04 Largest Right Triangle of Given Hypotenuse

$b = \sqrt{c^2 - a^2}$
 

$A = \frac{1}{2}ab$

$A = \frac{1}{2}a\sqrt{c^2 - a^2}$

$\dfrac{dA}{da} = \dfrac{1}{2}\left( a \times \dfrac{-2a}{2\sqrt{c^2 - a^2}} + \sqrt{c^2 - a^2} \right) = 0$

$-\dfrac{a^2}{\sqrt{c^2 - a^2}} + \sqrt{c^2 - a^2} = 0$

$\sqrt{c^2 - a^2} = \dfrac{a^2}{\sqrt{c^2 - a^2}}$

$c^2 - a^2 = a^2$

$2a^2 = c^2$

$a^2 = \frac{1}{2}c^2$

$a = \frac{1}{\sqrt{2}}c$
 

$b = \sqrt{c^2 - \frac{1}{2}c^2}$

$b = \sqrt{\frac{1}{2}c^2}$

$b = \frac{1}{\sqrt{2}}c$
 

$A = \frac{1}{2}ab$

$A_{max} = \frac{1}{2}\left( \frac{1}{\sqrt{2}}c \right) \left( \frac{1}{\sqrt{2}}c \right)$

$A_{max} = \frac{1}{4}c^2$           answer

$b = c \cos \theta$
 

$A = \frac{1}{2}ab$

$A = \frac{1}{2}(c \sin \theta)(c \cos \theta)$

$A = \frac{1}{2}c^2 \sin \theta \, \cos \theta$

$A = \frac{1}{4}c^2 (2\sin \theta \, \cos \theta)$

$A = \frac{1}{4}c^2 \sin 2\theta$

$\dfrac{dA}{d\theta} = \frac{1}{4}c^2 (\cos 2\theta)(2) = 0$

$\cos 2\theta = 0$

$2\theta = 90^\circ$

$\theta = 45^\circ$
 

$A = \frac{1}{4}c^2 \sin 2\theta$

$A_{max} = \frac{1}{4}c^2 \sin 2(45^{\circ})$

$A_{max} = \frac{1}{4}c^2 \sin 90^\circ$

$A_{max} = \frac{1}{4}c^2$           answer