Examples on Maxima and Minima Set 5

Go back to  'Applications of Derivatives'

Example – 29

A point P is given on the circumference of a circle of radius r. The chord QR is parallel to the tangent line at P. Find the maximum area of \(\Delta PQR\).

Solution: Observe that since QR is parallel to the tangent at P, the triangle PQR must be isosceles. This will become more clear upon carefully observing the following figure:

We can assume either \(\angle OQS\) or the length OS as the variable on which the area A will be dependent. We let the variable be\(\angle OQS = \theta \).

\[\begin{align} &\Rightarrow \qquad OS = r\sin \theta \\\\&\Rightarrow \qquad PS = r + r\sin \theta \\
\text{and}\;QR &= 2r cos \theta\\\\&\Rightarrow \qquad A = \frac{1}{2} \times QR \times PS\\\\ & \qquad  \qquad= {r^2}\cos \theta \left( {1 + \sin \theta } \right)\end{align}\]

\(\begin{align}\text{For maximum area,}  \qquad \frac{{dA}}{{d\theta }} &= 0\;\text{and}\;\frac{{{d^2}A}}{{d{\theta ^2}}} < 0\\\\
\frac{{dA}}{{d\theta }} &= {r^2}\left\{ { - \sin \theta \left( {1 + \sin \theta } \right) + \cos \theta .\cos \theta } \right\}\\\\ &= {r^2}\left\{ {{{\cos }^2}\theta  - {{\sin }^2}\theta  - \sin \theta } \right\}\\\\ &= {r^2}\left\{ {1 - \sin \theta  - 2{{\sin }^2}\theta } \right\}\\\\&= {r^2}\left( {1 + \sin \theta } \right)\left( {1 - 2\sin \theta } \right)\end{align}\)

\(\begin{align}&\text{This is 0 when} \qquad \qquad \sin \theta = \frac{1}{2} \Rightarrow \,\,\,\theta = \frac{\pi }{6}\\\\&\quad\qquad\qquad\qquad\qquad\text{or sin} \theta = \,–1 \qquad \text{(not a possible case)}\\&\text{Verify that}\; \frac{{{d^2}A}}{{d{\theta ^2}}} < 0 \;\; \text{for}\;\; \theta = \frac{\pi }{6}\\\\&\text{Therefore, area is maximum for}\;\theta = \frac{\pi }{6}.\\&\qquad \qquad \qquad {\left. A_\max= {r^2}\cos \theta \left( {1 + \sin \theta } \right) \right|_{\theta = \pi /6}}\\\\ &\qquad \qquad \qquad \qquad \;= \frac{{3\sqrt 3 \,{r^2}}}{4}\end{align}\)

Example - 30

Find the point on  the curve \(4{x^2} + {a^2}{y^2} = {\rm{ }}4{a^2},{\rm{ }}4{\rm{ }} < {a^2} < {\rm{ }}8\), that is farthest from the point \(\left( {0,{\rm{ }}-2} \right).\)

Solution: Upon rearrangement, the equation of the curve gives \(\begin{align}\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{4} = 1\end{align}\)  which is the equation of an ellipse. A general variable point P on this ellipse can be taken as (a cos\(\theta \), 2 sin\(\theta \)).

Let r represent the distance of P from (0, –2).

\[\begin{align}{r^2} &= {{\rm{ }}{{\left( {acos\theta {\rm{ }}-0} \right)}^2} + {\rm{ }}{{\left( {2{\rm{ }}sin\theta + {\rm{ }}2} \right)}^2}}\\\\&= {{a^2}co{s^2}\theta {\rm{ }} + {\rm{ }}4{\rm{ }}{{\left( {1{\rm{ }} + {\rm{ }}sin{\rm{ }}\theta } \right)}^2}} \end{align}\]

For \({r^2}\) to be maximum, \(\begin{align}\frac{{d\left( {{r^2}} \right)}}{{d\theta }} = 0\end{align}\)  and \(\begin{align}\frac{{{d^2}\left( {{r^2}} \right)}}{{d{\theta ^2}}} < 0\end{align}\)

\[\frac{{d\left( {{r^2}} \right)}}{{d\theta }} =  - 2{a^2}\sin \theta \cos \theta  + 8\left( {1 + \sin \theta } \right)\cos \theta \]

\[ = \cos \theta \left\{ {\left( {8 - 2{a^2}} \right)\sin \theta  + 8} \right\}\]

This is 0 when:

\[\begin{align}&\text{cos}\,\theta = 0 \qquad  \Rightarrow \qquad \theta  = \frac{\pi }{2}\\&\qquad \qquad \qquad\text{or}\\&(8{\rm{ }}-{\rm{ }}2{a^2}){\rm{ }}\,\text{sin}\;\theta = {\rm{ }}-{\rm{ }}8  \Rightarrow \,\,\sin \theta  = \frac{4}{{{a^2} - 4}}\end{align}\]

But \(\begin{align}\frac{4}{{{a^2} - 4}} > 1\end{align}\)  (verify) and sin\(\theta \)  cannot be greater than 1. Hence, this case does not given any valid value of  \(\theta \).

\[ \Rightarrow  \,\theta  = \frac{\pi }{2}\\\text{Verify that}\; {\left. {\frac{{{d^2}\left( {{r^2}} \right)}}{{d{\theta ^2}}}} \right|_{\theta  = \pi /2}} < 0\\ \Rightarrow  \theta  = \frac{\pi }{2} \text{ is a local maximum for } r^2\]

The required point is \(\begin{align}{\left. {\left( {a\cos \theta ,2\sin \theta } \right)} \right|_{\theta  = \pi /2}}\end{align}\) or \(\left( {0,{\rm{ }}2} \right).\)

TRY YOURSELF - IV

Q. 1     Find the minimum distance of a point on the curve \(\begin{align}\frac{{{a^2}}}{{{x^2}}} + \frac{{{b^2}}}{{{y^2}}} = 1\end{align}\) from the origin.

Q. 2     For a given curved surface area of a right circular cone, find the semi-vertical angle for which the volume of the cone is maximum.

Q. 3     Find the local maxima/minima values for the following funtions:

(a) \(\begin{align}\frac{{40}}{{3{x^4} + 8{x^3} - 18{x^2} + 60}}\end{align}\)

(b) \(\begin{align}\frac{x}{{1 + x\tan x}}\end{align}\)

(c) \(\begin{align}2x - {\tan ^{ - 1}}x - \log \left( {x + \sqrt {1 + {x^2}} } \right)\end{align}\)

(d) \(\begin{align}x{\left( {\log x} \right)^2}\end{align}\)

(e) \(\begin{align}{x^2}\log x\;{\rm{on}}\;[1,e]\end{align}\)

(h) \(\begin{align}\sqrt {\left( {1 - {x^2}} \right)\left( {2{x^2} + 1} \right)} \;{\rm{on}}\;\left[ { - 1,\;1} \right]\end{align}\)

Q. 4     A point P is given on the circumference of a circle of radius r. The chord QR is parallel to the tangent at P. Find the maximum area of \(\Delta PQR\)

Q. 5     A chord of length 2l divides a circular area of radius r into two segments. Find the sides of the rectangle with the largest area that can be inscribed in the smaller segment.

Download SOLVED Practice Questions of Examples on Maxima and Minima Set 5 for FREE
Applications of Derivatives
grade 11 | Questions Set 1
Applications of Derivatives
grade 11 | Answers Set 1
Applications of Derivatives
grade 11 | Questions Set 2
Applications of Derivatives
grade 11 | Answers Set 2