bound on area of right triangle

We may bound the area of a right triangleMathworldPlanetmath in terms of its perimeterPlanetmathPlanetmath. 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


with equality when one has an isosceles right triangle.


Suppose a triangle has legs of length x and y. Then its hypotenuseMathworldPlanetmath has length x2+y2, so the perimeter is given as


The area, of course, is


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 λ, we will have

Ax = λPx
Ay = λPy

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

12y = λ(1+xx2+y2)
12x = λ(1+yx2+y2)
P = x+y+x2+y2

Not that we cannot have λ=0 because that would mean that all sides of our triangle would have zero length. Hence, we may eliminate λ between the first two equations to obtain


which may be manipulated to yield


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


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


Since we want x0 and y0 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


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


so we have the solution

x = P2+2=2-22P
y = P2+2=2-22P.

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