrained by $y(x_i) = 0$ and $y(x_f) = 0$, such that the ratio between the arc length of the curve and the area below the curve was minimized.
This setup cannot produce all possible shapes, for example a square, since the square would be a vertical line followed by a horizontal line then another vertical line, which is not smooth. However, it can produce a semi-circle which is the answer I'm trying to arrive at
ie. $$y = sqrt{r^2 - x^2}$$
My first approach was to use calculus of variations. However, I'm not sure where the constraints of $y(x_i) = 0$ and $y(x_f) = 0$ can fit in (Perhaps I'm required to use parametric equations). These constraints are so that I get a "closed" curve.
I can define arc length as $$S = int_{x_i}^{x_f}sqrt{1 + (dy/dx)^2}dx$$
I can then define area as $$A = int_{x_i}^{x_f}{ydx}$$
Thus, the function I want to minimize is $$(S/A)^2$$. I have to square it so that the function doesn't "cheat" and go below the x axis, creating an unbounded negative ratio.
The Euler–Lagrange equation gives us:
$${displaystyle {frac {partial L}{partial f}}-{frac {d}{dx}}{frac {partial L}{partial f'}}=0}$$
If I was simply trying to minimize arc length, I could have $$ L = int_{0}^{1}{sqrt{1 + (dy/dx)^2}}$$
However, my $L$ is not a simple function, its actually the division of two integrals. Perhaps I can simplify the division into a single integral, but I haven't been able to figure out how. I would appreciate any advice on how to proceed, thanks!
calculus optimization calculus-of-variations
asked Nov 11 '18 at 21:04
user2045279user2045279
1726
$endgroup$
add a comment |
$begingroup$
I was trying to find the curve, $y = f(x)$, from $x = x_i$ to $x = x_f$, constrained by $y(x_i) = 0$ and $y(x_f) = 0$, such that the ratio between the arc length of the curve and the area below the curve was minimized.
This setup cannot produce all possible shapes, for example a square, since the square would be a vertical line followed by a horizontal line then another vertical line, which is not smooth. However, it can produce a semi-circle which is the answer I'm trying to arrive at
ie. $$y = sqrt{r^2 - x^2}$$
My first approach was to use calculus of variations. However, I'm not sure where the constraints of $y(x_i) = 0$ and $y(x_f) = 0$ can fit in (Perhaps I'm required to use parametric equations). These constraints are so that I get a "closed" curve.
I can define arc length as $$S = int_{x_i}^{x_f}sqrt{1 + (dy/dx)^2}dx$$
I can then define area as $$A = int_{x_i}^{x_f}{ydx}$$
Thus, the function I want to minimize is $$(S/A)^2$$. I have to square it so that the function doesn't "cheat" and go below the x axis, creating an unbounded negative ratio.
The Euler–Lagrange equation gives us:
$${displaystyle {frac {partial L}{partial f}}-{frac {d}{dx}}{frac {partial L}{partial f'}}=0}$$
If I was simply trying to minimize arc length, I could have $$ L = int_{0}^{1}{sqrt{1 + (dy/dx)^2}}$$
However, my $L$ is not a simple function, its actually the division of two integrals. Perhaps I can simplify the division into a single integral, but I haven't been able to figure out how. I would appreciate any advice on how to proceed, thanks!
calculus optimization calculus-of-variations
asked Nov 11 '18 at 21:04
user2045279user2045279
1726
$endgroup$
add a comment |
$begingroup$
I was trying to find the curve, $y = f(x)$, from $x = x_i$ to $x = x_f$, constrained by $y(x_i) = 0$ and $y(x_f) = 0$, such that the ratio between the arc length of the curve and the area below the curve was minimized.
This setup cannot produce all possible shapes, for example a square, since the square would be a vertical line followed by a horizontal line then another vertical line, which is not smooth. However, it can produce a semi-circle which is the answer I'm trying to arrive at
ie. $$y = sqrt{r^2 - x^2}$$
My first approach was to use calculus of variations. However, I'm not sure where the constraints of $y(x_i) = 0$ and $y(x_f) = 0$ can fit in (Perhaps I'm required to use parametric equations). These constraints are so that I get a "closed" curve.
I can define arc length as $$S = int_{x_i}^{x_f}sqrt{1 + (dy/dx)^2}dx$$
I can then define area as $$A = int_{x_i}^{x_f}{ydx}$$
Thus, the function I want to minimize is $$(S/A)^2$$. I have to square it so that the function doesn't "cheat" and go below the x axis, creating an unbounded negative ratio.
