# Miscellaneous Exercise Application of Derivatives Solution - NCERT Class 12

## Chapter 6 Ex.6.ME Question 1

Using differentials, find the approximate value of each of the following.

(i) \begin{align}{\left( {\frac{{17}}{{81}}} \right)^{\frac{1}{4}}} \end{align}

(ii) \begin{align}{\left( {33} \right)^{ - \frac{1}{5}}}\end{align}

### Solution

(i) \begin{align}{\left( {\frac{{17}}{{81}}} \right)^{\frac{1}{4}}}\end{align}

Consider $$y = {\left( x \right)^{\frac{1}{4}}}$$

Let $$x = \frac{{16}}{{81}}$$ and $$\Delta x = \frac{1}{{81}}$$

Then,

\begin{align} \Delta y &= {\left( {x + \Delta x} \right)^{\frac{1}{4}}} - {\left( x \right)^{\frac{1}{4}}}\\ &= {\left( {\frac{{17}}{{81}}} \right)^{\frac{1}{4}}} - {\left( {\frac{{16}}{{81}}} \right)^{\frac{1}{4}}}\\& = {\left( {\frac{{17}}{{81}}} \right)^{\frac{1}{4}}} - \frac{2}{3}\\ \frac{2}{3} + \Delta y &= {\left( {\frac{{17}}{{81}}} \right)^{\frac{1}{4}}} \end{align}

Now, $$dy$$ is approximately equal to $$\Delta y$$ and is given by,

\begin{align} dy &= \left( {\frac{{dy}}{{dx}}} \right)\Delta x \\ &= \frac{1}{{4{{\left( x \right)}^{\frac{3}{4}}}}}\left( {\Delta x} \right) \qquad \left[ {\because y = {{\left( x \right)}^{\frac{1}{4}}}} \right] \\ & = \frac{1}{{4{{\left( {\frac{{16}}{{81}}} \right)}^{\frac{3}{4}}}}}\left( {\frac{1}{{81}}} \right) \\ & = \frac{{27}}{{4 \times 8}} \times \frac{1}{{81}} \\ & = \frac{1}{{32 \times 3}} \\ &= \frac{1}{{96}} \\ & = 0.010 \\ \end{align}

Hence,

\begin{align} {\left( {\frac{{17}}{{81}}} \right)^{\frac{1}{4}}} &= \frac{2}{3} + 0.010\\ & = 0.667 + 0.010\\ &= 0.677 \end{align}

Thus, the approximate value of$${\left( {\frac{{17}}{{81}}} \right)^{\frac{1}{4}}} = 0.677$$.

(ii) \begin{align}{\left( {33} \right)^{ - \frac{1}{5}}} \end{align}

Consider $$y = {\left( x \right)^{ - \frac{1}{5}}}$$

Let $$x = 32$$ and $$\Delta x = 1$$

Then,

\begin{align}\Delta y &= {\left( {x + \Delta x} \right)^{ - \frac{1}{5}}} - {\left( x \right)^{ - \frac{1}{5}}}\\ &= {\left( {33} \right)^{ - \frac{1}{5}}} - {\left( {32} \right)^{ - \frac{1}{5}}}\\ &= {\left( {33} \right)^{ - \frac{1}{5}}} - \frac{1}{2}\\\frac{1}{2} + \Delta y &= {\left( {33} \right)^{ - \frac{1}{5}}}\end{align}

Now, $$dy$$ is approximately equal to $$\Delta y$$ and is given by,

\begin{align} dy &= \left( {\frac{{dy}}{{dx}}} \right)\Delta x \\ & = \frac{{ - 1}}{{5{{\left( x \right)}^{\frac{6}{5}}}}}\left( {\Delta x} \right) \qquad \left[ {\because y = {{\left( x \right)}^{ - \frac{1}{5}}}} \right] \\& = - \frac{1}{{5{{\left( 2 \right)}^6}}}\left( 1 \right) \\ &= - \frac{1}{{320}} \\ &= - 0.003 \\ \end{align}

Hence,

\begin{align}{\left( {33} \right)^{ - \frac{1}{5}}} &= \frac{1}{2} + \left( { - 0.003} \right)\\ &= 0.5 - 0.003\\ &= 0.497\end{align}

Thus, the approximate value of $${\left( {33} \right)^{ - \frac{1}{5}}} = 0.497$$.

## Chapter 6 Ex.6.ME Question 2

Show that the function given by $$f\left( x \right) = \frac{{\log x}}{x}$$ has maximum at $$x = e$$.

### Solution

The given function is $$f\left( x \right) = \frac{{\log x}}{x}$$

Therefore,

\begin{align}f'\left( x \right) &= \frac{{x\left( {\frac{1}{x}} \right) - \log x}}{{{x^2}}}\\ &= \frac{{1 - \log x}}{{{x^2}}}\end{align}

Now,

\begin{align}&f'\left( x \right) = 0\\& \Rightarrow 1 - \log x = 0\\ &\Rightarrow \log x = 1\\ &\Rightarrow \log x = \log e\\ &\Rightarrow x = e\end{align}

Also,

\begin{align}f''\left( x \right)& = \frac{{{x^2}\left( { - \frac{1}{x}} \right) - \left( {1 - \log x} \right)\left( {2x} \right)}}{{{x^4}}}\\ &= \frac{{ - x - 2x\left( {1 - \log x} \right)}}{{{x^4}}}\\& = \frac{{ - 3 + 2\log x}}{{{x^3}}}\end{align}

Now,

\begin{align}f''\left( e \right) &= \frac{{ - 3 + 2\log e}}{{{e^3}}}\\ &= \frac{{ - 3 + 2}}{{{e^3}}}\\ &= \frac{{ - 1}}{{{e^3}}} < 0\end{align}

Therefore, by second derivative test, $$f$$ is the maximum at $$x = e$$.

## Chapter 6 Ex.6.ME Question 3

The two equal sides of an isosceles triangle with fixed base $$b$$ are decreasing at the rate of $$3 \,\rm{cm}$$ per second. How fast is the area decreasing when the two equal sides are equal to the base?

### Solution

Let $$\Delta ABC$$ be isosceles where $$BC$$ is the base of fixed length $$b$$.

Also, let the length of the two equal sides of  $$\Delta ABC$$ be $$a$$.

Draw $$AD \bot BC$$.

Now, in $$\Delta ADC$$, by applying the Pythagoras theorem, we have:

$AD = \sqrt {{a^2} - \frac{{{b^2}}}{4}}$

Area of triangle,

$A = \frac{1}{2}b\sqrt {{a^2} - \frac{{{b^2}}}{4}}$

The rate of change of the area with respect to time $$\left( t \right)$$ is given by,

\begin{align}\frac{{dA}}{{dt}} &= \frac{1}{2}b.\frac{{2a}}{{2\sqrt {{a^2} - \frac{{{b^2}}}{4}} }}\frac{{da}}{{dt}}\\ &= \frac{{ab}}{{\sqrt {4{a^2} - {b^2}} }}\frac{{da}}{{dt}}\end{align}

It is given that the two equal sides of the triangle are decreasing at the rate of $$3 \rm{cm}$$

Therefore,

$\frac{{da}}{{dt}} = - 3\,\rm{cm/s}$

Hence,

$\Rightarrow \frac{{dA}}{{dt}} = \frac{{ - 3ab}}{{\sqrt {4{a^2} - {b^2}} }}$

When, $$a = b$$ we have:

\begin{align}\frac{{dA}}{{dt}} &= \frac{{ - 3{b^2}}}{{\sqrt {4{a^2} - {b^2}} }}\\ &= \frac{{ - 3{b^2}}}{{\sqrt {3{b^2}} }}\\ &= - \sqrt 3 b\end{align}

Hence, if the two equal sides are equal to the base, then the area of the triangle is decreasing at the rate of $$- \sqrt 3 bc{m^2}/s$$.

## Chapter 6 Ex.6.ME Question 4

Find the equation of the normal to curve $${y^2} = 4x$$ at the point $$\left( {{\rm{1}},{\rm{2}}} \right)$$.

