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Examples on Maxima and Minima Set 4

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Example - 27

 Find the area of the greatest isosceles triangle that can be inscribed in a given ellipse having its vertex coincident with one end of the major axis.

Solution: Assuming the equation of the ellipse to be \(\begin{align}\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1,\end{align}\)  let one vertex of the isosceles triangle be coincident with \(\left( -a,\text{ }0 \right)\). The other two vertices are variable (though related to each other as mirror reflections).

The area of  \(\Delta PQR\) is

\[\begin{align}A &= \frac{1}{2} \times QR \times PS\\\\&= \frac{1}{2} \times 2b\sin \theta  \times \left( {a + a\cos \theta } \right)\\\\&= ab\sin \theta \left( {1 + \cos \theta } \right)\end{align}\]

For maximum area,

\[\begin{align}\frac{{dA}}{{d\theta }}&= 0\,\,{\rm{and }}\frac{{{d^2}A}}{{d{\theta ^2}}} < 0\\\\
\Rightarrow \qquad \frac{{dA}}{{d\theta }}&= ab\left\{ {\cos \theta \left( {1 + \cos \theta } \right) - \sin \theta  \cdot \sin \theta } \right\}\\\\&= ab\left\{ {\cos \theta  + {{\cos }^2}\theta  - {{\sin }^2}\theta } \right\}\\\\&= ab\left\{ {\cos \theta  + {{\cos }^2}\theta  - 1 + {{\cos }^2}\theta } \right\}\\\\&= ab\left\{ {2{{\cos }^2}\theta  + \cos \theta  - 1} \right\}\\\\&= ab\left( {2\cos \theta  - 1} \right)\left( {\cos \theta  + 1} \right)\end{align}\]

\(\begin{align}\text{This is 0 when} & \qquad \qquad \cos \theta = \frac{1}{2}\quad\; \Rightarrow \quad \theta = \frac{\pi }{3}\\\text{or} & \qquad \qquad \cos \theta = - 1\quad \Rightarrow \quad \theta = \pi \qquad\text{(obviously a non-valid solution)}\\\end{align}\)

Verify that \(\begin{align}{\left. {\frac{{{d^2}A}}{{d{\theta ^2}}}} \right|_{\theta  = \pi /3}} < 0\end{align}\)  so that \(\begin{align}\theta  = \frac{\pi }{3}\end{align}\)   is a point of local maximum for A.

\[\begin{align}A_\max &= ab\sin \theta \left( {1 + \cos \theta } \right) |_{\theta  = \pi /3}\\ &= ab \times \frac{{\sqrt 3 }}{2} \times \frac{3}{2}\\&= \frac{{3\sqrt 3 ab}}{4}\end{align}\]

Example – 28

Find the points on the curve \(a{x^2} + {\rm{ }}2bxy + a{y^2} = c, \;0 < \,a\, <\, b\, < c, \) whose distance from the origin is minimum.

Solution: In some cases, the form of a variable point  on a given curve is obvious from the equation of the curve. For example, we can take a variable point on \(\begin{align}\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\end{align}\)  as ( a cos \(\theta \), b sin\(\theta \)) and on    \(y = {x^2}\)  as  \(\left( {t,{t^2}} \right)\) and so on.

However, if this form is not obvious from the curve, we can take it to be \(({\text{r cos}}\;\theta ,\,\text{r sin}\;\theta )\) and make this point (which could represent any point on the plane) satisfy the equation of the given curve. We will do this for the current example.

Let (r cos \(\theta \), r sin \(\theta \)) be a point on the given curve. The co-ordinates therefore must satisfy the equation of the curve:

\[\begin{align}& a{\left( {rcos{\rm{ }}\theta } \right)^2} + {\rm{ }}2b\left( {rcos{\rm{ }}\theta } \right)\left( {rsin{\rm{ }}\theta } \right){\rm{ }} + a{\left( {rsin{\rm{ }}\theta } \right)^2} = c\\\\ &\Rightarrow  \qquad ar{^2} + b{r^2}sin{\rm{ }}2\theta {\rm{ }} = c\\\\ &\Rightarrow   \qquad{r^2} = \frac{c}{{a + b\sin 2\theta }}\qquad  \qquad \qquad \qquad \rm{(i)} \end{align}\]

Any point on the curve must satisfy (i). From (i), it is immediately obvious that \({r^2}\) has a minimum value of \(\begin{align}\frac{c}{{a + b}}\end{align}\)  (when \({\rm{sin}}2\theta  = 1\)). We were able to obtain the answer without differentiation.

\[\begin{align}\text{sin 2}\theta &=1\\\\\Rightarrow \qquad 2\theta &= \frac{\pi }{2},\frac{{5\pi }}{2}\\\\\Rightarrow \qquad \,\theta &= \frac{\pi }{4},\frac{{5\pi }}{4} \qquad ( \theta \;\text{must lie in} [0, 2\pi ] \text{so we obtain only two possible values)}\\\\{r_{\min }} &= \sqrt {\frac{c}{{a + b}}} \end{align}\]

For \(\begin{align}\theta  = \frac{\pi }{4},\end{align}\)  the required point is \((\text{r cos}\,\theta, \text{r sin}\,\theta )\) or \(\begin{align}\left( {\sqrt {\frac{c}{{2\left( {a + b} \right)}},} \,\,\sqrt {\frac{c}{{2\left( {a + b} \right)}}} } \right)\end{align}\)

For  \(\begin{align}\theta  = \frac{{5\pi }}{4}\end{align},\) the required point is \((\text{r cos}\;\theta, \text{r sin}\,\theta)\) or \(\begin{align}\left( { - \sqrt {\frac{c}{{2\left( {a + b} \right)}}} , - \sqrt {\frac{c}{{2\left( {a + b} \right)}}} } \right)\end{align}\)

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