An isoperimetric problem

Question

How can I find a triangle which encloses the largest area with given (total) length using caculus of variation? I know the direct method but I can't find one using variational method.

Answer 1

Let the sides be of length

$$ (2x,2y,2L) $$

where $L$ is a constant and $x,y$ are variables.

Semi-perimeter constraint function is $$ p=(x+y+L) $$

Square of largest area objective function is $$ q=A^2=(-L+x+y)(L+x+y)(L-x+y)(L+x-y) $$

Apply Euler-Lagrange equation and simplify algebraically using symmetry in variables ($x,y),$

Area square $q$ $$-L^4 + 2 L^2 x^2 - x^4 + 2 L^2 y^2 + 2 x^2 y^2 - y^4$$ resulting in relation between partial derivatives $$\dfrac{p_x}{p_y} =\dfrac{q_x}{q_y} \text{ i.e.,}$$

$$\frac{\frac{\partial p}{ \partial x}}{\frac{\partial p}{ \partial y}}= \frac{\frac{\partial q}{ \partial x}}{\frac{\partial q}{ \partial y}}$$

$$\dfrac{4L^2-4x^3+4xy^2}{4L^2y+4x^2y-4y^3}=1$$

Simplifying, $(x-y)$ is found to be a factor. So

$$ x=y=L,$$

i.e., maximum area is enclosed for an equilateral triangle.