### Solution

The equation of the given curve is $${y^2} = 4x$$

Differentiating with respect to $$x$$, we have:

\begin{align}&2y\frac{{dy}}{{dx}} = 4\\& \Rightarrow \frac{{dy}}{{dx}} = \frac{4}{{2y}}\\& \Rightarrow \frac{{dy}}{{dx}} = \frac{2}{y}\\&{\left. { \Rightarrow \frac{{dy}}{{dx}}} \right]_{\left( {1,2} \right)}} = \frac{2}{2} = 1\end{align}

Now, the slope of the normal at point $$\left( {{\rm{1}},{\rm{2}}} \right)$$ is

$\frac{{ - 1}}{{{{\left. {\frac{{dy}}{{dx}}} \right]}_{\left( {1,2} \right)}}}} = \frac{{ - 1}}{1} = - 1$

Equation of the normal at $$\left( {{\rm{1}},{\rm{2}}} \right)$$ is

\begin{align} &\Rightarrow y - 2 = - x + 1\\ &\Rightarrow x + y - 3 = 0\end{align}

## Chapter 6 Ex.6.ME Question 5

Show that the normal at any point θ to the curve $$x = a\cos \theta + a\theta \sin \theta$$, $$y = a\sin \theta - a\theta \cos \theta$$ is at a constant distance from the origin.

### Solution

We have $$x = a\cos \theta + a\theta \sin \theta$$

Therefore,

\begin{align}\frac{{dx}}{{d\theta }}& = - a\sin \theta + a\sin \theta + a\theta \cos \theta \\ &= a\theta \cos \theta \end{align}

Also, $$y = a\sin \theta - a\theta \cos \theta$$

Hence,

\begin{align}\frac{{dy}}{{d\theta }}& = a\cos \theta - a\cos \theta + a\theta \cos \theta \\& = a\theta \cos \theta \end{align}

Thus,

\begin{align}\frac{{dy}}{{dx}} &= \frac{{dy}}{{d\theta }}.\frac{{d\theta }}{{dx}}\\& = \frac{{a\theta \sin \theta }}{{a\theta \cos \theta }}\\& = \tan \theta \end{align}

Slope of the normal at any point $$θ$$ is $$\frac{{ - 1}}{{\tan \theta }}$$.

The equation of the normal at a given point $$\left( {x,y} \right)$$ is given by,

\begin{align}&y - a\sin \theta + a\theta \cos \theta = \frac{{ - 1}}{{\tan \theta }}\left( {x - a\cos \theta - a\theta \sin \theta } \right)\\ &\Rightarrow y\sin \theta - a{\sin ^2}\theta + a\theta \sin \theta \cos \theta = - x\cos \theta + a{\cos ^2}\theta + a\theta \sin \theta \cos \theta \\ &\Rightarrow x\cos \theta + y\sin \theta - a\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right) = 0\\& \Rightarrow x\cos \theta + y\sin \theta - a = 0\end{align}

Now, the perpendicular distance of the normal from the origin is

$$\frac{{\left| { - a} \right|}}{{\sqrt {{{\cos }^2}\theta + {{\sin }^2}\theta } }} = \frac{{\left| { - a} \right|}}{{\sqrt 1 }} = \left| { - a} \right|$$, which is independent of $$θ$$

Hence, the perpendicular distance of the normal from the origin is constant.

## Chapter 6 Ex.6.ME Question 6

Find the intervals in which the function $$f$$ given by \begin{align}f\left( x \right) = \frac{{4\sin x - 2x - x\cos x}}{{2 + \cos x}} \end{align} is

(i) Increasing

(ii) Decreasing

### Solution

We have \begin{align}f\left( x \right) = \frac{{4\sin x - 2x - x\cos x}}{{2 + \cos x}}\end{align}

Hence,

\begin{align}f'\left( x \right) &= \frac{{\left( {2 + \cos x} \right)\left( {4\cos x - 2 - \cos x + x\sin x} \right) - \left( {4\sin x - 2x - x\cos x} \right)\left( { - \sin x} \right)}}{{{{\left( {2 + \cos x} \right)}^2}}}\\ &= \frac{{6\cos x - 4 + 2x\sin x + 3{{\cos }^2}x - 2\cos x + x\sin x\cos x + 4{{\sin }^2}x - 2x\sin x - x\sin x\cos x}}{{{{\left( {2 + \cos x} \right)}^2}}}\\& = \frac{{4\cos x - 4 + 3{{\cos }^2}x + 4{{\sin }^2}x}}{{{{\left( {2 + \cos x} \right)}^2}}}\\& = \frac{{4\cos x - 4 + 3{{\cos }^2}x + 4 - 4{{\cos }^2}x}}{{{{\left( {2 + \cos x} \right)}^2}}}\\& = \frac{{4\cos x - {{\cos }^2}x}}{{{{\left( {2 + \cos x} \right)}^2}}}\end{align}

$$f'\left( x \right) = \frac{{\cos x\left( {4 - \cos x} \right)}}{{{{\left( {2 + \cos x} \right)}^2}}}$$

Now,

\begin{align}&f'\left( x \right) = 0\\& \Rightarrow \cos x = 0{\rm{ or }}\cos x = 4\end{align}

But $$\cos x \ne 4$$

Hence,

\begin{align}&\cos x = 0\\ &\Rightarrow x = \frac{\pi }{2},\frac{{3\pi }}{2}\end{align}

Now, $$x = \frac{\pi }{2}$$ and $$x = \frac{{3\pi }}{2}$$ divide $$\left( {0,2\pi } \right)$$ into three disjoint intervals i.e., $$\left( {0,\frac{\pi }{2}} \right)$$, $$\left( {\frac{\pi }{2},\frac{{3\pi }}{2}} \right)$$ and $$\left( {\frac{{3\pi }}{2},2\pi } \right)$$.

In intervals, $$\left( {0,\frac{\pi }{2}} \right)$$ and $$\left( {\frac{{3\pi }}{2},2\pi } \right)$$, $$f'\left( x \right) > 0$$

Thus, $$f\left( x \right)$$ is increasing for $$0 < x < \frac{x}{2}$$ and $$\frac{{3x}}{2} < x < 2\pi$$.

In the interval $$\left( {\frac{\pi }{2},\frac{{3\pi }}{2}} \right)$$, $$f'\left( x \right) < 0$$

Thus, $$f\left( x \right)$$ is decreasing for $$\frac{\pi }{2} < x < \frac{{3\pi }}{2}$$.

## Chapter 6 Ex.6.ME Question 7

Find the intervals in which the function $$f$$ given by \begin{align}f\left( x \right) = {x^3} + \frac{1}{{{x^3}}},x \ne 0 \end{align} is

(i) Increasing

(ii) Decreasing

### Solution

We have \begin{align}f\left( x \right) = {x^3} + \frac{1}{{{x^3}}}\end{align}

Therefore,

\begin{align}f'\left( x \right) &= 3{x^2} - \frac{3}{{{x^4}}}\\& = \frac{{3{x^6} - 3}}{{{x^4}}}\end{align}

Now,

\begin{align}&f'\left( x \right) = 0\\ &\Rightarrow 3{x^6} - 3 = 0\\ &\Rightarrow {x^6} = 1\\& \Rightarrow x = \pm 1\end{align}

Now, the points $$x = 1$$ and $$x = - 1$$ divide the real line into three disjoint intervals i.e., $$\left( { - \infty , - 1} \right)$$, $$\left( { - 1,1} \right)$$ and $$\left( {1,\infty } \right)$$.

In intervals $$\left( { - \infty , - 1} \right)$$ and $$\left( {1,\infty } \right)$$ i.e., when $$x < - 1$$ and $$x > 1$$, $$f'\left( x \right) > 0$$.

Thus, when $$x < - 1$$ and $$x > 1$$, $$f$$ is increasing.

In interval $$\left( { - 1,1} \right)$$ i.e., when $$-\text{1}<~x~<\text{1}$$, $$f'\left( x \right) < 0$$.

Thus, when $$-\text{1}<~x~<\text{1}$$, $$f$$ is decreasing.

## Chapter 6 Ex.6.ME Question 8

Find the maximum area of an isosceles triangle inscribed in the ellipse $$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$$ with its vertex at one end of the major axis.

### Solution

The given ellipse is $$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$$

Let the major axis be along the $$x$$-axis.

Let $$ABC$$ be the triangle inscribed in the ellipse where vertex $$C$$ is at $$\left( {a,0} \right)$$.

