Examples (Revised) - Chapter 6 - Applications Of Derivatives - Ncert Solutions class 12 - Maths

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NCERT Class 12 Maths Solutions: Chapter 6 - Applications Of Derivatives

Example 1

Find the rate of change of the area of a circle per second with respect to its radius $r$ when $r=5 \mathrm{~cm}$.

Solution

The area $\mathrm{A}$ of a circle with radius $r$ is given by $\mathrm{A}=\pi r^2$. Therefore, the rate of change of the area A with respect to its radius $r$ is given by $\frac{d \mathrm{~A}}{d r}=\frac{d}{d r}\left(\pi r^2\right)=2 \pi r$.

When $r=5 \mathrm{~cm}, \frac{d \mathrm{~A}}{d r}=10 \pi$. Thus, the area of the circle is changing at the rate of $10 \pi \mathrm{cm}^2 / \mathrm{s}$.

Example 2

The volume of a cube is increasing at a rate of 9 cubic centimetres per second. How fast is the surface area increasing when the length of an edge is 10 centimetres?

Solution

Let $x$ be the length of a side, $\mathrm{V}$ be the volume and $\mathrm{S}$ be the surface area of the cube. Then, $\mathrm{V}=x^3$ and $\mathrm{S}=6 x^2$, where $x$ is a function of time $t$.

Now
$
\frac{d \mathrm{~V}}{d t}=9 \mathrm{~cm}^3 / \mathrm{s} \text { (Given) }
$

Therefore
$
\begin{aligned}
9 & =\frac{d V}{d t}=\frac{d}{d t}\left(x^3\right)=\frac{d}{d x}\left(x^3\right) \cdot \frac{d x}{d t} \quad \text { (By Chain Rule) } \\
& =3 x^2 \cdot \frac{d x}{d t}
\end{aligned}
$
or
$
\begin{aligned}
\frac{d x}{d t} & =\frac{3}{x^2} \\
\frac{d S}{d t} & =\frac{d}{d t}\left(6 x^2\right)=\frac{d}{d x}\left(6 x^2\right) \cdot \frac{d x}{d t} \\
& =12 x \cdot\left(\frac{3}{x^2}\right)=\frac{36}{x}
\end{aligned}
$
(By Chain Rule)
(Using (1))

Hence, when
$
x=10 \mathrm{~cm}, \frac{d S}{d t}=3.6 \mathrm{~cm}^2 / \mathrm{s}
$

Example 3

$\mathrm{~A}$ stone is dropped into a quiet lake and waves move in circles at a speed of $4 \mathrm{~cm}$ per second. At the instant, when the radius of the circular wave is $10 \mathrm{~cm}$, how fast is the enclosed area increasing?

Solution

The area $\mathrm{A}$ of a circle with radius $r$ is given by $\mathrm{A}=\pi r^2$. Therefore, the rate of change of area $\mathrm{A}$ with respect to time $t$ is
$
\frac{d \mathrm{~A}}{d t}=\frac{d}{d t}\left(\pi r^2\right)=\frac{d}{d r}\left(\pi r^2\right) \cdot \frac{d r}{d t}=2 \pi r \frac{d r}{d t}
$
(By Chain Rule)

It is given that
$
\frac{d r}{d t}=4 \mathrm{~cm} / \mathrm{s}
$

Therefore, when $r=10 \mathrm{~cm}, \quad \frac{d \mathrm{~A}}{d t}=2 \pi(10)(4)=80 \pi$
Thus, the enclosed area is increasing at the rate of $80 \pi \mathrm{cm}^2 / \mathrm{s}$, when $r=10 \mathrm{~cm}$.

Example 4

The length $x$ of a rectangle is decreasing at the rate of $3 \mathrm{~cm} /$ minute and the width $y$ is increasing at the rate of $2 \mathrm{~cm} /$ minute. When $x=10 \mathrm{~cm}$ and $y=6 \mathrm{~cm}$, find the rates of change of (a) the perimeter and (b) the area of the rectangle.

Solution

Since the length $x$ is decreasing and the width $y$ is increasing with respect to time, we have
$
\frac{d x}{d t}=-3 \mathrm{~cm} / \mathrm{min} \quad \text { and } \quad \frac{d y}{d t}=2 \mathrm{~cm} / \mathrm{min}
$
(a) The perimeter $\mathrm{P}$ of a rectangle is given by
$
\begin{aligned}
\mathrm{P} & =2(x+y) \\
\frac{d \mathrm{P}}{d t} & =2\left(\frac{d x}{d t}+\frac{d y}{d t}\right)=2(-3+2)=-2 \mathrm{~cm} / \mathrm{min}
\end{aligned}
$

Therefore
(b) The area $\mathrm{A}$ of the rectangle is given by
$
\mathrm{A}=x \cdot y
$

