# 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

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

with equality when one has an isosceles right triangle.

###### Proof.

Suppose a triangle has legs of length $x$ and $y$. Then its 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

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

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

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

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

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

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

Not that we cannot have $\lambda =0$ because that would mean that 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 $x\ge 0$ and $y\ge 0$ 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

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

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

∎

Title | bound on area of right triangle |
---|---|

Canonical name | BoundOnAreaOfRightTriangle |

Date of creation | 2013-03-22 16:30:41 |

Last modified on | 2013-03-22 16:30:41 |

Owner | rspuzio (6075) |

Last modified by | rspuzio (6075) |

Numerical id | 17 |

Author | rspuzio (6075) |

Entry type | Theorem |

Classification | msc 51-00 |