Since the ellipse is symmetrical with respect to the $$x$$-axis and $$y$$-axis, we can assume the coordinates of A to be $$\left( { - {x_{\rm{1}}},{y_{\rm{1}}}} \right)$$ and the coordinates of B to be $$\left( { - {x_{\rm{1}}}, - {y_{\rm{1}}}} \right)$$

Now, we have $${y_1} = \pm \frac{b}{a}\sqrt {{a^2} - {x_1}^2}$$

Coordinates of $$A$$ are $$\left( { - {x_1},\frac{b}{a}\sqrt {{a^2} - {x_1}^2} } \right)$$ and

the coordinates of $$B$$ are $$\left( {{x_1}, - \frac{b}{a}\sqrt {{a^2} - {x_1}^2} } \right)$$

As the point $$\left( {{x_{\rm{1}}},{y_{\rm{1}}}} \right)$$ lies on the ellipse, the area of triangle $$ABC$$ $$\left( A \right)$$ is given by,

\begin{align}A &= \frac{1}{2}\left| {a\left( {\frac{{2b}}{a}\sqrt {{a^2} - {x_1}^2} } \right) + \left( { - {x_1}} \right)\left( { - \frac{b}{a}\sqrt {{a^2} - {x_1}^2} } \right) + \left( { - {x_1}} \right)\left( { - \frac{b}{a}\sqrt {{a^2} - {x_1}^2} } \right)} \right|\\ &= b\sqrt {{a^2} - {x_1}^2} + {x_1}\frac{b}{a}\sqrt {{a^2} - {x_1}^2} \qquad \ldots \left( 1 \right)\end{align}

Therefore,

\begin{align}\frac{{dA}}{{d{x_1}}} &= \frac{{ - 2{x_1}b}}{{2\sqrt {{a^2} - {x_1}^2} }} + \frac{b}{a}\sqrt {{a^2} - {x_1}^2} - \frac{{ - 2b{x_1}^2}}{{2a\sqrt {{a^2} - {x_1}^2} }}\\ &= \frac{b}{{a\sqrt {{a^2} - {x_1}^2} }}\left[ { - {x_1}a + \left( {{a^2} - {x_1}^2} \right) - {x_1}^2} \right]\\ &= \frac{{b\left( { - 2{x_1}^2 - {x_1}a + {a^2}} \right)}}{{a\sqrt {{a^2} - {x_1}^2} }}\end{align}

Now, $$\frac{{dA}}{{d{x_1}}} = 0$$

Hence,

\begin{align} &\Rightarrow - 2{x_1}^2 - {x_1}a + {a^2} = 0\\ &\Rightarrow {x_1} = \frac{{a \pm \sqrt {{a^2} - 4\left( { - 2} \right)\left( {{a^2}} \right)} }}{{2\left( { - 2} \right)}}\\ &\Rightarrow {x_1} = \frac{{a \pm \sqrt {9{a^2}} }}{{ - 4}}\\ &\Rightarrow {x_1} = \frac{{a \pm 3a}}{{ - 4}}\\ &\Rightarrow {x_1} = - a,\frac{a}{2}\end{align}

But $${x_1} \ne - a$$

Therefore,

\begin{align}{x_1} &= \frac{a}{2}\\{y_1} &= \frac{b}{a}\sqrt {{a^2} - \frac{{{a^2}}}{4}} \\ &= \frac{{ba}}{{2a}}\sqrt 3 \\ &= \frac{{\sqrt 3 b}}{2}\end{align}

Now,

\begin{align}\frac{{{d^2}A}}{{d{x_1}^2}} &= \frac{b}{a}\left\{ {\frac{{\sqrt {{a^2} - {x_1}^2} \left( { - 4{x_1} - a} \right) - \left( { - 2{x_1}^2 - {x_1}a + {a^2}} \right)\frac{{\left( { - 2{x_1}} \right)}}{{2\sqrt {{a^2} - {x_1}^2} }}}}{{{a^2} - {x_1}^2}}} \right\}\\ &= \frac{b}{a}\left\{ {\frac{{\left( {{a^2} - {x_1}^2} \right)\left( { - 4{x_1} - a} \right) + {x_1}\left( { - 2{x_1}^2 - {x_1}a + {a^2}} \right)}}{{{{\left( {{a^2} - {x_1}^2} \right)}^{\frac{3}{2}}}}}} \right\}\\ &= \frac{b}{a}\left\{ {\frac{{2{x^3} - 3{a^2}x - {a^3}}}{{{{\left( {{a^2} - {x_1}^2} \right)}^{\frac{3}{2}}}}}} \right\}\end{align}

Also, when, $${x_1} = \frac{a}{2}$$

Then,

\begin{align}\frac{{{d^2}A}}{{d{x_1}^2}} &= \frac{b}{a}\left\{ {\frac{{2\left( {{{\frac{a}{8}}^3}} \right) - 3{a^2}\left( {\frac{a}{2}} \right) - {a^3}}}{{{{\left( {{a^2} - {{\left( {\frac{a}{2}} \right)}^2}} \right)}^{\frac{3}{2}}}}}} \right\}\\ &= \frac{b}{a}\left\{ {\frac{{\frac{{{a^3}}}{4} - \frac{3}{2}{a^3} - {a^3}}}{{{{\left( {\frac{{3{a^2}}}{4}} \right)}^{\frac{3}{2}}}}}} \right\}\\& = - \frac{b}{a}\left\{ {\frac{{\frac{9}{4}{a^3}}}{{{{\left( {\frac{{3{a^2}}}{4}} \right)}^{\frac{3}{2}}}}}} \right\} < 0\end{align}

Thus, the area is the maximum when $${x_1} = \frac{a}{2}$$ .

Hence, Maximum area of the triangle is given by,

\begin{align}A &= b\sqrt {{a^2} - \frac{{{a^2}}}{4}} + \left( {\frac{a}{2}} \right)\frac{b}{a}\sqrt {{a^2} - \frac{{{a^2}}}{4}} \\ &= ab\frac{{\sqrt 3 }}{2} + \left( {\frac{a}{2}} \right)\frac{b}{a} \times \frac{{a\sqrt 3 }}{2}\\ &= \frac{{ab\sqrt 3 }}{2} + \frac{{ab\sqrt 3 }}{4}\\& = \frac{{3\sqrt 3 }}{4}ab\end{align}

## Chapter 6 Ex.6.ME Question 9

A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is $$2 \,\rm{m}$$ and volume is $$8{m^3}$$. If building of tank costs $$₹ \,70$$ per sq. meters for the base and $$₹ \,45$$ per square metres for sides. What is the cost of least expensive tank?

### Solution

Let $$l$$, $$b$$ and $$h$$ represent the length, breadth, and height of the tank respectively.

Then, we have height, $$h = 2m$$ and volume of the tank, $$V = 8{m^3}$$

Volume of the tank

\begin{align} & V=lbh \\ & \Rightarrow \text{8}=~l~\times ~b~\times ~\text{2} \\ & \Rightarrow lb=4 \\ & \Rightarrow b=\frac{4}{l} \\ \end{align}

Now, area of the base, $$lb = 4$$

Area of the 4 walls,

\begin{align}A &= 2h\left( {l + b} \right)\\ &= 4\left( {l + \frac{4}{l}} \right)\end{align}

Hence,

$\frac{{dA}}{{dl}} = 4\left( {l - \frac{4}{{{l^2}}}} \right)$

Now,

\begin{align}&\frac{{dA}}{{dl}} = 0\\& \Rightarrow \left( {l - \frac{4}{{{l^2}}}} \right) = 0\\ &\Rightarrow {l^2} = 4\\ &\Rightarrow l = \pm 2\end{align}

However, the length cannot be negative.

Therefore, we have $$l = 4$$

Hence,

\begin{align}b& = \frac{4}{l}\\ &= \frac{4}{2}\\& = 2\end{align}

Now,

$\frac{{{d^2}A}}{{d{l^2}}} = \frac{{32}}{{{l^3}}}$

When, $$l = 2$$

Then,

\begin{align}\frac{{{d^2}A}}{{d{l^2}}} &= \frac{{32}}{8}\\ &= 4 > 0\end{align}

Thus, by second derivative test, the area is the minimum when $$l = 2$$

We have $$l = b = h = 2$$

Therefore,

Cost of building the base in is

\begin{align}70\times ~\left( lb \right) & =70\left( 4 \right) \\ & =280 \end{align}

Cost of building the walls in is

\begin{align} 2h\left( l~+~b \right)~\times ~45&=2\times 2\left( 2+2 \right)\times 45 \\ & =720 \end{align}

Required total cost is is

$280 + 720 = 1000$

Thus, the total cost of the tank will be ₹ $$1000.$$

## Chapter 6 Ex.6.ME Question 10

The sum of the perimeter of a circle and square is $$k$$, where $$k$$ is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle.

