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Question

Mathematics Question on Maxima and Minima

The maximum area of a right angled triangle with hypotenuse hh is :

A

h222\frac{h^{2}}{2\sqrt{2}}

B

h22\frac{h^{2}}{2}

C

h22\frac{h^{2}}{\sqrt{2}}

D

h24\frac{h^{2}}{4}

Answer

h24\frac{h^{2}}{4}

Explanation

Solution

Let base =b= b Altitude (or perpendicular) p→(q → p)p \rightarrow \left(q\,\rightarrow \,p\right) =h2−b2=\sqrt{h^{2}-b^{2}} Area, A=12×base×A=\frac{1}{2}\times base \times altitude =12×b×h2−b2=\frac{1}{2}\times b\times\sqrt{h^{2}-b^{2}} ⇒dAdb=12[h2−b2+b.−2b2h2−b2]\Rightarrow \frac{dA}{db}=\frac{1}{2}\left[\sqrt{h^{2}-b^{2}}+b. \frac{-2b}{2\sqrt{h^{2}-b^{2}}}\right] =12[h2−2b2h2−b2]=\frac{1}{2}\left[\frac{h^{2}-2b^{2}}{\sqrt{h^{2}-b^{2}}}\right] Put dAdb=0,⇒b=h2\frac{dA}{db}=0, \Rightarrow b=\frac{h}{\sqrt{2}} Maximum area =12×h2×h2−h22=h24=\frac{1}{2}\times\frac{h}{\sqrt{2}}\times\sqrt{h^{2}-\frac{h^{2}}{2}=\frac{h^{2}}{4}}