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You have a straight line of length $b$. You want to connect the ends of this fence so as to enclose a maximum area. You have a cost constraint. In the area between $x=0$ and $x=\frac{b}{2}$ costs $1$ dollar per ft and from $x=\frac{b}{2}$ to $x=b$ costs $2$ dollars per ft.

I realize I need to use lagrange multipliers and arc length. could someone get me started on this problem. any help or links would be appreciated..

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The first step is to write the area of the curve, it will give the integral you want to maximize. The second step is to write the cost, in this case, it is the sum of two integrals that equals something. From here, you already said the method you have to use. –  juanrapha Apr 17 '13 at 21:45
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An illustration would be helpful. At its current state, it's unclear whether you want to enclose the area using onl a this fence, or you want to build another fence. –  TZakrevskiy Aug 9 '13 at 17:13

1 Answer 1

Assume your fence runs from $O=(0,0)$ to $P=\bigl({b\over2},h\bigr)$ and then to $Q=(b,0)$ and that the arc length between $O$ and $P$ is $s_1$, between $P$ and $Q$ is $s_2$. Then the solution to Dido's problem tells you that the enclosed area between the base $OQ$ and the fence is largest when the two arcs are circular arcs of the given lengths connecting $O$ and $P$, resp., $P$ and $Q$. In this way your variational problem is reduced to a problem in a finite number of variables.

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