The Euler–Lagrange equation gives us:
$${displaystyle {frac {partial L}{partial f}}-{frac {d}{dx}}{frac {partial L}{partial f'}}=0}$$
If I was simply trying to minimize arc length, I could have $$ L = int_{0}^{1}{sqrt{1 + (dy/dx)^2}}$$
However, my $L$ is not a simple function, its actually the division of two integrals. Perhaps I can simplify the division into a single integral, but I haven't been able to figure out how. I would appreciate any advice on how to proceed, thanks!
calculus optimization calculus-of-variations
asked Nov 11 '18 at 21:04
user2045279user2045279
1726
$endgroup$
I was trying to find the curve, $y = f(x)$, from $x = x_i$ to $x = x_f$, constrained by $y(x_i) = 0$ and $y(x_f) = 0$, such that the ratio between the arc length of the curve and the area below the curve was minimized.
This setup cannot produce all possible shapes, for example a square, since the square would be a vertical line followed by a horizontal line then another vertical line, which is not smooth. However, it can produce a semi-circle which is the answer I'm trying to arrive at
ie. $$y = sqrt{r^2 - x^2}$$
My first approach was to use calculus of variations. However, I'm not sure where the constraints of $y(x_i) = 0$ and $y(x_f) = 0$ can fit in (Perhaps I'm required to use parametric equations). These constraints are so that I get a "closed" curve.
I can define arc length as $$S = int_{x_i}^{x_f}sqrt{1 + (dy/dx)^2}dx$$
I can then define area as $$A = int_{x_i}^{x_f}{ydx}$$
Thus, the function I want to minimize is $$(S/A)^2$$. I have to square it so that the function doesn't "cheat" and go below the x axis, creating an unbounded negative ratio.
The Euler–Lagrange equation gives us:
$${displaystyle {frac {partial L}{partial f}}-{frac {d}{dx}}{frac {partial L}{partial f'}}=0}$$
If I was simply trying to minimize arc length, I could have $$ L = int_{0}^{1}{sqrt{1 + (dy/dx)^2}}$$
However, my $L$ is not a simple function, its actually the division of two integrals. Perhaps I can simplify the division into a single integral, but I haven't been able to figure out how. I would appreciate any advice on how to proceed, thanks!
calculus optimization calculus-of-variations
calculus optimization calculus-of-variations
asked Nov 11 '18 at 21:04
user2045279user2045279
1726
asked Nov 11 '18 at 21:04
user2045279user2045279
1726
asked Nov 11 '18 at 21:04
user2045279user2045279
1726
asked Nov 11 '18 at 21:04
user2045279user2045279
1726
asked Nov 11 '18 at 21:04
user2045279user2045279
1726
1726
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This the oldest known problem in Calculus of Variations called the Dido's Problem.
In modern times, this is known as an Isoperimetric Problem. I understand that you want to minimize the ratio between the perimeter and the area. The standard way I know to solve this is by making the area constant$(A)$ and then try to find the shortest curve that has this constant area $A$.
begin{align}
&int_{x_i}^{x_f}sqrt{1 + dot{y}^2}dxrightarrow min\
&int_{x_i}^{x_f}{ydx}=A
end{align}
With end constrains $y(x_i) = 0$ and $y(x_f) = 0$. Then you can solve it as a bolza problem as follows.
begin{align}
&L=lambda _0sqrt{1 + dot{y}^2}+lambda_1y\
&dot{y}L_{dot{y}}-L=constant
end{align}
Where the second equation is given by the Euler-Lagrange equation and $lambda_0$ is positive for a minimization problem and $lambda$'s are unique upto a multiple. So it would be sufficient to check for $lambda_0=0,1$.
answered Dec 6 '18 at 9:22
mm-crjmm-crj
425213
$endgroup$
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%2f2994461%2fmaximizing-perimeter-to-area-ratio-of-a-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
$begingroup$
This the oldest known problem in Calculus of Variations called the Dido's Problem.