### Solution

Let $$r$$ be the radius of the circle and $$a$$ be the side of the square.

Then, we have:

\begin{align}&2\pi r + 4a = k\\ &\Rightarrow a = \frac{{k - 2\pi r}}{4}\end{align}

The sum of the areas of the circle and the square $$\left( A \right)$$ is given by,

\begin{align}A &= \pi {r^2} + {a^2}\\ &= \pi {r^2} + \frac{{{{\left( {k - 2\pi r} \right)}^2}}}{{16}}\end{align}

Hence,

\begin{align}\frac{{dA}}{{dr}}& = 2\pi r + \frac{{2\left( {k - 2\pi r} \right)\left( { - 2\pi } \right)}}{{16}}\\& = 2\pi r - \frac{{\pi \left( {k - 2\pi r} \right)}}{4}\end{align}

Now,

\begin{align}&\frac{{dA}}{{dr}} = 0\\& \Rightarrow 2\pi r - \frac{{\pi \left( {k - 2\pi r} \right)}}{4} = 0\\& \Rightarrow 2\pi r = \frac{{\pi \left( {k - 2\pi r} \right)}}{4}\\& \Rightarrow 8r = k - 2\pi r\\& \Rightarrow \left( {8 + 2\pi } \right)r = k\\& \Rightarrow r = \frac{k}{{\left( {8 + 2\pi } \right)}}\\& \Rightarrow r = \frac{k}{{2\left( {4 + \pi } \right)}}\end{align}

When, $$r = \frac{k}{{2\left( {4 + \pi } \right)}}$$, $$\Rightarrow \frac{{{d^2}A}}{{d{r^2}}} > 0$$

The sum of the areas is least when, $$r = \frac{k}{{2\left( {4 + \pi } \right)}}$$

When, $$r = \frac{k}{{2\left( {4 + \pi } \right)}}$$

Then,

\begin{align}a &= \frac{{k - 2\pi \left[ {\frac{k}{{2\left( {4 + \pi } \right)}}} \right]}}{4}\\ &= \frac{{k\left( {4 + \pi } \right) - \pi k}}{{4\left( {4 + \pi } \right)}}\\ &= \frac{{4k}}{{4\left( {4 + \pi } \right)}}\\& = \frac{k}{{4 + \pi }}\\ &= 2r\end{align}

Hence, it has been proved that the sum of their areas is least when the side of the square is double the radius of the circle.

## Chapter 6 Ex.6.ME Question 11

A window is in the form of rectangle surmounted by a semicircular opening. The total perimeter of the window is $$10 \,\rm{m}$$. Find the dimensions of the window to admit maximum light through the whole opening.

### Solution

Let $$x$$ and $$y$$ be the length and breadth of the rectangular window.

Radius of the semicircular opening be $$\frac{x}{2}$$

It is given that the perimeter of the window is $$10\,\rm{m}$$.

Therefore,

\begin{align}&x + 2y + \frac{{\pi x}}{2} = 10\\ &\Rightarrow x\left( {1 + \frac{\pi }{2}} \right) + 2y = 10\\& \Rightarrow 2y = 10 - x\left( {1 + \frac{\pi }{2}} \right)\\& \Rightarrow y = 5 - x\left( {\frac{1}{2} + \frac{\pi }{4}} \right)\end{align}

Area of the window $$\left( A \right)$$ is given by,

\begin{align}A &= xy + \frac{\pi }{2}{\left( {\frac{x}{2}} \right)^2}\\ &= x\left[ {5 - x\left( {\frac{1}{2} + \frac{\pi }{4}} \right)} \right] + \frac{\pi }{8}{x^2}\\ &= 5x - {x^2}\left( {\frac{1}{2} + \frac{\pi }{4}} \right) + \frac{\pi }{8}{x^2}\end{align}

Therefore,

\begin{align}\frac{{dA}}{{dx}} &= 5 - 2x\left( {\frac{1}{2} + \frac{\pi }{4}} \right) + \frac{\pi }{4}x\\ &= 5 - x\left( {1 + \frac{\pi }{2}} \right) + \frac{\pi }{4}x\\\frac{{{d^2}A}}{{d{x^2}}} &= - \left( {1 + \frac{\pi }{2}} \right) + \frac{\pi }{4}\\& = - 1 - \frac{\pi }{4}\end{align}

Now,

\begin{align}&\frac{{dA}}{{dx}} = 0\\& \Rightarrow 5 - x\left( {1 + \frac{\pi }{2}} \right) + \frac{\pi }{4}x = 0\\ &\Rightarrow 5 - x - \frac{\pi }{4}x = 0\\ &\Rightarrow x\left( {1 + \frac{\pi }{4}} \right) = 5\\ &\Rightarrow x = \frac{5}{{\left( {1 + \frac{\pi }{4}} \right)}}\\& \Rightarrow x = \frac{{20}}{{\pi + 4}}\end{align}

When, $$x = \frac{{20}}{{\pi + 4}}$$

Then, $$\frac{{{d^2}A}}{{d{x^2}}} < 0$$

Therefore, by second derivative test, the area is the maximum when length is $$\frac{{20}}{{\pi + 4}}\,\rm{m}$$

Now,

\begin{align}y &= 5 - \frac{{20}}{{\pi + 4}}\left( {\frac{{2 + \pi }}{4}} \right)\\& = 5 - \frac{{5\left( {2 + \pi } \right)}}{{\pi + 4}}\\& = \frac{{10}}{{\pi + 4}}\end{align}

Hence, the required dimensions of the window to admit maximum light is given by length $$\frac{{20}}{{\pi + 4}}\,\rm{m}$$ and breadth $$\frac{{10}}{{\pi + 4}}\,\rm{m}$$.

## Chapter 6 Ex.6.ME Question 12

A point on the hypotenuse of a triangle is at distance $$a$$ and $$b$$ from the sides of the triangle.

Show that the minimum length of the hypotenuse is \begin{align}{\left( {{a^{\frac{2}{3}}} + {b^{\frac{2}{3}}}} \right)^{\frac{3}{2}}} \end{align}.

### Solution

Let $$\Delta ABC$$ be right-angled at $$B,$$ $$AB = x$$, $$BC = y$$ and $$\angle C = \theta$$.

Also, let $$P$$  be a point on the hypotenuse of the triangle such that $$P$$ is at a distance of $$a$$ and $$b$$ from the sides $$AB$$ and $$BC$$ respectively.

We have, $$AC = \sqrt {{x^2} + {y^2}}$$

Now,

\begin{align} & PC=~b~\text{cosec}~\theta \\ & AP=a\sec ~\theta \\ \end{align}

Hence,\begin{align} AC&=AP+PC \\ AC&=~b~\cos \text{ec}~\theta +a\sec ~\theta \ \ \ \ \ \ \ \ \ \ \ \ \ldots \left( 1 \right) \\ \end{align}

Therefore,

$\frac{d\left( AC \right)}{d\theta }=-b~\cos \text{ec}~\theta \cot \theta +a\sec ~\theta \tan \theta$

Now,

\begin{align} & \frac{d\left( AC \right)}{d\theta }=0 \\ & \Rightarrow -b~\cos ec~\theta \cot \theta +a\sec ~\theta \tan \theta =0 \\ & \Rightarrow a\sec ~\theta \tan \theta =b~\cos ec~\theta \cot \theta \\ & \Rightarrow \frac{a}{\cos \theta }.\frac{\sin \theta }{\cos \theta }=\frac{b}{\sin \theta }.\frac{\cos \theta }{\sin \theta } \\ & \Rightarrow a{{\sin }^{3}}\theta =b{{\cos }^{3}}\theta \\ & \Rightarrow {{\left( a \right)}^{\frac{1}{3}}}\sin \theta ={{\left( b \right)}^{\frac{1}{3}}}\cos \theta \\ & \Rightarrow \tan \theta ={{\left( \frac{b}{a} \right)}^{\frac{1}{3}}} \\ \end{align}

Thus,

$$\sin \theta = \frac{{{{\left( b \right)}^{\frac{1}{3}}}}}{{\sqrt {{a^{\frac{2}{3}}} + {b^{\frac{2}{3}}}} }}$$ and $$\cos \theta = \frac{{{{\left( a \right)}^{\frac{1}{3}}}}}{{\sqrt {{a^{\frac{2}{3}}} + {b^{\frac{2}{3}}}} }}\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)$$

It can be clearly shown that $$\frac{{{d^2}\left( {AC} \right)}}{{d{\theta ^2}}} < 0$$, when $$\tan \theta = {\left( {\frac{b}{a}} \right)^{\frac{1}{3}}}$$ .

