Find the curve(i.e. the segment of a standard curve like circle, ellipse etc.) amongst all curves(segments) that have fixed total length, passes through $ (a,b)$ and $(c,d)$ and has maximum area between itself and the line joining the two points. I think calculus of variations is the way, but what is the functional to be extremized? Also whether double integrals have a role to play?
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duplicate - http://math.stackexchange.com/questions/1834585/curve-enclosing-the-maximum-area?noredirect=1#comment3750653_1834585 - please don't post same questions twice – gt6989b Jun 21 '16 at 20:44
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What area should be maximazed? – Virtuoz Jun 21 '16 at 20:44
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@gt6989b thanks. I modified the question slightly – vidyarthi Jun 21 '16 at 20:47
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Is this a closed curve? As the question stands right now, there is no specification of a boundary for an "area". (This differs from the "duplicate" linked, as that question used the $ \ x-$ axis as part of the boundary, and a curve segment passing through the indicated points for the rest.) – colormegone Jun 21 '16 at 21:02
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I think the answer is circle and is a direct consequence of the isoperimetric problem. – vidyarthi Jun 21 '16 at 21:02
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It's definitely a circle: http://www.cut-the-knot.org/Generalization/isop.shtml – Hrhm Jun 21 '16 at 23:25
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If $y=f(x)$ is the equation of the curve, the functional to be extremized (supposing $a<c$) is the one giving the area between the curve and the straight line: $$ \int_a^c \left(f(x)-{d-b\over c-a}(x-a)+b\right)dx, $$ with the constraint: $$ \int_a^c \sqrt{1+[f'(x)]^2}\,dx=l. $$
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