In modern times, this is known as an Isoperimetric Problem. I understand that you want to minimize the ratio between the perimeter and the area. The standard way I know to solve this is by making the area constant$(A)$ and then try to find the shortest curve that has this constant area $A$.
begin{align}
&int_{x_i}^{x_f}sqrt{1 + dot{y}^2}dxrightarrow min\
&int_{x_i}^{x_f}{ydx}=A
end{align}
With end constrains $y(x_i) = 0$ and $y(x_f) = 0$. Then you can solve it as a bolza problem as follows.
begin{align}
&L=lambda _0sqrt{1 + dot{y}^2}+lambda_1y\
&dot{y}L_{dot{y}}-L=constant
end{align}
Where the second equation is given by the Euler-Lagrange equation and $lambda_0$ is positive for a minimization problem and $lambda$'s are unique upto a multiple. So it would be sufficient to check for $lambda_0=0,1$.
answered Dec 6 '18 at 9:22
mm-crjmm-crj
425213
$endgroup$
add a comment |
$begingroup$
This the oldest known problem in Calculus of Variations called the Dido's Problem.
In modern times, this is known as an Isoperimetric Problem. I understand that you want to minimize the ratio between the perimeter and the area. The standard way I know to solve this is by making the area constant$(A)$ and then try to find the shortest curve that has this constant area $A$.
begin{align}
&int_{x_i}^{x_f}sqrt{1 + dot{y}^2}dxrightarrow min\
&int_{x_i}^{x_f}{ydx}=A
end{align}
With end constrains $y(x_i) = 0$ and $y(x_f) = 0$. Then you can solve it as a bolza problem as follows.
begin{align}
&L=lambda _0sqrt{1 + dot{y}^2}+lambda_1y\
&dot{y}L_{dot{y}}-L=constant
end{align}
Where the second equation is given by the Euler-Lagrange equation and $lambda_0$ is positive for a minimization problem and $lambda$'s are unique upto a multiple. So it would be sufficient to check for $lambda_0=0,1$.
answered Dec 6 '18 at 9:22
mm-crjmm-crj
425213
$endgroup$
add a comment |
$begingroup$
This the oldest known problem in Calculus of Variations called the Dido's Problem.
In modern times, this is known as an Isoperimetric Problem. I understand that you want to minimize the ratio between the perimeter and the area. The standard way I know to solve this is by making the area constant$(A)$ and then try to find the shortest curve that has this constant area $A$.
begin{align}
&int_{x_i}^{x_f}sqrt{1 + dot{y}^2}dxrightarrow min\
&int_{x_i}^{x_f}{ydx}=A
end{align}
With end constrains $y(x_i) = 0$ and $y(x_f) = 0$. Then you can solve it as a bolza problem as follows.
begin{align}
&L=lambda _0sqrt{1 + dot{y}^2}+lambda_1y\
&dot{y}L_{dot{y}}-L=constant
end{align}
Where the second equation is given by the Euler-Lagrange equation and $lambda_0$ is positive for a minimization problem and $lambda$'s are unique upto a multiple. So it would be sufficient to check for $lambda_0=0,1$.
answered Dec 6 '18 at 9:22
mm-crjmm-crj
425213
$endgroup$
This the oldest known problem in Calculus of Variations called the Dido's Problem.
In modern times, this is known as an Isoperimetric Problem. I understand that you want to minimize the ratio between the perimeter and the area. The standard way I know to solve this is by making the area constant$(A)$ and then try to find the shortest curve that has this constant area $A$.
begin{align}
&int_{x_i}^{x_f}sqrt{1 + dot{y}^2}dxrightarrow min\
&int_{x_i}^{x_f}{ydx}=A
end{align}
With end constrains $y(x_i) = 0$ and $y(x_f) = 0$. Then you can solve it as a bolza problem as follows.
begin{align}
&L=lambda _0sqrt{1 + dot{y}^2}+lambda_1y\
&dot{y}L_{dot{y}}-L=constant
end{align}
Where the second equation is given by the Euler-Lagrange equation and $lambda_0$ is positive for a minimization problem and $lambda$'s are unique upto a multiple. So it would be sufficient to check for $lambda_0=0,1$.
answered Dec 6 '18 at 9:22
mm-crjmm-crj
425213
answered Dec 6 '18 at 9:22
mm-crjmm-crj
425213
answered Dec 6 '18 at 9:22
mm-crjmm-crj
425213
answered Dec 6 '18 at 9:22
mm-crjmm-crj
425213
425213
add a comment |
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%2f2994461%2fmaximizing-perimeter-to-area-ratio-of-a-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