By second derivative test the length of the hypotenuse is the maximum when $$\tan \theta = {\left( {\frac{b}{a}} \right)^{\frac{1}{3}}}$$

When, $$\tan \theta = {\left( {\frac{b}{a}} \right)^{\frac{1}{3}}}$$ , we have:

\begin{align}AC &= \frac{{b\sqrt {{a^{\frac{2}{3}}} + {b^{\frac{2}{3}}}} }}{{{{\left( b \right)}^{\frac{1}{3}}}}} + \frac{{a\sqrt {{a^{\frac{2}{3}}} + {b^{\frac{2}{3}}}} }}{{{{\left( a \right)}^{\frac{1}{3}}}}}\;\;\;\;\;\;\;\;\;\;\;\left[ {{\rm{Using }}\left( 1 \right){\text{ and }}\left( 2 \right)} \right]\\& = \sqrt {{a^{\frac{2}{3}}} + {b^{\frac{2}{3}}}} \left( {{b^{\frac{2}{3}}} + {a^{\frac{2}{3}}}} \right)\\& = {\left( {{a^{\frac{2}{3}}} + {b^{\frac{2}{3}}}} \right)^{\frac{3}{2}}}\end{align}

Thus, the maximum length of the hypotenuses is \begin{align}{\left( {{a^{\frac{2}{3}}} + {b^{\frac{2}{3}}}} \right)^{\frac{3}{2}}}\end{align} proved.

## Chapter 6 Ex.6.ME Question 13

Find the points at which the function $$f$$ given by $$f\left( x \right) = {\left( {x - 2} \right)^4}{\left( {x + 1} \right)^3}$$ has

(i) local maxima

(ii) local minima

(iii) point of inflexion

### Solution

The given function is $$f\left( x \right) = {\left( {x - 2} \right)^4}{\left( {x + 1} \right)^3}$$

Thereofore,

\begin{align}f'\left( x \right) &= 4{\left( {x - 2} \right)^3}{\left( {x + 1} \right)^3} + 3{\left( {x + 1} \right)^2}{\left( {x - 2} \right)^4}\\& = {\left( {x - 2} \right)^3}{\left( {x + 1} \right)^2}\left[ {4\left( {x + 1} \right) + 3\left( {x - 2} \right)} \right]\\& = {\left( {x - 2} \right)^3}{\left( {x + 1} \right)^2}\left( {7x - 2} \right)\end{align}

Now,

\begin{align}&f'\left( x \right) = 0\\& \Rightarrow x = - 1,x = \frac{2}{7},x = 2\end{align}

For values of $$x$$ close to $$\frac{2}{7}$$ and to the left of $$\frac{2}{7},f'\left( x \right) > 0$$

Also, for values of $$x$$ close to $$\frac{2}{7}$$ and to the right of $$\frac{2}{7},f'\left( x \right) < 0$$.

Thus, $$x = \frac{2}{7}$$ is the point of local maxima.

Now, for values of $$x$$ close to $$2$$ and to the left of $$2,f'\left( x \right) < 0$$

Also, for values of $$x$$ close to $$2$$ and to the right of $$2,f'\left( x \right) > 0$$.

Thus, $$x = 2$$ is the point of local minima.

Now, as the value of $$x$$ varies through $$- 1$$, $$f'\left( x \right)$$ does not change its sign.

Thus, $$x = - 1$$ is the point of inflexion.

## Chapter 6 Ex.6.ME Question 14

Find the absolute maximum and minimum values of the function $$f$$ given by $$f\left( x \right) = {\cos ^2}x + \sin x,x \in \left[ {0,\pi } \right]$$.

### Solution

We have $$f\left( x \right) = {\cos ^2}x + \sin x$$

Therefore,

\begin{align}f'\left( x \right) &= 2\cos x\left( { - \sin x} \right) + \cos x\\ &= - 2\sin x\cos x + \cos x\end{align}

Now,

\begin{align}& f'\left( x \right) = 0 \hfill \\& \Rightarrow - 2\sin x\cos x + \cos x = 0 \hfill \\ &\Rightarrow \cos x = 2\sin x\cos x \hfill \\& \Rightarrow \cos x\left( {2\sin x - 1} \right) = 0 \hfill \\& \Rightarrow \sin x = \frac{1}{2}{\text{ or }}\cos x = 0 \hfill \\& \Rightarrow x = \frac{\pi }{6}{\text{ or }}\frac{\pi }{2}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\because x \in \left[ {0,\pi } \right] \hfill \\ \end{align}

Now, evaluating the value of $$f$$ at critical points $$x = \frac{\pi }{6},\frac{\pi }{2}$$ and at the end points of the interval $$\left[ {0,\pi } \right]$$ i.e., at $$x = 0$$ and $$x = \pi$$, we have:

\begin{align}f\left( {\frac{\pi }{6}} \right) &= {\cos ^2}\left( {\frac{\pi }{6}} \right) + \sin \left( {\frac{\pi }{6}} \right)\\& = {\left( {\frac{{\sqrt 3 }}{2}} \right)^2} + \frac{1}{2}\\& = \frac{5}{4}\\f\left( 0 \right) &= {\cos ^2}\left( 0 \right) + \sin \left( 0 \right)\\& = 1 + 0\\& = 1\\f\left( \pi \right) &= {\cos ^2}\left( \pi \right) + \sin \left( \pi \right)\\ &= {\left( { - 1} \right)^2} + 0\\ &= 1\\f\left( {\frac{\pi }{2}} \right) &= {\cos ^2}\left( {\frac{\pi }{2}} \right) + \sin \left( {\frac{\pi }{2}} \right)\\ &= 0 + 1\\& = 1\end{align}

Hence, the absolute maximum value of $$f$$ is $$\frac{5}{4}$$ occurring at $$x = \frac{\pi }{6}$$ and the absolute minimum value of $$f$$ is 1 occurring at $$x = 0,\frac{\pi }{2},\pi$$.

## Chapter 6 Ex.6.ME Question 15

Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius $$r$$ is $$\frac{{4r}}{3}$$.

### Solution

A sphere of fixed radius $$\left( r \right)$$ is given.

Let $$R$$ and $$h$$ be the radius and the height of the cone respectively.

The volume $$\left( V \right)$$ of the cone is given by

$$V = \frac{1}{3}\pi {R^2}h$$

Now, from the right $$\Delta BCD$$, we have:

\begin{align}&BC = \sqrt {{r^2} - {R^2}} \\ &\Rightarrow h = r + \sqrt {{r^2} - {R^2}}\end{align}

Hence,

\begin{align}V &= \frac{1}{3}\pi {R^2}\left( {r + \sqrt {{r^2} - {R^2}} } \right)\\& = \frac{1}{3}\pi {R^2}r + \frac{1}{3}\pi {R^2}\sqrt {{r^2} - {R^2}} \end{align}

Therefore,

\begin{align}\frac{{dV}}{{dR}}& = \frac{2}{3}\pi Rr + \frac{2}{3}\pi R\sqrt {{r^2} - {R^2}} + \frac{{\pi {R^2}}}{3}.\frac{{\left( { - 2R} \right)}}{{2\sqrt {{r^2} - {R^2}} }}\\& = \frac{2}{3}\pi Rr + \frac{2}{3}\pi R\sqrt {{r^2} - {R^2}} - \frac{{\pi {R^3}}}{{3\sqrt {{r^2} - {R^2}} }}\\ &= \frac{2}{3}\pi Rr + \frac{{2\pi R\left( {{r^2} - {R^2}} \right) - \pi {R^3}}}{{3\sqrt {{r^2} - {R^2}} }}\\ &= \frac{2}{3}\pi Rr + \frac{{2\pi R{r^2} - 3\pi {R^3}}}{{3\sqrt {{r^2} - {R^2}} }}\end{align}

