Suppose we are given two points P = ($2$,$0$), and Q($4$,$0$). I want to find the curve of length 6 connecting P to Q, which encloses the largest area above the horizontal axis. The constraint for this problem is $6 = \int_{2}^{4}\sqrt{1+\dot{y}^2}dx$ and our objective is $J = -\int_{2}^{4}y(x)dx$. At this point I should construct the Hamiltonian but I have no idea how to proceed.
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Do you have some extremal theorems that you're expected to use? – Al.G. Apr 15 '23 at 13:04
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@Al.G. After writing the Hamiltonian, I will show that Pontryagin's minimum principle holds. – eet Apr 15 '23 at 13:13
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2There is no Pontryagin here. No optimal control in there. This is a calculus of variations problem. – KBS Apr 15 '23 at 14:02
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1This may help – D S Apr 15 '23 at 14:58
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Does this answer your question? Maximum area under a curve by calculus of variations – Kurt G. May 15 '23 at 04:41