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the curve of fixed length $l$ that joins the points $(0,0)$ and $(1,0)$ lies above the $x-axis$ and encloses the maximum area between itself and the $x-axis$, is a segment of

  1. A straight line
  2. A parabola
  3. An ellipse
  4. A circle

I don't know exactly how to solve it but it seems Circle is the right Answer

slow but keen learner
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  • I don't understand the question. The line and any other curve through $(0,0)$ and $(1,0)$ will have a different length. How can you search for the curve of fixed length $l$? – gt6989b Jun 21 '16 at 14:56
  • I think we have to find curve(enclosing the maximum area) of fixed length joining point (0,0)and (1,0).Curve is also lying above x axis. This question is from a exam – slow but keen learner Jun 21 '16 at 15:09
  • you did not answer my question, just restated yours. A line, a parabola, a circle, etc. above $x$-axis, passing through these 2 points will be of different length. How can you fix the length if all these curves are of different length? – gt6989b Jun 21 '16 at 15:14
  • See https://en.wikipedia.org/wiki/Isoperimetric_inequality , particularly early on with the reference to "Dido's Problem". (Basically, you are correct.) – colormegone Jun 21 '16 at 20:57
  • @RecklessReckoner means right answer is circle? – slow but keen learner Jun 22 '16 at 01:32
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    The curve is a "segment of a circle", to follow the question statement by the appropriate choice (4) . – colormegone Jun 22 '16 at 01:37
  • The answer is a circle: http://www.cut-the-knot.org/Generalization/isop.shtml – Hrhm Jun 23 '16 at 00:56

4 Answers4

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Here l is the fixed perimeter of plane curve passing through two given points (0,0) and (1,0). Let S be the area enclosed by the plane curve and x-axis

Then, we are maximize $S=\int _0^1 y\;dx$ with boundary conditions y(0)=y(1)=0 subject to constraint $\int_0^1(1+y'^2)^\frac{1}{2}\;dx=l$

Let $F(x,y,y')=y+ \lambda(1+y'^2)^\frac{1}{2} $

Where $\lambda$ is the lagrange multiplier . Then the required extremal satiesfied the Euler's equation

$\frac{\partial F}{\partial y}-\frac{d}{dx}(\frac{\partial F}{\partial y'})=0\\ \Rightarrow 1-\frac{d}{dx}\{\frac{\lambda y'}{(1+y'^2)^\frac{1}{2}}\}=0\\ Integrating \;w\;r\;to\;x, x-\{\frac{\lambda y'}{(1+y'^2)^\frac{1}{2}}\}=a\\ \Rightarrow (x-a)^2=\{\frac{\lambda^2 y'^2}{(1+y'^2)}\}\\ \Rightarrow \frac{1+y'^2}{y'^2}=\frac{\lambda^2}{(x-a)^2}\\ \Rightarrow y'^2=\frac{(x-a)^2}{\lambda^2-(x-a)^2}\\ \frac{dy}{dx}=\pm \frac{(x-a)}{(\lambda^2-(x-a)^2)^\frac{1}{2}}\\ Integrating, \; y=b\pm\{\lambda^2-(x-a)^2\}^\frac{1}{2}\\ \Rightarrow (y-b)^2=\lambda^2-(x-a)^2\\ \Rightarrow (x-a)^2+(y-b)^2=\lambda^2$
this is equation of a circle with radius $\lambda$

kimchi lover
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  • thanx very much for the detailed solution – slow but keen learner Jun 24 '16 at 15:14
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    So what do you do if $l>\pi/2$ ? I think a good solution should provide boundaries when it is valid ! – yarchik Apr 12 '17 at 10:31
  • A circular segment is the answer for all $l>1$. It's just that for $l>\pi / 2$ the curve parameterization here (y as a function of x) wouldn't work, so you'd have to switch to polar or parametric coordinates for the proof. Intuitively though, as $l$ gets large the answer approaches a circle. – Quantum7 Jan 04 '20 at 22:03
  • Why is lambda needed here? Can't we say lambda = 1? Because what we need is just the difference between the area and the perimeter to be minimum and hence the functional could be just simple difference between them. Why a multiplier? – Apoorv Mishra Apr 06 '22 at 14:34
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The above problem once again:

A curve of fixed length l lies above the x-axis, joins the points (0,0) and (1,0), and encloses the maximum area between itself and the x-axis is --- a circle.

The constraints on l:

1) l>1 (obvious)

2) l=< π/2

Let's look at constraint 2:

2) : As l approaches π/2 the centre of the circle moves up, from below the x-axis( negative y-values) along a line parallel to.the x-axis through (1/2,0).

For l= π/2 the centre of the circle is at (1/2,0), maximal area is a semi circle.

Assume l>π/2 .

Problem: As before, this time with l>π/2. Find the curve that encloses the maximal area between itself and the x-axis.