Now,

\begin{align}&\frac{{dV}}{{d{R^2}}} = 0\\ &\Rightarrow \frac{2}{3}\pi Rr + \frac{{2\pi R{r^2} - 3\pi {R^3}}}{{3\sqrt {{r^2} - {R^2}} }} = 0\\ &\Rightarrow \frac{{2\pi Rr}}{3} = \frac{{3\pi {R^3} - 2\pi R{r^2}}}{{3\sqrt {{r^2} - {R^2}} }}\\ &\Rightarrow 2r\sqrt {{r^2} - {R^2}} = 3{R^2} - 2{r^2}\\& \Rightarrow 4{r^2}\left( {{r^2} - {R^2}} \right) = {\left( {3{R^2} - 2{r^2}} \right)^2}\\ &\Rightarrow 4{r^4} - 4{r^2}{R^2} = 9{R^4} + 4{r^4} - 12{R^2}{r^2}\\& \Rightarrow 9{R^4} - 8{r^2}{R^2} = 0\\& \Rightarrow 9{R^2} = 8{r^2}\\ &\Rightarrow {R^2} = \frac{{8{r^2}}}{9}\end{align}

Also,

\begin{align}\frac{{{d^2}V}}{{d{R^2}}}& = \frac{{2\pi r}}{3} + \frac{{3\sqrt {{r^2} - {R^2}} \left( {2\pi {r^2} - 9\pi {R^2}} \right) - \left( {2\pi R{r^2} - 3\pi {R^3}} \right)\left( { - 6R} \right)\frac{1}{{2\sqrt {{r^2} - {R^2}} }}}}{{9\left( {{r^2} - {R^2}} \right)}}\\ &= \frac{{2\pi r}}{3} + \frac{{3\sqrt {{r^2} - {R^2}} \left( {2\pi {r^2} - 9\pi {R^2}} \right) + \left( {2\pi R{r^2} - 3\pi {R^3}} \right)\left( {3R} \right)\frac{1}{{\sqrt {{r^2} - {R^2}} }}}}{{9\left( {{r^2} - {R^2}} \right)}}\end{align}

When, $${R^2} = \frac{{8{r^2}}}{9}$$, $$\Rightarrow \frac{{{d^2}V}}{{d{R^2}}} < 0$$ .

Thus, the volume is the maximum when $${R^2} = \frac{{8{r^2}}}{9}$$.

When, $${R^2} = \frac{{8{r^2}}}{9}$$

Then, height of the cone

\begin{align}h &= r + \sqrt {{r^2} - \frac{{8{r^2}}}{9}} \\& = r + \sqrt {\frac{{{r^2}}}{9}} \\ &= r + \frac{r}{3}\\ &= \frac{{4r}}{3}\end{align}

Hence, it can be seen that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius $$\frac{{4r}}{3}$$.

## Chapter 6 Ex.6.ME Question 16

Let $$f$$ be a function defined on $$\left[ {a,b} \right]$$ such that $$f'\left( x \right) > 0$$, for all $$x~\in \left( a,b \right)$$. Then prove that $$f$$ is an increasing function on $$\left( {a,b} \right)$$.

### Solution

Let $${{x}_{1}},{{x}_{2}}~\in \left( a,b \right)$$ such that $${x_1} > {x_2}$$

Consider the sub-interval $$\left[ {{x_1},{x_2}} \right]$$

Since $$f\left( x \right)$$ is differentiable on $$\left( {a,b} \right)$$ and $$\left[ {{x_1},{x_2}} \right] \subset \left( {a,b} \right)$$.

Therefore, $$f\left( x \right)$$ is continuous on $$\left[ {{x_1},{x_2}} \right]$$ and differentiable on $$\left( {{x_1},{x_2}} \right)$$.

By the Lagrange's mean value theorem, there exists $$c \in \left( {{x_1},{x_2}} \right)$$ such that

$f\left( c \right) = \frac{{f\left( {{x_2}} \right) - f\left( {{x_1}} \right)}}{{{x_1} - {x_2}}} \qquad \ldots \left( 1 \right)$

Since, $$f'\left( x \right) > 0$$ for all $$x~\in \left( a,b \right)$$, so in particular,

\begin{align}&f'\left( c \right) > 0\\& \Rightarrow \frac{{f\left( {{x_2}} \right) - f\left( {{x_1}} \right)}}{{{x_1} - {x_2}}} > 0\;\;\;\;\;\;\;\;\;\;\left[ {{\text{Using }}\left( 1 \right)} \right]\\ &\Rightarrow f\left( {{x_2}} \right) - f\left( {{x_1}} \right) > 0\\& \Rightarrow f\left( {{x_2}} \right) > f\left( {{x_1}} \right)\\ &\Rightarrow f\left( {{x_1}} \right) < f\left( {{x_2}} \right)\end{align}

Since, $${x_1},{x_2}$$ are arbitrary points in $$\left( {a,b} \right)$$.

Therefore, $${x_1} < {x_2} \Rightarrow f\left( {{x_1}} \right) < f\left( {{x_2}} \right)$$ for all $${{x}_{1}},{{x}_{2}}~\in \left( a,b \right)$$.

Hence, $$f\left( x \right)$$ is increasing on $$\left( {a,b} \right)$$.

## Chapter 6 Ex.6.ME Question 17

Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius $$R$$ is $$\frac{{2R}}{{\sqrt 3 }}$$. Also find the maximum volume.

### Solution

A sphere of fixed radius $$\left( R \right)$$ is given.

Let $$r$$ and $$h$$ be the radius and the height of the cylinder respectively.

From the given figure, we have $$h = 2\sqrt {{R^2} - {r^2}}$$

The volume $$\left( V \right)$$ of the cylinder is given by,

\begin{align}V & = \pi {r^2}h\\ &= 2\pi {r^2}\sqrt {{R^2} - {r^2}} \end{align}

Therefore,

\begin{align}V &= \pi {r^2}h = 2\pi {r^2}\sqrt {{R^2} - {r^2}} \\\frac{{dV}}{{dr}} &= 4\pi r\sqrt {{R^2} - {r^2}} + \frac{{2\pi {r^2}\left( { - 2r} \right)}}{{2\sqrt {{R^2} - {r^2}} }}\\ &= 4\pi r\sqrt {{R^2} - {r^2}} - \frac{{2\pi {r^3}}}{{\sqrt {{R^2} - {r^2}} }}\\ &= \frac{{4\pi r\sqrt {{R^2} - {r^2}} - 2\pi {r^3}}}{{\sqrt {{R^2} - {r^2}} }}\\& = \frac{{4\pi r{R^2} - 6\pi {r^3}}}{{\sqrt {{R^2} - {r^2}} }}\end{align}

Now,

\begin{align}&\frac{{dV}}{{dr}} = 0\\ &\Rightarrow \frac{{4\pi r{R^2} - 6\pi {r^3}}}{{\sqrt {{R^2} - {r^2}} }} = 0\\& \Rightarrow 4\pi r{R^2} = 6\pi {r^3}\\& \Rightarrow {r^2} = \frac{{2{R^2}}}{3}\end{align}

Also,

\begin{align}\frac{{{d^2}V}}{{d{r^2}}} &= \frac{{\sqrt {{R^2} - {r^2}} \left( {4\pi {R^2} - 18\pi {r^2}} \right) - \left( {4\pi r{R^2} - 6\pi {r^3}} \right)\frac{{\left( { - 2r} \right)}}{{2\sqrt {{R^2} - {r^2}} }}}}{{\left( {{R^2} - {r^2}} \right)}}\\ &= \frac{{\left( {{R^2} - {r^2}} \right)\left( {4\pi {R^2} - 18\pi {r^2}} \right) + r\left( {4\pi r{R^2} - 6\pi {r^3}} \right)}}{{{{\left( {{R^2} - {r^2}} \right)}^{\frac{3}{2}}}}}\\ &= \frac{{\left( {4\pi {R^4} - 22\pi {r^2}{R^2} + 12\pi {r^4} + 4\pi {r^2}{R^2}} \right)}}{{{{\left( {{R^2} - {r^2}} \right)}^{\frac{3}{2}}}}}\end{align}

Now, it can be observed that at $${r^2} = \frac{{2{R^2}}}{3}$$, $$\Rightarrow \frac{{{d^2}V}}{{d{r^2}}} < 0$$

Thus, the volume is the maximum when $${r^2} = \frac{{2{R^2}}}{3}$$ .

