Geometrical optimization problem.

Question

Sum of areas of 2 equilateral triangles is equal 2. Find the length of their sides for which sum of their circumferences is the smallest.

I noticed that if 3x+3y is the smallest then so is x+y and that the formula for an area of the equilateral triangle is $Area = \frac{(Side)^2\sqrt{3}}{4}$ so I constructed the Lagrangian which looks like this:

$\mathcal{L}(x,y,\lambda) = x+y-\lambda(\frac{x^2\sqrt{3}}{4}+\frac{y^2\sqrt{3}}{4}-2)$

and $\nabla\mathcal{L}=0$ if

$ \left\{ \begin{array}{ll} \frac{\lambda x\sqrt{3}}{2}=1\\ \frac{\lambda y\sqrt{3}}{2}=1\\ \frac{x^2\sqrt{3}}{4}+\frac{y^2\sqrt{3}}{4}-2=0 \end{array} \right.$

but then the only extremum we get for x=y. What for me is quite surprising because if you can minimize the circumference you should also be able to maximize (even if one of the pair x or y is set to 0 forgetting about geometrical background of this problem) and in this case when I have one extremum only how should I know if it is maximum or minimum?

Answer 1

Indeed it is not a minimum, to check let compare with the solution for $y=0$ that is

$$\frac{x^2\sqrt{3}}{4}-2=0\implies x^2=\frac 8 {\sqrt 3} \implies x=\frac{2\sqrt 2}{\sqrt[4] 3}$$

thus what you are obtaining is the maximum value for $x+y$.

Note that the minimum should be reached for $x=y=-\frac{2}{\sqrt[4]3}$ but we are excluding that solution since we also need that $x\ge 0$ and $y\ge 0$ which are two others constraints to be considered in the original problem.

Answer 2

You can state the problem also as:

Given a circle

$$ C \to x^2+y^2=\frac{2}{\mu} $$

and a line

$$ L\to x + y = \lambda $$

find the minimum/maximum values for $\lambda$ such that $C$ and $L$ are tangent but with the restrictions $x > 0$ and $y > 0$

The Lagrangian formulation then can be stated as

$$ L(x,y,\lambda) = x + y + \lambda_1(x^2+y^2-\frac{2}{\mu})+\lambda_2(x-\epsilon_1^2)+\lambda_3(y-\epsilon_2^2) $$

Here $\epsilon_1, \epsilon_2$ are slack variables to avoid the inequalities. Now the stationary points can be found by solving

$$ \nabla L = (2\lambda_1x+\lambda_2+1,2\lambda_1 y +\lambda_3+1,x^2+y^2-\frac{2}{\mu},x-\epsilon_1^2,y-\epsilon_2^2,\epsilon_1\lambda_2,\epsilon_2\lambda_3) = 0 $$

and the solution is at

$$ x = y = \frac{1}{\sqrt{\mu}}\ \ \mbox{giving}\ \ x+y = \frac{2}{\sqrt{\mu}} $$