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Find curve of given length $l<2 R$ that maximizes area enclosed between it and arc of circle radius $R$ while passing through two circle points $A,B$ not lying on the x-axis.

enter image description here

Narasimham
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  • A circle doesn't have an x-axis, it has a center and a radius. If you need an axis, define one, please! –  Jul 21 '17 at 07:43
  • Is the changed grammar ok? – Narasimham Jul 21 '17 at 07:56
  • It's not a problem of grammar. Again: There is no x-axis in a circle, unless you define one. –  Jul 21 '17 at 08:07
  • I guess the problem is finding a curve such that: 1) it intersects a circle of radius $R$ in exactly two points $A,B$; 2) said two points are not diametrally opposite; 3) the length of the curve is $l<2R$; 4) the area of the smallest region enclosed by the curve and an arc wih extremal points $A,B$ is maximum. –  Jul 21 '17 at 08:40
  • Does the curve also have to be external to the circle? –  Jul 21 '17 at 08:43
  • Required to extremize the area.. $ , l < 2R $ constrains the curve to be inside the circle, although this would be automatically included in the solution. Added an image. – Narasimham Jul 21 '17 at 10:42

1 Answers1

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Too long for a comment, I would like to present my "line of attack". My best guess is that the supremum is attained for an arc of circumference from $A$ to $B$, long $2R$.

First note that the area between the segment $AB$ and the arc of circumference of radius $R$ does not depend on our choice of the candidate maximiser if the trial curve is not contained in the region between the segment $AB$ and the arc of radius $R$ : if we could satisfy this condition, the problem would boil then down to maximizing the area between the trial curve (of length $<2R$) and the segment $AB$.

Let us then consider a circumference going through $A$ and $B$, such that the length of one arc from $A$ to $B$ (named in the sequel “good arc”, to distinguish it from the other arc from $A$ to $B$, the “bad” arc) equals $2R$. enter image description hereBy symmetry we can always place the "good" arc on the opposite side, with respect to the segment $AB$ and the arc from $A$ to $B$ of radius $R$ you draw on your diagram.

This circumference will have a radius $\hat{R}$, and it maximizes the contained areas among all curves of length $2\pi \hat{R}$, by Dido’s argument.

Let us now consider a trial curve, such that the “bad” arc is unaffected, and the “good” arc is deformed such that its length remains constant. Clearly the total area contained within the varied curve decreases. As the area between the segment $AB$ and the “bad” arc is unvaried by construction, we conclude that the area between the segment $AB$ and the deformed “good” arc decreased. Should be straightforward to prove that the "good arc" does not intersect the region between the segment $AB$ and the arc of radius $R$.

This proves the “good” arc is a maximiser. Your constraints $l <2R$ excludes it, so that it is just a supremum.

An aedonist
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  • @Narasimham, if my answer is unclear or different from what you are looking for, I would be glad to amend it – An aedonist Jul 23 '17 at 17:10
  • Apologies for delay. For one thing I felt after posting that I had inadvertently cheated because the entire set up can Euclidaen shift so $A,B$ are on x-axis shoreline when your argument holds... by satisfying both arcs of length $< 2R$ both are valid and and equally good. Secondly (it was down voted, even if that is ignored) was hoping someone would come in with a response... – Narasimham Jul 28 '17 at 20:26