When, $${r^2} = \frac{{2{R^2}}}{3}$$

Then, the height of the cylinder is

\begin{align}2\sqrt {{R^2} - \frac{{2{R^2}}}{3}} &= 2\sqrt {\frac{{{R^2}}}{3}} \\ &= \frac{{2R}}{{\sqrt 3 }}\end{align}

Hence, the volume of the cylinder is the maximum when the height of the cylinder is $$\frac{{2R}}{{\sqrt 3 }}$$.

## Chapter 6 Ex.6.ME Question 18

Show that height of the cylinder of greatest volume which can be inscribed in a right circular cone of height $$h$$ and semi vertical angle $$\alpha$$ is one-third that of the cone and the greatest volume of cylinder is $$\frac{4}{{27}}\pi {h^3}{\tan ^2}\alpha$$.

### Solution

The given right circular cone of fixed height $$h$$ and semi-vertical angle $$\alpha$$ can be drawn as:

Here, a cylinder of radius $$R$$ and height $$H$$ is inscribed in the cone.

Then, $$\angle GAO=~\alpha$$, $$OG=~r$$, $$OA=~h$$, $$OE=~R$$ and $$CE=~H$$.

We have, $$r~=~h~\text{tan}~\alpha$$

Now, since $$\Delta AOG$$ is similar to $$\Delta CEG$$, we have:

\begin{align}& \frac{{AO}}{{OG}} = \frac{{CE}}{{EG}} \hfill \\ &\Rightarrow \frac{h}{r} = \frac{H}{{r - R}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ {\because EG = OG - OE} \right] \hfill \\& \Rightarrow H = \frac{h}{r}\left( {r - R} \right) \hfill \\& \Rightarrow H = \frac{h}{{h\tan \alpha }}\left( {h\tan \alpha - R} \right) \hfill \\& \Rightarrow H = \frac{1}{{\tan \alpha }}\left( {h\tan \alpha - R} \right) \hfill \\ \end{align}

Now, the volume $$\left( V \right)$$ of the cylinder is given by,

\begin{align}V &= \pi {R^2}H\\ &= \frac{{\pi {R^2}}}{{\tan \alpha }}\left( {h\tan \alpha - R} \right)\\& = \pi {R^2}h - \frac{{\pi {R^3}}}{{\tan \alpha }}\end{align}

Therefore,

$\frac{{dV}}{{dR}} = 2\pi Rh - \frac{{3\pi {R^2}}}{{\tan \alpha }}$

Now,

\begin{align}&\frac{{dV}}{{dR}} = 0\\ &\Rightarrow 2\pi Rh - \frac{{3\pi {R^2}}}{{\tan \alpha }} = 0\\& \Rightarrow 2\pi Rh = \frac{{3\pi {R^2}}}{{\tan \alpha }}\\ &\Rightarrow 2h\tan \alpha = 3R\\ &\Rightarrow R = \frac{{2h}}{3}\tan \alpha \end{align}

Also

$$\frac{{{d^2}V}}{{d{R^2}}} = 2\pi h - \frac{{6\pi R}}{{\tan \alpha }}$$

For $$R = \frac{{2h}}{3}\tan \alpha$$, we have:

\begin{align}\frac{{{d^2}V}}{{d{R^2}}} &= 2\pi h - \frac{{6\pi }}{{\tan \alpha }}\left( {\frac{{2h}}{3}\tan \alpha } \right)\\& = 2\pi h - 4\pi h\\ &= - 2\pi h < 0\end{align}.

By second derivative test, the volume of the cylinder is the greatest when $$R = \frac{{2h}}{3}\tan \alpha$$.

When, $$R = \frac{{2h}}{3}\tan \alpha$$

Then,

\begin{align}H &= \frac{1}{{\tan \alpha }}\left( {h\tan \alpha - \frac{{2h}}{3}\tan \alpha } \right)\\& = \frac{1}{{\tan \alpha }}\left( {\frac{{h\tan \alpha }}{3}} \right)\\& = \frac{h}{3}\end{align}

Thus, the height of the cylinder is one-third the height of the cone when the volume of the cylinder is the greatest.

Thus, the maximum volume of the cylinder can be obtained as:

\begin{align}\pi {\left( {\frac{{2h}}{3}\tan \alpha } \right)^2}\left( {\frac{h}{3}} \right)& = \pi \left( {\frac{{4{h^2}}}{9}{{\tan }^2}\alpha } \right)\left( {\frac{h}{3}} \right)\\& = \frac{4}{{27}}\pi {h^3}{\tan ^2}\alpha \end{align}

Hence, the given result is proved.

## Chapter 6 Ex.6.ME Question 19

A cylindrical tank of radius $$10\, \rm{m}$$ is being filled with wheat at the rate of $$314$$ cubic metre per hour. Then the depth of the wheat is increasing at the rate of

(A)  $$1m/h$$

(B)  $$0.1m/h$$

(C)  $$1.1m/h$$

(D)  $$0.5m/h$$

### Solution

Let $$r$$ be the radius of the cylinder.

Then, volume $$\left( V \right)$$ of the cylinder is given by,

\begin{align}V &= \pi {r^2}h\\ &= \pi {\left( {10} \right)^2}h\\ &= 100\pi h\end{align}

Differentiating with respect to time $$\left( t \right)$$, we have:

$\frac{{dV}}{{dt}} = 100\pi \frac{{dh}}{{dt}}$

The tank is being filled with wheat at the rate of $$314{m^3}/h$$

$\frac{{dV}}{{dt}} = 314{m^3}/h$

Thus, we have:

\begin{align}100\pi \frac{{dh}}{{dt}} &= 314\\\frac{{dh}}{{dt}} &= \frac{{314}}{{100\left( {3.14} \right)}}\\ &= 1\end{align}

Hence, the depth of wheat is increasing at the rate of $$1m/h$$.

Thus, the correct option is A.

## Chapter 6 Ex.6.ME Question 20

The slope of the tangent to the curve $$x = {t^2} + 3t - 8$$, $$y = 2{t^2} - 2t - 5$$ at the point $$\left( {2, - 1} \right)$$ is

(A) \begin{align}\frac{{22}}{7} \end{align}

(B) \begin{align}\frac{6}{7}\end{align}

(C) \begin{align}\frac{7}{6}\end{align}

(D) \begin{align}\frac{{ - 6}}{7}\end{align}

### Solution

The given curve is $$x = {t^2} + 3t - 8$$ and $$y = 2{t^2} - 2t - 5$$.

Therefore,

\begin{align}\frac{{dx}}{{dt}} &= 2t + 3\\\frac{{dy}}{{dt}} &= 4t - 2\end{align}

Hence,

$\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}} = \frac{{4t - 2}}{{2t + 3}}$

The given point is $$\left( {2, - 1} \right)$$

At $$x = 2$$, we have:

\begin{align}&{t^2} + 3t - 8 = 2\\ &\Rightarrow {t^2} + 3t - 10 = 0\\& \Rightarrow \left( {t - 2} \right)\left( {t + 5} \right) = 0\\& \Rightarrow t = 2{\text{ or }}t = - 5\end{align}

At $$y = - 1$$ ,we have:

\begin{align}&2{t^2} - 2t - 5 = - 1\\& \Rightarrow 2{t^2} - 2t - 4 = 0\\ &\Rightarrow 2\left( {{t^2} - t - 2} \right) = 0\\& \Rightarrow \left( {t - 2} \right)\left( {t + 1} \right) = 0\\& \Rightarrow t = 2{\text{ or }}t = - 1\end{align}

The common value is $$t = 2$$

Hence, the slope of the tangent to the given curve at point $$\left( {2, - 1} \right)$$ is

\begin{align}{\left. {\frac{{dy}}{{dx}}} \right]_{t = 2}}& = \frac{{4\left( 2 \right) - 2}}{{2\left( 2 \right) + 3}}\\& = \frac{{8 - 2}}{{4 + 3}}\\ &= \frac{6}{7}\end{align}

Thus, the correct option is B.

