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NDA Math Application of Derivatives

Consider the following information for the next three (03) items that follow:

Three sides of a trapezium are each equal to

6 \text{cm}

. Let

\alpha \in \left(0, \frac{\pi}{2}\right)

be the angle between a pair of adjacent sides.

[NDA I 2018]

Question : 39 of 73

Marks: +1, -0

\alpha

Solution:

A=Area

= \; \; \frac{1}{2}(6+6+2x)\sqrt{36-x^{2}}

= (6+x)\sqrt{36-x^{2}}

\frac{dA}{dx} = (6+x)\left( \; \; \frac{-2x}{2\sqrt{36-x^{2}}} \right) + \sqrt{36-x^{2}}

= \sqrt{36-x^{2}} - \; \; \frac{x(6+x)}{\sqrt{36-x^{2}}} = \; \; \frac{36-6x-2x^{2}}{\sqrt{36-x^{2}}}

\frac{dA}{dx} = 0 \Rightarrow 36-6x-2x^{2}=0 \Rightarrow x=3

\frac{d^{2}A}{dx^{2}} < 0

Now ,

\cos \; \alpha = \frac{x}{6} = \frac{3}{6} = \frac{1}{2} \Rightarrow \alpha = 3

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