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Homebound on area of right triangle

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# 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 constrained optimization using Lagrange multipliers.

###### Theorem 1.

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

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

with equality when one has an isosceles right triangle.

###### Proof.

Suppose a triangle has legs of length $x$ and $y$. Then its hypotenuse has 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 gradient of $P$. That is to say, for some constant $\lambda$, we will have

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

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

Together with the constraint, these form a system of three equations for the three quantities $x$, $y$, and $\lambda$. Writing them out explicitly,

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

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

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

Not that we cannot have $\lambda=0$ because that would mean that all sides 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 second factor 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-multiply to obtain

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

Since we want $x\geq 0$ and $y\geq 0$ but the right-hand side is non-positive, the only option would be to have a trianagle of zero 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

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

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

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