Name: Local Maxima and Minima of a Function
Uploaded: 2019-07-19
Duration: 14 min 17 s
Description: Do you know what is minima or maxima of a function? This video also explains how to find local maxima and minima by differentiation in a detailed manner.

Question

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

Accepted Answer

Let $h$ be the hypotenuse of the right-angled triangle and let $x$ be its altitude.Then,

Base of triangle $= \sqrt{h^{2} - x^{2}} . . . (1)$

Let $A$ be the area of triangle.

So, $A = \frac{1}{2} x \sqrt{h^{2} - x^{2}}$

$\Rightarrow \frac{d A}{d x} = \frac{1}{2} \{1 . \sqrt{h^{2} - x^{2}} + x \cdot \frac{1}{2} {(h^{2} - x^{2})}^{- \frac{1}{2}} \frac{d}{d x} (h^{2} - x^{2})\}$

$\Rightarrow \frac{d A}{d x} = \frac{1}{2} \{\sqrt{h^{2} - x^{2}} - \frac{x^{2}}{\sqrt{h^{2} - x^{2}}}\}$

$\Rightarrow \frac{d A}{d x} = \frac{1}{2} \{\frac{h^{2} - 2 x^{2}}{\sqrt{h^{2} - x^{2}}}\}$

For maximum or minimum we have, $\frac{d A}{d x} = 0$

$\Rightarrow \frac{1}{2} \{\frac{h^{2} - 2 x^{2}}{\sqrt{h^{2} - x^{2}}}\} = 0$

$\Rightarrow h^{2} = 2 x^{2}$

$\Rightarrow x = \frac{h}{\sqrt{2}}$

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 $x = \frac{h}{\sqrt{2}}$

So, base $= \sqrt{h^{2} - (\frac{h^{2}}{2})} = \frac{h}{\sqrt{2}}$ [Using $(1)$ ]

Hence, $A$ is maximum when triangle is isosceles.

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

Important Questions on Applications of Derivatives

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