Comments welcome.

Peter Szilas
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  • Not clear how do you treat the case $l>\pi/2$. Could you please elaborate. – yarchik Apr 12 '17 at 10:27
  • @yarchik. Consider the centers of the circles at (1/2.y), y > 0, mirror images of the circles discussed in the answer above - x-axis is axis of symmetry - passing through the two fixed points. So for l > π/2 the length of the circles' part of the circumference above the x-axis is l , the length given . These circles pass through the fixed points and have lengths l > π/2. Does this make sense? – Peter Szilas Apr 13 '17 at 01:14
  • Your solution is, of course, correct because you are talking about the second part of the problem. Namely, it is known that a circle is the sought curve, while its radius and center still need to be found. You perfectly demonstrated the solution. I have a problem with the variational part. Minimizing $\int_{0}^{1}(y(x)+\sqrt{1+y'^2(x)}) dx$ we do so on the $[0,,1]$ interval. Your circle, however, extends beyond $[0,,1]$ for $l>\pi/2$ ! This contradiction I addressed explicitly in my question. Ok, if one formulates the min. problem parametrically $x(t), y(t)$. – yarchik Apr 13 '17 at 07:20
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    @yarchik. Share your concerns. I have not seen them addressed in a textbook. The derivation of the solution is problematic. Another member, Rafael , you may want to check the following link, seems to think likewise. http://math.stackexchange.com/questions/2122751/calculus-of-variations-when-y-doesnt-exist-example-the-isoperimetric-pro – Peter Szilas Apr 13 '17 at 13:35
  • Thank you, sometimes such tiny details are missing in textbooks. By the way, your link is relevant. I made some search but obviously overlooked it. – yarchik Apr 13 '17 at 14:50
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A. little more to the above problem, maybe of interest.

(x-a)^2 + (y-b)^2 = (Lambda)^2.

Centre of the circle at (a,b) , radius Lambda.

1)Curve passes through (0,0) ,

2) Curve passes through (1,0),

3)Length of curve is l.

3 equations for 3 unknowns, OK.

The centre of any circle passing through 2 points lies on the perpendicular bisector of the segment joining the 2 points, in this case (0,0) and (1,0), I.e. a line parallel to the y-axis passing through (1/2,0).

So the x coordinate of the centre of the circle : a = 1/2.

Check: Setting x=0,y=0 and x=1,y=0 in the equation of the above circle gives

1) a^2 +b^2 = (Lambda)^2 and

2) (1-a)^2 + b^2 = (Lambda)^2

which gives a = 1/2.

Now to b and Lambda ( radius ).

Consider the line parallel to the y-axis through (1/2,0). The centres of all possible circles passing through (0,0) and (1,0) are on this line.

Pick a point on this line below the x -axis, call it M, and join it to (0,0) and (1,0): a segment of a circle.

Call the angle subtended by straight line joining M and (0,0) and M and (1/2,0) Phi.

1) Lambda × sin (Phi) = 1/2

2) 2× Lambda × (Phi) = l ( length)

2 equations for two unknowns, Lambda (radius) and Phi, angle subtended.

( b = - Lambda × cos(Phi) )

Dividing eq. 1) by eq.2) gives:

3) sin (Phi) = (1/l) × (Phi)

Left hand side is a sin fct., right hand side a linear fct. of (Phi).

There is, with the constraints on l mentioned below, exactly one solution 0 < (Phi) <= π/2, I.e. there is exactly one point of intersection of the sin curve with the straight line (See reference below).

Constraints for l:

1) l >= 1 , the length of the curve must be greater than the length of the straight line joining (0,0) and (1,0) (shortest distance)

2) l =< π/2 , largest l for centre of circle located at (1/2,0).

Comments are welcome.

Ref.

Thomas Mueller, Mathe-Seiten, Isoperimetrisches Problem.

Peter Szilas
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By reflecting the curve across the $x$-axis we can make it in to a closed loop, which doubles both the length and the area. Thus this is equivalent to the usual isoperimetric problem (maximize area inside a loop of fixed length) with the constraint of reflection symmetry. Since the circle is the unconstrained optimum and also has the reflection symmetry, it provides the solution to your problem: the curve is a semicircle.

Anthony Carapetis
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    This is not quite right. Consider the length $l = 1.00001$. It's intuitively clear that $l$ cannot make a semicircle with the given line segment. Indeed, there is only one possible length $l$ that works for the semicircle, namely the length of half of the circumference of circle of radius $1/2$. Notice also that the isoperimetric problem is for a regular curve; we cannot make a priori assumption that the loop you made will be regular at the endpoints of the line segment. – Squirrel-Power Oct 26 '18 at 05:17
  • You're right, apparently I somehow missed the fact that the distance between the two intersection points is fixed. – Anthony Carapetis Oct 27 '18 at 00:17