# classical isoperimetric problem

The points $a$ and $b$ on the $x$-axis have to be by an arc with a given length (http://planetmath.org/ArcLength) $l$ such that the area between the $x$-axis and the arc is as great as possible.

Denote the equation of the searched arc by $y=y(x)$. The task, which belongs to the isoperimetric problems^{} (http://planetmath.org/IsoperimetricProblem), can be formulated as

$\text{to maximise}\mathit{\hspace{1em}}{\displaystyle {\int}_{a}^{b}}y\mathit{d}x$ | (1) |

under the constraint condition

${\int}_{a}^{b}}\sqrt{1+{y}^{\prime \mathrm{\hspace{0.17em}2}}}\mathit{d}x=l.$ | (2) |

We have the integrands

$$f(x,y,{y}^{\prime})\equiv y,g(x,y,{y}^{\prime})\equiv \sqrt{1+{y}^{\prime \mathrm{\hspace{0.17em}2}}}.$$ |

The variation problem for the functional in (1) may be considered as a free variation problem (without conditions) for the functional ${\int}_{a}^{b}(f-\lambda g)\mathit{d}x$ where $\lambda $ is a Lagrange multiplier. For this end we need the Euler–Lagrange differential equation^{} (http://planetmath.org/EulerLagrangeDifferentialEquation)

$\frac{\partial}{\partial y}}(f-\lambda g)-{\displaystyle \frac{d}{dx}}{\displaystyle \frac{\partial}{\partial {y}^{\prime}}}(f-\lambda g)=\mathrm{\hspace{0.33em}0}.$ | (3) |

Since the expression $f-\lambda g$ does not depend explicitly on $x$, the differential equation (3) has, by the Beltrami identity^{}, a first integral of the form

$$(f-\lambda g)-{y}^{\prime}\cdot ({f}_{{y}^{\prime}}^{\prime}-\lambda {g}_{{y}^{\prime}}^{\prime})={C}_{2},$$ |

which reads simply

$$y-\frac{\lambda}{\sqrt{1+{y}^{\prime \mathrm{\hspace{0.17em}2}}}}={C}_{2}.$$ |

This differential equation may be written

$${y}^{\prime}\equiv \frac{dy}{dx}=\frac{\sqrt{{\lambda}^{2}-{(y-{C}_{2})}^{2}}}{y-{C}_{2}},$$ |

where one can separate the variables (http://planetmath.org/SeparationOfVariables) and integrate, obtaining the equation

$${(x-{C}_{1})}^{2}+{(y-{C}_{2})}^{2}={\lambda}^{2}$$ |

of a circle. Here, the parametres ${C}_{1},{C}_{2},\lambda $ may be determined from the conditions

$$y(a)=y(b)=\mathrm{\hspace{0.33em}0},\text{arc length}=l.$$ |

Thus the extremal of this variational problem is a circular arc (http://planetmath.org/CircularSegment) connecting the given points.

Note that in every point of the arc, the angle of view of the line segment between the given points is constant.

Title | classical isoperimetric problem |
---|---|

Canonical name | ClassicalIsoperimetricProblem |

Date of creation | 2013-03-22 19:10:26 |

Last modified on | 2013-03-22 19:10:26 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 22 |

Author | pahio (2872) |

Entry type | Example |

Classification | msc 47A60 |

Classification | msc 49K22 |

Classification | msc 49K05 |

Related topic | LagrangeMultiplierMethod |

Related topic | CircularSegment |

Related topic | AngleOfViewOfALineSegment |