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

Find the largest possible area of a right angled triangle whose hypotenuse is

5 cm

long.

Accepted Answer

Let the base of the right angled triangle be $x$ and its height be $y$ . Then,

$x^{2} + y^{2} = 5^{2}$

$\Rightarrow y^{2} = 25 - x^{2}$

$\Rightarrow y = \sqrt{25 - x^{2}}$

As, the area of the triangle, $A = \frac{1}{2} \times x \times y$

$\Rightarrow A (x) = \frac{1}{2} \times x \times \sqrt{25 - x^{2}}$

$\Rightarrow A (x) = \frac{x \sqrt{25 - x^{2}}}{2}$

$\Rightarrow A^{'} (x) = \frac{\sqrt{25 - x^{2}}}{2} + \frac{x (- 2 x)}{4 \sqrt{25 - x^{2}}}$

$\Rightarrow A^{'} (x) = \frac{\sqrt{25 - x^{2}}}{2} - \frac{x^{2}}{2 \sqrt{25 - x^{2}}}$

$\Rightarrow A^{'} (x) = \frac{25 - x^{2} - x^{2}}{2 \sqrt{25 - x^{2}}}$

$\Rightarrow A^{'} (x) = \frac{25 - 2 x^{2}}{2 \sqrt{25 - x^{2}}}$

For maxima or minima, we must have $f^{'} (x) = 0$

$A^{'} (x) = 0$

$\Rightarrow \frac{25 - 2 x^{2}}{2 \sqrt{25 - x^{2}}} = 0$

$\Rightarrow 25 - 2 x^{2} = 0$

$\Rightarrow 2 x^{2} = 25$

$\Rightarrow x = \frac{5}{\sqrt{2}}$

So, $y = \sqrt{25 - \frac{25}{2}}$

$= \sqrt{\frac{50 - 25}{2}}$

$= \sqrt{\frac{25}{2}}$

$= \frac{5}{\sqrt{2}}$

Also, $A^{''} (x) = \frac{[- 4 x \sqrt{25 - x^{2}} - \frac{(25 - 2 x^{2}) (- 2 x)}{2 \sqrt{25 - x^{2}}}]}{25 - x^{2}}$

$= \frac{[\frac{- 4 x (25 - x^{2}) + (25 x - 2 x^{3})}{\sqrt{25 - x^{2}}}]}{25 - x^{2}}$

$= \frac{- 100 x + 4 x^{3} + 25 x - 2 x^{3}}{(25 - x^{2}) \sqrt{25 - x^{2}}}$

$= \frac{- 75 x + 2 x^{3}}{(25 - x^{2}) \sqrt{25 - x^{2}}}$

$\Rightarrow A^{''} (\frac{5}{\sqrt{2}}) = \frac{- 75 (\frac{5}{\sqrt{2}}) + 2 {(\frac{5}{\sqrt{2}})}^{3}}{{(25 - {(\frac{5}{\sqrt{2}})}^{2})}^{\frac{3}{2}}} < 0$

So, $x = (\frac{5}{\sqrt{2}})$ is point of maxima.

$∴$ The largest possible area of the triangle $= \frac{1}{2} \times (\frac{5}{\sqrt{2}}) \times (\frac{5}{\sqrt{2}}) = \frac{25}{4} square units$ .

Find the largest possible area of a right angled triangle whose hypotenuse is $5 cm$ long.

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