Therefore
$
\begin{aligned}
\frac{d \mathrm{~A}}{d t} & =\frac{d x}{d t} \cdot y+x \cdot \frac{d y}{d t} \\
& =-3(6)+10(2) \quad(\text { as } x=10 \mathrm{~cm} \text { and } y=6 \mathrm{~cm}) \\
& =2 \mathrm{~cm}^2 / \mathrm{min}
\end{aligned}
$

Example 5

The total cost $C(x)$ in Rupees, associated with the production of $x$ units of an item is given by
$
\mathrm{C}(x)=0.005 x^3-0.02 x^2+30 x+5000
$

Find the marginal cost when 3 units are produced, where by marginal cost we mean the instantaneous rate of change of total cost at any level of output.

Solution

Since marginal cost is the rate of change of total cost with respect to the output, we have

Marginal
$
\operatorname{cost}(\mathrm{MC})=\frac{d C}{d x}=0.005\left(3 x^2\right)-0.02(2 x)+30
$

When
$
\begin{aligned}
x=3, \mathrm{MC} & =0.015\left(3^2\right)-0.04(3)+30 \\
& =0.135-0.12+30=30.015
\end{aligned}
$

Hence, the required marginal cost is ₹ 30.02 (nearly).

Example 6

The total revenue in Rupees received from the sale of $x$ units of a product is given by $\mathrm{R}(x)=3 x^2+36 x+5$. Find the marginal revenue, when $x=5$, where by marginal revenue we mean the rate of change of total revenue with respect to the number of items sold at an instant.

Solution

Since marginal revenue is the rate of change of total revenue with respect to the number of units sold, we have

Marginal Revenue
$
\begin{aligned}
(\mathrm{MR}) & =\frac{d \mathrm{R}}{d x}=6 x+36 \\
x & =5, \mathrm{MR}=6(5)+36=66
\end{aligned}
$

When
Hence, the required marginal revenue is ₹ 66 .

Example 7

Show that the function given by $f(x)=7 x-3$ is increasing on $\mathbf{R}$.
Solution

Let $x_1$ and $x_2$ be any two numbers in $\mathbf{R}$. Then
$
x_1 $$

Thus, by Definition 1 , it follows that $f$ is strictly increasing on $\mathbf{R}$.
We shall now give the first derivative test for increasing and decreasing functions. The proof of this test requires the Mean Value Theorem studied in Chapter 5.

Example 8

Show that the function $f$ given by
$
f(x)=x^3-3 x^2+4 x, x \in \mathbf{R}
$
is increasing on $\mathbf{R}$.
Solution

Note that
$
\begin{aligned}
f^{\prime}(x) & =3 x^2-6 x+4 \\
& =3\left(x^2-2 x+1\right)+1 \\
& =3(x-1)^2+1>0, \text { in every interval of } \mathbf{R}
\end{aligned}
$

Therefore, the function $f$ is increasing on $\mathbf{R}$.
Example 9

Prove that the function given by $f(x)=\cos x$ is
(a) decreasing in $(0, \pi)$
(b) increasing in $(\pi, 2 \pi)$, and
(c) neither increasing nor decreasing in $(0,2 \pi)$.

Solution

Note that $f^{\prime}(x)=-\sin x$
(a) Since for each $x \in(0, \pi), \sin x>0$, we have $f^{\prime}(x)<0$ and so $f$ is decreasing in $(0, \pi)$.
(b) Since for each $x \in(\pi, 2 \pi), \sin x<0$, we have $f^{\prime}(x)>0$ and so $f$ is increasing in $(\pi, 2 \pi)$.
(c) Clearly by (a) and (b) above, $f$ is neither increasing nor decreasing in $(0,2 \pi)$.

Example 10

Find the intervals in which the function $f$ given by $f(x)=x^2-4 x+6$ is
(a) increasing
(b) decreasing

Solution

We have
$
\begin{array}{ll} 
& f(x)=x^2-4 x+6 \\
\text { or } \quad & f^{\prime}(x)=2 x-4
\end{array}
$

Therefore, $f^{\prime}(x)=0$ gives $x=2$. Now the point $x=2$ divides the real line into two disjoint intervals namely, $(-\infty, 2)$ and ( $2, \infty$ ) (Fig 6.3). In the interval $(-\infty, 2), f^{\prime}(x)=2 x$ $-4<0$.

Therefore, $f$ is decreasing in this interval. Also, in the interval $(2, \infty), f^{\prime}(x)>0$ and so the function $f$ is increasing in this interval.

