bound on area of right triangle

We may bound the area of a right triangle in terms of its perimeter.The derivation of this bound is a good exercise in constrainedoptimization using Lagrange multipliers.

Theorem 1.

If a right triangle has perimeter $P$, then its area is bounded as

$$A\le \frac{3-2\sqrt{2}}{4}{P}^2$$

with equality when one has an isosceles right triangle.

Proof.

Suppose a triangle has legs of length $x$ and $y$. Then its hypotenusehas length $\sqrt{x^2+y^2}$, so the perimeter is given as

$$P=x+y+\sqrt{x^2+y^2}.$$

The area, of course, is

$$A=\frac{1}{2}xy.$$

We want to maximize $A$ subject to the constraint that $P$ be constant.This means that the gradient of $A$ will be proportional to the gradientof $P$. That is to say, for some constant $\lambda$, we will have

${\displaystyle \frac{\partial A}{\partial x}}$$=$$\lambda {\displaystyle \frac{\partial P}{\partial x}}$

${\displaystyle \frac{\partial A}{\partial y}}$$=$$\lambda {\displaystyle \frac{\partial P}{\partial y}}$

Together with the constraint, these form a system of three equationsfor the three quantities $x$, $y$, and $\lambda$. Writing them outexplicitly,

${\displaystyle \frac{1}{2}}y$$=$$\lambda \left(1+{\displaystyle \frac{x}{\sqrt{x^2+y^2}}}\right)$

${\displaystyle \frac{1}{2}}x$$=$$\lambda \left(1+{\displaystyle \frac{y}{\sqrt{x^2+y^2}}}\right)$

$P$$=$$x+y+\sqrt{x^2+y^2}$

Not that we cannot have $\lambda =0$ because that would mean that allsides of our triangle would have zero length. Hence, we may eliminate$\lambda$ between the first two equations to obtain

$$x\left(1+\frac{x}{\sqrt{x^2+y^2}}\right)=y\left(1+\frac{y}{\sqrt{x^2+y^2}}\right),$$

which may be manipulated to yield

$$(x-y)\left(1+\frac{x+y}{\sqrt{x^2+y^2}}\right)=0.$$

We have two case to consider — either the first factor or the secondfactor may equal zero. If the second factor equals zero,

$$1+\frac{x+y}{\sqrt{x^2+y^2}}=0,$$

move the “1” to the other side of the equation and cross-multiplyto obtain

$$x+y=-\sqrt{x^2+y^2}.$$

Since we want $x\ge 0$ and $y\ge 0$ but the right-hand side isnon-positive, the only option would be to have a trianagle ofzero area.The other possibility was to have the second factor equal zero,which would give

$$x-y=0.$$

In this case, $x$ equals $y$. Imposing this condition on the constraint,we see that

$$P=(2+\sqrt{2})x,$$

so we have the solution

$x$$=$${\displaystyle \frac{P}{2+\sqrt{2}}}={\displaystyle \frac{2-\sqrt{2}}{2}}P$

$y$$=$${\displaystyle \frac{P}{2+\sqrt{2}}}={\displaystyle \frac{2-\sqrt{2}}{2}}P.$

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