## Chapter 6 Ex.6.ME Question 21

The line $$y = mx + 1$$ is a tangent to the curve $${y^2} = 4x$$ if the value of $$m$$ is

(A) $$1$$

(B) $$2$$

(C) $$3$$

(D) $$\frac{1}{2}$$

### Solution

The equation of the tangent to the given curve is $$y = mx + 1$$

Now, substituting $$y = mx + 1$$ in $${y^2} = 4x$$, we get:

\begin{align}&{\left( {mx + 1} \right)^2} = 4x\\ &\Rightarrow {m^2}{x^2} + 1 + 2mx - 4x = 0\\ &\Rightarrow {m^2}{x^2} + \left( {2m - 4} \right)x + 1 = 0\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\end{align}

Since a tangent touches the curve at one point, the roots of equation (1) must be equal.

Therefore, we have:

Discriminant $$= 0$$

\begin{align} &\Rightarrow {\left( {2m - 4} \right)^2} - 4\left( {{m^2}} \right)\left( 1 \right) = 0\\& \Rightarrow 4{m^2} + 16 - 16m - 4{m^2} = 0\\ &\Rightarrow 16 - 16m = 0\\& \Rightarrow m = 1\end{align}

Hence, the required value of $$m$$ is 1.

Thus, the correct option is A.

## Chapter 6 Ex.6.ME Question 22

The normal at the point $$\left( {1,1} \right)$$ on the curve $$2y + {x^2} = 3$$ is

(A) $$x + y = 0$$

(B) $$x - y = 0$$

(C) $$x + y + 1 = 0$$

(D) $$x - y = 1$$

### Solution

The equation of the given curve is $$2y + {x^2} = 3$$

Differentiating with respect to $$x$$, we have:

\begin{align}&\frac{{2dy}}{{dx}} + 2x = 0\\& \Rightarrow \frac{{dy}}{{dx}} = - x\\ &\Rightarrow {\left. {\frac{{dy}}{{dx}}} \right]_{\left( {1,1} \right)}} = - 1\end{align}

The slope of the normal to the given curve at point $$\left( {1,1} \right)$$ is

\begin{align}\frac{{ - 1}}{{{{\left. {\frac{{dy}}{{dx}}} \right]}_{\left( {1,1} \right)}}}} = 1 \end{align}

Hence, the equation of the normal to the given curve at $$\left( {1,1} \right)$$ is given as:

\begin{align}& \Rightarrow y - 1 = 1\left( {x - 1} \right)\\ &\Rightarrow y - 1 = x - 1\\& \Rightarrow x - y = 0\end{align}

Thus, the correct option is B.

## Chapter 6 Ex.6.ME Question 23

The normal to the curve $${x^2} = 4y$$ passing $$\left( {1,2} \right)$$ is

(A) $$x + y = 3$$

(B) $$x - y = 3$$

(C) $$x + y = 1$$

(D) $$x - y = 1$$

### Solution

The equation of the given curve is $${x^2} = 4y$$

Differentiating with respect to $$x$$, we have:

\begin{align}&2x = 4.\frac{{dy}}{{dx}}\\& \Rightarrow \frac{{dy}}{{dx}} = \frac{x}{2}\end{align}

The slope of the normal to the given curve at point $$\left( {h,k} \right)$$ is

\begin{align}\frac{{ - 1}}{{{{\left. {\frac{{dy}}{{dx}}} \right]}_{\left( {h,k} \right)}}}} = - \frac{2}{h}\end{align}

Hence, the equation of the normal to the given curve at $$\left( {h,k} \right)$$ is given as:

$y - k = - \frac{2}{h}\left( {x - h} \right)$

Now, it is given that the normal passes through the point $$\left( {1,2} \right)$$

Therefore, we have:

\begin{align}&2 - k = - \frac{2}{h}\left( {1 - h} \right)\\& \Rightarrow k = 2 + \frac{2}{h}\left( {1 - h} \right)\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\end{align}

Since $$\left( {h,k} \right)$$ lies on the curve $${x^2} = 4y$$, we have $${h^2} = 4k$$

$\Rightarrow k = \frac{{{h^2}}}{4}$

From equation (1), we have:

\begin{align}&\frac{{{h^2}}}{4} = 2 + \frac{2}{h}\left( {1 - h} \right)\\& \Rightarrow \frac{{{h^3}}}{4} = 2h + 2 - 2h\\ &\Rightarrow \frac{{{h^3}}}{4} = 2\\& \Rightarrow {h^3} = 8\\& \Rightarrow h = 2\end{align}

Therefore,

\begin{align}&k = \frac{{{h^2}}}{4}\\ &\Rightarrow k = 1\end{align}

Hence, the equation of the normal is given as:

\begin{align} &\Rightarrow y - 1 = \frac{{ - 2}}{2}\left( {x - 2} \right)\\& \Rightarrow y - 1 = - x + 2\\ &\Rightarrow x + y = 3\end{align}

Thus, the correct option is A.

## Chapter 6 Ex.6.ME Question 24

The points on the curve $$9{y^2} = {x^3}$$, where the normal to the curve makes equal intercepts with the axes are

(A) $$\left( {4, \pm \frac{8}{3}} \right)$$

(B) $$4,\frac{{ - 8}}{3}$$

(C) $$\left( {4, \pm \frac{3}{8}} \right)$$

(D) $$\left( { \pm 4,\frac{3}{8}} \right)$$

### Solution

The equation of the given curve is $$9{y^2} = {x^3}$$

Differentiating with respect to $$x$$, we have:

\begin{align}&9\left( {2y} \right)\frac{{dy}}{{dx}} = 3{x^2}\\& \Rightarrow \frac{{dy}}{{dx}} = \frac{{{x^2}}}{{6y}}\end{align}

The slope of the normal to the given curve at point $$\left( {{x_1},{y_1}} \right)$$ is

$\frac{{ - 1}}{{{{\left. {\frac{{dy}}{{dx}}} \right]}_{\left( {{x_1},{y_1}} \right)}}}} = - \frac{{6{y_1}}}{{{x_1}^2}}$

The equation of the normal to the curve at $$\left( {{x_1},{y_1}} \right)$$ is

\begin{align}&y - {y_1} = - \frac{{6{y_1}}}{{{x_1}^2}}\left( {x - {x_1}} \right)\\ &\Rightarrow {x_1}^2y - {x_1}^2{y_1} = - 6x{y_1} + 6{x_1}{y_1}\\& \Rightarrow 6x{y_1} + {x_1}^2y = 6{x_1}{y_1} + {x_1}^2{y_1}\\ &\Rightarrow \frac{{6x{y_1} + {x_1}^2y}}{{6{x_1}{y_1} + {x_1}^2{y_1}}} = 1\\& \Rightarrow \frac{{6x{y_1}}}{{6{x_1}{y_1} + {x_1}^2{y_1}}} + \frac{{{x_1}^2y}}{{6{x_1}{y_1} + {x_1}^2{y_1}}} = 1\\ &\Rightarrow \frac{x}{{\frac{{{x_1}\left( {6 + {x_1}} \right)}}{6}}} + \frac{y}{{\frac{{{y_1}\left( {6 + {x_1}} \right)}}{{{x_1}}}}} = 1\end{align}

It is given that the normal makes equal intercepts with the axes.

Therefore, we have:

\begin{align}& \Rightarrow \frac{{{x_1}\left( {6 + {x_1}} \right)}}{6} = \frac{{{y_1}\left( {6 + {x_1}} \right)}}{{{x_1}}}\\& \Rightarrow \frac{{{x_1}}}{6} = \frac{{{y_1}}}{{{x_1}}}\\& \Rightarrow {x_1}^2 = 6{y_1}\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\end{align}

Also, the point $$\left( {{x_1},{y_1}} \right)$$ lies on the curve, so we have

$9{y_1}^2 = {x_1}^3 \qquad \ldots \left( 2 \right)$

From (1) and (2), we have:

\begin{align} &\Rightarrow 9{\left( {\frac{{{x_1}^2}}{6}} \right)^2} = {x_1}^3\\& \Rightarrow \frac{{{x_1}^4}}{4} = {x_1}^3\\ &\Rightarrow {x_1} = 4\end{align}

From (2), we have:

\begin{align} &\Rightarrow 9{y_1}^2 = {4^3}\\& \Rightarrow {y_1}^2 = \frac{{64}}{9}\\ &\Rightarrow {y_1} = \pm \frac{8}{3}\end{align}

Hence, the required points are $$\left( {4, \pm \frac{8}{3}} \right)$$

Thus, the correct option is A.

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