ThmDex – An index of mathematical definitions, results, and conjectures.

Result R5677 on D5407: Positive real number

Of all right triangles with a given perimeter, the isosceles triangle maximizes the area

Formulation 0

Let $a, b, \lambda \in (0, \infty)$ each be a D5407: Positive real number such that

(i)	\begin{equation} a + b + \sqrt{a^2 + b^2} = \lambda \end{equation}

Then

(1)	\begin{equation} \frac{1}{2} a b \leq \left( \frac{\lambda - \sqrt{a^2 + b^2}}{\sqrt{8}} \right)^2 \end{equation}
(2)	\begin{equation} \frac{1}{2} a b = \left( \frac{\lambda - \sqrt{a^2 + b^2}}{\sqrt{8}} \right)^2 \quad \iff \quad a = b \end{equation}

Proofs

Proof 0

Let $a, b, \lambda \in (0, \infty)$ each be a D5407: Positive real number such that

(i)	\begin{equation} a + b + \sqrt{a^2 + b^2} = \lambda \end{equation}

Since we have \begin{equation} \left( \frac{\lambda - \sqrt{a^2 + b^2}}{\sqrt{8}} \right)^2 = \frac{1}{2} \left( \frac{a + b}{2} \right)^2 \end{equation} this result is a direct corollary to R5209: Of all rectangles with a given perimeter, the square maximizes the area. $\square$