Finding minimum using lagrange

Question

Let $x$, $y$ and $z$ be the length of a perpendicular drawn to each side from a point inside the equilateral triangle with one side length of $2$.

At this time $$I = x ^ 2 + y ^ 2 + z ^ 2 - xy - yz - zx.$$

Find the minimum of it.

Since I have to find the condition or the constraint. My question is why the constrain have to be area of the triangle

Here the constrain is the area of ​​the triangle that is $x + y + z = \sqrt{3}$(area)

I dont know how to get it. I know the area of triangle is $\sqrt 3$ but why it is $x+y+z$?

Thank you.

Answer 1

$$I=\frac{1}{2}\sum_{cyc}(x-y)^2\geq0.$$ The equality occurs for $x=y=z$, which says that $0$ is a minimal value of $I$.

I think using of Lagrange multipliers here is unnecessary, but if you wish...

Since $$C=\{(x.y,z)|x\geq0,y\geq0,z\geq0\}$$ is a compact and $I$ is a continuous function on $C$,

we see that $I$ gets on $C$ the minimal value.

Let $(x,y,z)$ is a minimum point of $I$.

There are two cases.

1. $xyz=0$.

Let $x=0$.

Hence, $$I=y^2-yz+z^2=\left(y-\frac{z}{2}\right)^2+\frac{3z^2}{4}\geq0.$$ The equality occurs for $x=y=z=0$;

2. $xyz\neq0$.

Hence, we need $$\frac{\partial I}{\partial x}=2x-y-z=0,$$ $$\frac{\partial I}{\partial y}=2y-x-z=0$$ and $$\frac{\partial I}{\partial z}=2z-x-y=0,$$ which gives $x=y=z$ and since $I(x,x,x)=0$ again, we got that $0$ is a minimal value.

If you want to get a value of $x+y+z$ then it's just a length of an altitude of the triangle, id est, $\frac{2\sqrt3}{2}=\sqrt3.$

Because you can calculate the area of the triangle twice: $$\frac{2\cdot x}{2}+\frac{2\cdot y}{2}+\frac{2\cdot z}{2}=\frac{2\cdot \sqrt3}{2}$$

Answer 2

The sum of $x+y+z$ is constant and is equal to the height of the triangle $h=\sqrt{2^2-1^2}=\sqrt 3$. Nothing to do with the area. Indeed you can join the internal point with the vertices of the triangle and get $3$ triangles (see the picture below)

The sum of the areas of the three triangles is obviously the area of the equilateral triangle $\dfrac{fm}{2}+\dfrac{gn}{2}+\dfrac{ph}{2}=\dfrac{f\cdot CH}{2}$

as $f=g=h$ we can collect in the LHS

$\dfrac{f}{2}(m+n+p)=\dfrac{f\cdot CH}{2}$

and therefore $m+n+p=CH$ as I stated above.

The minimum is achieved when the point $D$ is the baricenter of the triangle. Setting the gradient to null

$I_x=2 x - y - z=0$

$I_y=-x + 2 y - z=0$

$I_z=-x - y + 2 z=0$

we get $x=y=z$

and as $x+y+z=\sqrt 3$ we have the minimum when

$x=y=z=\dfrac{\sqrt 3}{3}$

and $I=0$