By Area between Smooth Curve and Line with Fixed Endpoints is Maximized by Arc of Circle the maximizing curve is an arc of a circle. It is described as follows: :If $\dfrac l \pi \le \lambda < \infty$ then: ::$y = \sqrt {\lambda^2 - x^2} - \sqrt {\lambda^2 - a^2}$ :where: ::$l = 2 \lambda \, \map \arctan {\dfrac a {\sqrt {\lambda^2 - a^2} } }$ :If $\dfrac l {2 \pi} \le \lambda < \dfrac l \pi$ then: ::$y = \sqrt{\lambda^2 - a^2} - \sqrt{\lambda^2 - x^2}$ :where: ::$l = 2 \lambda \paren {\pi - \map \arctan {\dfrac a {\sqrt {\lambda^2 - a^2} } } }$ In the first case the area is a definite integral between $x = -a$ and $x = a$: {{begin-eqn}} {{eqn | l = A       | r = \int_{-a}^a \paren {\sqrt {\lambda^2 - x^2} - \sqrt {\lambda^2 - a^2} } \rd x }} {{eqn | r = \intlimits {\frac 1 2 \paren {x \sqrt {\lambda^2 - a^2} + \lambda^2 \arctan {\dfrac x {\sqrt {\lambda^2 - x^2} } } } - x \sqrt {\lambda^2 - a^2} } {-a} a       | c = Primitive of Root of a squared minus x squared }} {{eqn | r = \frac 1 2 \paren {\lambda^2 \arctan \frac a {\sqrt {\lambda^2 - a^2} } - a \sqrt {\lambda^2 - a^2} } }} {{end-eqn}} Solve the length constraint for $a$. The solution, satisfying strict positivity of $l$ and the allowed values of $\lambda$, is: :$a = \dfrac {\lambda \map \tan {\frac l {2 \lambda} } } {\sqrt {1 + \map {\tan^2} {\frac l {2 \lambda} } } }$ Substitution of this into the expression for the area leads to: :$A = \dfrac \lambda 2 \paren {l - \lambda \sin \dfrac l \lambda}$ $l$ is a constant, while $\lambda$ is a free variable. For the allowed values of $\lambda$, the area function is a continuous real function. By Definition of Local Maximum, the maximum is either in a subinterval of domain of $A$ {{WRT}} $\lambda$ or at one of the boundary values. By Derivative at Maximum or Minimum, we have to find the extremum of $A$: :$\dfrac {\d A} {\d \lambda} = \dfrac \lambda 2 \paren {\dfrac l \lambda + \dfrac l \lambda \cos \dfrac l \lambda - 2 \sin \dfrac l \lambda}$ This vanishes at $\dfrac l \lambda = \paren {2 k + 1} \pi$ and $\dfrac l \lambda = 0$, where $k \in \Z$. Then the area at these extremums acquires the following values: :$\map A {\dfrac l \lambda = 0} = 0$; :$\map A {\dfrac l \lambda = \paren {2 k + 1} \pi} = \dfrac {l^2} {\paren {1 + 2 k} 2 \pi}$ This is maximized for $k = 0$, or $\dfrac l \lambda = \pi$. Incidentally, these are also boundary values of the area function. This concludes the maximization of area when the arc is being varied from a straight line to a semicirle. The second part considers the variation of the curve from a semicirle to a circle. In this case the area is that of a semicirle plus a part of the lower semicirle: {{begin-eqn}} {{eqn | l = A        | r = \pi \frac {\lambda^2} 2 + \int_{-\lambda}^{-a} \sqbrk {\sqrt {\lambda^2 - a^2} - \paren {\sqrt {\lambda^2 - a^2} - \sqrt {\lambda^2 - x^2} } } \rd x + \int_{-a}^a \sqbrk {\sqrt {\lambda^2 - a^2} } \rd x + \int_a^\lambda \sqbrk {\sqrt {\lambda^2 - a^2} - \paren {\sqrt {\lambda^2 - a^2} - \sqrt {\lambda^2 - x^2} } } \rd x }}  {{eqn | r = \pi \lambda^2 + a \sqrt {\lambda^2 - a^2} - \lambda^2 \arctan \paren {\frac a {\sqrt {\lambda^2 - a^2} } }       | c = Primitive of Root of a squared minus x squared }} {{end-eqn}} Like in the previous case, solve the length constraint for $a$, while satisfying positivity and range conditions: :$a = \dfrac {\tan {\frac {2 \pi \lambda - l} {2 \lambda} } } {\sqrt {1 + \tan^2 {\frac {2 \pi \lambda - l} {2 \lambda} } } }$ Substitution into the area expression leads to: :$A = \dfrac {\lambda} 2 \paren {l - \lambda \sin \dfrac l \lambda}$ To find the extremum, compute its derivative {{WRT}} $\lambda$: :$\dfrac {\d A} {\d \lambda} = \dfrac 1 2 \paren {l + l \cos \dfrac l \lambda - 2 \lambda \sin \dfrac l \lambda}$ It vanishes if $\dfrac l \lambda = 0$ or $\dfrac l \lambda = \pi \paren {1 + 2 k}$, with $k \in \Z$. From these solutions the one satisfying the range of $\lambda$ is $\dfrac l \lambda = \pi$. The area for this value is $\dfrac {\pi \lambda^2} 2$. For completeness we have to check the other boundary value in this range, namely, $\lambda = \dfrac l {2 \pi}$. :$\map A {\lambda = \dfrac l {2\pi} } = \pi \lambda^2$. Since we have length as an input for this problem, express both areas in terms of length. :$\map A {l = \pi \lambda} = \dfrac {l^2} {2 \pi}$ :$\map A {l = 2 \pi \lambda} = \dfrac {l^2} {4 \pi}$ Hence, the area is maximized when the curve $y$ is a semicirle. {{qed}} Category:Isoperimetrical Problems nxo4guskqb1siiysuw57hgyo75ow037