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

A right-angled triangle with constant area is given. Prove that the hypotenuse of the triangle is least when the triangle is isosceles.

Accepted Answer

Let $∆ A B C$ be right-angle triangle with constant area $P$ . Let $l$ be the length of the hypotenuse of $∆ A B C$ .

Therefore $A B = l \sin θ a n d B C = l \cos θ . . . . . . . . . . . (i)$

Also, $∠ A C B = θ$ .

$∴ area of triangle ∆ A B C = \frac{1}{2} \times B C \times A B$

$\Rightarrow A = \frac{1}{2} (l \cos θ) (l \sin θ) = \frac{l^{2}}{4} (\sin 2 θ) \Rightarrow l^{2} = 4 P (\cos e c 2 θ)$

Now, to prove that the hypotenuse of the triangle is least when the triangle is isosceles is, we need to consider product of sides as function.

Now, $l$ is minimise when $l^{2}$ is minimise, i.e. we want to find the value of $θ$ for which $l^{2}$ is least.

Let, $L = l^{2} = 4 P \cos e c 2 θ$

On differentiating both the sides w.r.t. $θ$ , we get

$\frac{d L}{d θ} = - 8 P cosec 2 θ \cot 2 θ$

and $\frac{d^{2} L}{d θ^{2}} = - 8 P [- 2 {cosec}^{3} 2 θ + \cot 2 θ (- 2 cosec 2 θ \cot 2 θ)]$ [using product rule]

$\frac{d^{2} L}{d θ^{2}} = 16 P [{cosec}^{3} 2 θ + cosec 2 θ \cot^{2} 2 θ]$

Now, For maxima/minima, $\frac{d L}{d θ} = 0$

$\Rightarrow - 8 P cosec 2 θ \cot 2 θ = 0 \Rightarrow cosec 2 θ \cot 2 θ = 0 \Rightarrow \frac{1}{\sin 2 θ} \frac{\cos 2 θ}{\sin 2 θ} = 0 \Rightarrow \frac{\cos 2 θ}{\sin^{2} 2 θ} = 0 \Rightarrow \cos 2 θ = 0 \Rightarrow 2 θ = \frac{π}{2} \Rightarrow θ = \frac{π}{4} (∵ θ \in 0, \frac{π}{2})$

Now, at $θ = \frac{π}{4}$ ;

$\frac{d^{2} L}{d θ^{2}} = 16 P [{cosec}^{3} 2 \frac{π}{4} + cosec 2 \frac{π}{4} \cot^{2} 2 \frac{π}{4}] = 16 P [1^{3} + 1 \times 0] = 16 P (∵ 16 P > 0)$

Thus, $θ = \frac{π}{4}$ is the point of minima.

Therefore, $L$ is minimum when $θ = \frac{π}{4}$ .

Therefore from (i), $A B = l \sin θ = l \sin \frac{π}{4} = \frac{l}{\sqrt{2}} a n d B C = l \cos θ = l \cos \frac{π}{4} = \frac{l}{\sqrt{2}}$ .

$∴ A B = B C$

Hence,the hypotenuse of the triangle is least when the triangle is isosceles . Therefore $∆ A B C$ is an isosceles triangle.

A right-angled triangle with constant area is given. Prove that the hypotenuse of the triangle is least when the triangle is isosceles.

Important Questions on Applications of Derivatives

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