Prove that the area of a right angled triangle of given hypotenuse is maximum when the triangle is isosceles - Maths - Application of Derivatives

Question

Prove that the area of a right angled triangle of given
hypotenuse is maximum when the
triangle is isosceles

Harleen Birdi · Accepted Answer

Let h be the hypotenuse of the right-angles triangle, and let x be it altitude Then, Base of the triangle $= \sqrt{h^{2} - x^{2}}$ Let A be the area of the triangle Then, $A = \frac{1}{2} x \sqrt{h^{2} - x^{2}} \Rightarrow \frac{d A}{d x} = \frac{1}{2} {1 \sqrt{h^{2} - x^{2}} + x \frac{1}{2} (h^{2} - x^{2})^{- \frac{1}{2}} \frac{d}{d x} (h^{2} - x^{2})} \Rightarrow \frac{d A}{d x} = \frac{1}{2} {\sqrt{h^{2} - x^{2}} - \frac{x^{2}}{\sqrt{h^{2} - x^{2}}}} = \frac{1}{2} {\frac{h^{2} - 2 x^{2}}{\sqrt{h^{2} - x^{2}}}}$

For maximum or minimum, we have $\frac{d A}{d x^{2}} = 0 \Rightarrow \frac{1}{2} {\frac{h^{2} - 2 x^{2}}{\sqrt{h^{2} - x^{2}}}} = 0 \Rightarrow = 2 x^{2} \Rightarrow x = \frac{h}{\sqrt{2}}$ Now, $\frac{d^{2} A}{d x^{2}} = \frac{1}{2} {(- 4 x) \frac{1}{\sqrt{h^{2} - x^{2}}} + (h^{2} - 2 x^{2}) (\frac{1}{2}) (h^{2} - x^{2})^{- \frac{3}{2}} \frac{d}{d x} (h^{2} - x^{2})} \Rightarrow \frac{d^{2} A}{d x^{2}} = \frac{1}{2} {\frac{- 4 x}{\sqrt{h^{2} - x^{2}}} + \frac{x (h^{2} - 2 x^{2})}{(h^{2} - x^{2})^{\frac{3}{2}}}} \Rightarrow (\frac{d^{2} A}{d x^{2}})_{x = \frac{h}{\sqrt{2}}} = - 2 < 0$ Thus, A is maximum when $x = \frac{h}{\sqrt{2}}$ $∴ Base = \sqrt{h^{2} - (\frac{h^{2}}{2})} = \frac{h}{\sqrt{2}}$ Hence, A maximum when the triangle is isosceles