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 least perimeter of an isosceles triangle in which a circle of radius $r$ can be inscribed is

6 \sqrt{3} r

.

Accepted Answer

Gvien radius of Inscribed circle is $6 \sqrt{3} r$
Let $∠ F A O = ∠ E A O = θ$

Now as we know that tangent and normal to a circle are perpendicular to each other

$\Rightarrow O F ⊥ A B, O D ⊥ B C, O E ⊥ A C$

From $∆ A O F$

$\tan θ = \frac{r}{A F} \Rightarrow A F = r \cot θ$

Similarly $A E = r \cot θ$

From $∆ A B D \Rightarrow ∠ A B D = 90 - θ = \frac{π}{2} - θ$

$O B$ bisects $∠ A B D \Rightarrow ∠ O B D = \frac{π}{4} - \frac{θ}{2}$

From $∆ B O D \Rightarrow \tan (\frac{π}{4} - \frac{θ}{2}) = \frac{r}{B D} \Rightarrow B D = r \cot (\frac{π}{4} - \frac{θ}{2})$

Similary $C D = r \cot (\frac{π}{4} - \frac{θ}{2})$

Now as we know that length of tangents to circle from a point are of equal length

$B D = B F = C D = C E = r \cot (\frac{π}{4} - \frac{θ}{2})$

Perimeter of the triangle is given by $P = A F + F B + B D + D C + C E + A E$

$\Rightarrow P = 2 r \cot θ + 4 r \cot (\frac{π}{4} - \frac{θ}{2})$

For the perimeter to be minimum $\frac{d P}{d θ} = 0$

$\Rightarrow \frac{d (2 r \cot θ + 4 r \cot (\frac{π}{4} - \frac{θ}{2}))}{d θ} = 0$

$\Rightarrow 2 r \frac{d (\cot θ)}{d θ} + 4 r \frac{d (\cot (\frac{π}{4} - \frac{θ}{2}))}{d θ} = 0$

We know that $\frac{d (\cot x)}{d x} = - {cosec}^{2} x$

$\Rightarrow 2 r (- {cosec}^{2} θ) + 4 r (- {cosec}^{2} (\frac{π}{4} - \frac{θ}{2})) (- \frac{1}{2}) = 0 \Rightarrow - 2 r ({cosec}^{2} θ) + 2 r ({cosec}^{2} (\frac{π}{4} - \frac{θ}{2})) = 0 \Rightarrow {cosec}^{2} θ = {cosec}^{2} (\frac{π}{4} - \frac{θ}{2}) \Rightarrow \sin^{2} θ = \sin^{2} (\frac{π}{4} - \frac{θ}{2}) \Rightarrow \sin θ = \pm \sin (\frac{π}{4} - \frac{θ}{2})$

But As $0 < θ < \frac{π}{2} \Rightarrow \sin (\frac{π}{4} - \frac{θ}{2}) \neq - \sin θ$

$∴ \sin θ = \sin (\frac{π}{4} - \frac{θ}{2})$

$\Rightarrow θ = (\frac{π}{4} - \frac{θ}{2}) \Rightarrow \frac{3 θ}{2} = \frac{π}{4} \Rightarrow θ = \frac{π}{6}$

Now finding second derviative

$\frac{d^{2} P}{d θ^{2}} = \frac{d (\frac{d P}{d θ})}{d θ} = \frac{d (- 2 r ({cosec}^{2} θ) + 2 r ({cosec}^{2} (\frac{π}{4} - \frac{θ}{2})))}{d θ} \Rightarrow \frac{d^{2} P}{d θ^{2}} = 2 r \frac{d (- {cosec}^{2} θ)}{d θ} + 2 r \frac{d ({cosec}^{2} (\frac{π}{4} - \frac{θ}{2}))}{d θ} \Rightarrow \frac{d^{2} P}{d θ^{2}} = 2 r (- 2 cosec θ (- cosec θ \cot θ)) + 2 r (2 cosec (\frac{π}{4} - \frac{θ}{2}) (- cosec (\frac{π}{4} - \frac{θ}{2}) \cot (\frac{π}{4} - \frac{θ}{2})) (- \frac{1}{2}))$

$\Rightarrow \frac{d^{2} P}{d θ^{2}} = 4 r {cosec}^{2} θ \cot θ - 2 r {cosec}^{2} (\frac{π}{4} - \frac{θ}{2}) \cot (\frac{π}{4} - \frac{θ}{2})$

Value of $\frac{d^{2} P}{d θ^{2}} > 0$ at $θ = \frac{π}{6}$

Therefore $P$ is minimum at $θ = \frac{π}{6}$

Hence Minimum perimeter $P = 2 r \cot \frac{π}{6} + 4 r \cot (\frac{π}{4} - \frac{π}{12}) \Rightarrow P = 2 r \cot \frac{π}{6} + 4 r \cot (\frac{π}{6}) \Rightarrow P = 6 r \cot \frac{π}{6} \Rightarrow P = 6 r (\sqrt{3})$

Therefore the least perimeter of an isosceles triangle in which a circle of radius $r$ can be inscribed is $6 \sqrt{3} r$

Prove that the least perimeter of an isosceles triangle in which a circle of radius $r$ can be inscribed is $6 \sqrt{3} r$ .

Important Questions on Maxima and Minima

Learn and Prepare for any exam you want !

Prove that the least perimeter of an isosceles triangle in which a circle of radius r can be inscribed is 63r.

Important Questions on Maxima and Minima

Learn and Prepare for any exam you want !

Prove that the least perimeter of an isosceles triangle in which a circle of radius $r$ can be inscribed is $6 \sqrt{3} r$ .