The problem of Dido

Joseph Mallord William Turner, 1815

Dido building Carthage (William Turner, 1815)


Dido was the legendary founder of Carthage (Tunisia). When she arrived in 814BC on the coast of Tunisia, she asked for a piece of land. Her request was satisfied provided that the land could be encompassed by an ox-hide. With a remarkable mathematical intuition, she cut the ox-hide into a long thin strip and used it to encircle the land. This land became Carthage and Dido became the Queen.


What is the closed curve which has the maximum area for a given perimeter ?


The curve can be described by position (x(t),y(t)) and velocity (\dot{x}(t),\dot{y}(t)). We need to compute the length of the curve and the area of the enclosed surface.

The length is obtained by integration of the line element ds along the curve. The line element can be intuitively computed as an infinitesimal Pythagorean formula ds^2 = dx^2+dy^2 or ds = \sqrt{\dot{x}^2+\dot{y}^2}dt.

\displaystyle L = \oint \sqrt{\dot{x}^2+\dot{y}^2}dt

The area is less intuitive to compute but it is simple if we use the Green’s theorem.

Green’s Theorem : Over a region D in the plane with boundary \partial D we have

\displaystyle \int_{\partial D } f dx+g dy = \int_D \left( \frac{\partial g}{\partial x} - \frac{\partial f}{\partial y} \right) dx dy

With f=-\frac{y}{2} and g=\frac{x}{2} we get \frac{1}{2} \int_{\partial D} x dy - y dx = \int_D dxdy. Thus the enclosed area is given by the elegant formula :

\displaystyle A = \frac{1}{2} \oint (x \dot{y} - y \dot{x})dt

Now, the problem can be stated as an optimization problem with constraints :

Determine the closed curve that maximizes A subject to the constraint L = p


There is a well established solution to this problem, namely an isoperimetric problem in the calculus of variations.

Isoperimetric problem : Optimize \oint F(t,z,\dot{z}) dt subject to the constraint \oint G(t,z,\dot{z}) dt = constant

To solve this problem, define the Lagrangian function

\displaystyle \Lambda(t,z,\dot{z}) = F(t,z,\dot{z}) + \lambda G(t,z,\dot{z})

where \lambda is a constant called a Lagrange multiplier. Then the solutions to the problem can be shown to satisfy the Euler-Lagrange equations :

\displaystyle \frac{d}{dt} \frac{\partial \Lambda}{\partial \dot{z}} - \frac{\partial \Lambda}{\partial z} = 0

The Lagrangian of the Dido problem is :

\displaystyle \Lambda = \frac{1}{2} (x \dot{y} - y \dot{x}) + \lambda \sqrt{\dot{x}^2+\dot{y}^2}

There are two Euler-Lagrange equations corresponding to x and y respectively :

\displaystyle \frac{d}{dt} \frac{\partial \Lambda}{\partial \dot{x}} - \frac{\partial \Lambda}{\partial x} = 0

\displaystyle \frac{d}{dt} \frac{\partial \Lambda}{\partial \dot{y}} - \frac{\partial \Lambda}{\partial y} = 0

Applying the differentiation :

\displaystyle \frac{d}{dt} \left(- \frac{1}{2}y+\frac{\lambda \dot{x}}{\sqrt{\dot{x}^2+\dot{y}^2}} \right) - \frac{1}{2} \dot{y} = 0

\displaystyle \frac{d}{dt} \left( \frac{1}{2}x+\frac{\lambda \dot{y}}{\sqrt{\dot{x}^2+\dot{y}^2}} \right) + \frac{1}{2} \dot{x} = 0

Integrating both sides :

\displaystyle y - \frac{\lambda \dot{x}}{\sqrt{\dot{x}^2+\dot{y}^2}} = b and \displaystyle x + \frac{\lambda \dot{y}}{\sqrt{\dot{x}^2+\dot{y}^2}} = a

\displaystyle y-b = \frac{\lambda \dot{x}}{\sqrt{\dot{x}^2+\dot{y}^2}} and \displaystyle x-a = - \frac{\lambda \dot{y}}{\sqrt{\dot{x}^2+\dot{y}^2}}

Squaring and summing :

\displaystyle (x-a)^2+(y-b)^2=\lambda^2

We recognize the equation of a circle of center (a,b) and radius \lambda. Since the imposed perimeter is p we can deduce \lambda = \frac{p}{2 \pi}. Thus, the answer to the problem is a circle of radius \frac{p}{2 \pi} and area \frac{p^2}{4 \pi}.


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