Discussion
The extremal of the isoperimetric problem :
\[ I(y(x)) = \int_{1}^{4} (y')^{2}\, dx \]
subject to the conditions :
\[ \int_{1}^{4} y\, dx = 36 \]
and \(y(1) = 3\), \(y(4) = 24\) is :
Answer: A
The extremal is a parabola.
For an isoperimetric problem, extremise \(\int (F + \lambda G)\,dx\) with the constraint integrand \(G = y\) and a Lagrange multiplier \(\lambda\). Here \(H = (y')^{2} + \lambda y\), and its Euler equation is
\[ \begin{aligned} \lambda - \frac{d}{dx}(2y') &= 0 \\ y'' &= \frac{\lambda}{2} \end{aligned} \]
So \(y = \tfrac{\lambda}{4}x^{2} + c_{1}x + c_{2}\), a quadratic, whose graph is a parabola. The three constants are fixed by \(y(1) = 3\), \(y(4) = 24\) and the area condition; they give \(y = x^{2} + 2x\), which indeed has \(\int_{1}^{4} y\,dx = 21 + 15 = 36\).
The circle is the answer to the classical isoperimetric problem of maximising area for a fixed arc length, where the functional is \(\int\sqrt{1 + y'^{2}}\); this functional is different.
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