Example 11

Find the intervals in which the function $f$ given by $f(x)=4 x^3-6 x^2-72 x$ +30 is (a) increasing (b) decreasing.
Solution

We have
$
f(x)=4 x^3-6 x^2-72 x+30
$
or
$
\begin{aligned}
f^{\prime}(x) & =12 x^2-12 x-72 \\
& =12\left(x^2-x-6\right) \\
& =12(x-3)(x+2)
\end{aligned}
$

Therefore, $f^{\prime}(x)=0$ gives $x=-2,3$. The points $x=-2$ and $x=3$ divides the real line into three disjoint intervals, namely, $(-\infty,-2),(-2,3)$ and $(3, \infty)$.

In the intervals $(-\infty,-2)$ and $(3, \infty), f^{\prime}(x)$ is positive while in the interval $(-2,3)$, $f^{\prime}(x)$ is negative. Consequently, the function $f$ is increasing in the intervals $(-\infty,-2)$ and $(3, \infty)$ while the function is decreasing in the interval $(-2,3)$. However, $f$ is neither increasing nor decreasing in $\mathbf{R}$.

Example 12

Find intervals in which the function given by $f(x)=\sin 3 x, x \in\left[0, \frac{\pi}{2}\right]$ is
(a) increasing (b) decreasing.

Solution

We have
or
$
\begin{aligned}
f(x) & =\sin 3 x \\
f^{\prime}(x) & =3 \cos 3 x
\end{aligned}
$

Therefore, $f^{\prime}(x)=0$ gives $\cos 3 x=0$ which in turn gives $3 x=\frac{\pi}{2}, \frac{3 \pi}{2}$ (as $x \in\left[0, \frac{\pi}{2}\right]$ implies $\left.3 x \in\left[0, \frac{3 \pi}{2}\right]\right)$. So $x=\frac{\pi}{6}$ and $\frac{\pi}{2}$. The point $x=\frac{\pi}{6}$ divides the interval $\left[0, \frac{\pi}{2}\right]$ $\text { into two disjoint intervals }\left[0, \frac{\pi}{6}\right) \text { and }\left(\frac{\pi}{6}, \frac{\pi}{2}\right] \text {. }$

Now, $f^{\prime}(x)>0$ for all $x \in\left[0, \frac{\pi}{6}\right)$ as $0 \leq x<\frac{\pi}{6} \Rightarrow 0 \leq 3 x<\frac{\pi}{2}$ and $f^{\prime}(x)<0$ for all $x \in\left(\frac{\pi}{6}, \frac{\pi}{2}\right)$ as $\frac{\pi}{6} Therefore, f$ is increasing in $\left[0, \frac{\pi}{6}\right)$ and decreasing in $\left(\frac{\pi}{6}, \frac{\pi}{2}\right)$.

Also, the given function is continuous at $x=0$ and $x=\frac{\pi}{6}$. Therefore, by Theorem 1 , $f$ is increasing on $\left[0, \frac{\pi}{6}\right]$ and decreasing on $\left[\frac{\pi}{6}, \frac{\pi}{2}\right]$.

Example 13

Find the intervals in which the function $f$ given by
$
f(x)=\sin x+\cos x, 0 \leq x \leq 2 \pi
$
is increasing or decreasing.
Solution

We have
or
$
\begin{aligned}
f(x) & =\sin x+\cos x \\
f^{\prime}(x) & =\cos x-\sin x
\end{aligned}
$

Now $f^{\prime}(x)=0$ gives $\sin x=\cos x$ which gives that $x=\frac{\pi}{4}, \frac{5 \pi}{4}$ as $0 \leq x \leq 2 \pi$
The points $x=\frac{\pi}{4}$ and $x=\frac{5 \pi}{4}$ divide the interval $[0,2 \pi]$ into three disjoint intervals, $\text { namely, }\left[0, \frac{\pi}{4}\right),\left(\frac{\pi}{4}, \frac{5 \pi}{4}\right) \text { and }\left(\frac{5 \pi}{4}, 2 \pi\right] \text {. }$

Note that $f^{\prime}(x)>0$ if $x \in\left[0, \frac{\pi}{4}\right) \cup\left(\frac{5 \pi}{4}, 2 \pi\right]$
or $\quad f$ is increasing in the intervals $\left[0, \frac{\pi}{4}\right)$ and $\left(\frac{5 \pi}{4}, 2 \pi\right]$

Also $\quad f^{\prime}(x)<0$ if $x \in\left(\frac{\pi}{4}, \frac{5 \pi}{4}\right)$
or $\quad f$ is decreasing in $\left(\frac{\pi}{4}, \frac{5 \pi}{4}\right)$

Example 14

Find the maximum and the minimum values, if any, of the function $f$ given by
$
f(x)=x^2, x \in \mathbf{R} .
$

Solution From the graph of the given function (Fig 6.8), we have $f(x)=0$ if $x=0$. Also
$
f(x) \geq 0, \text { for all } x \in \mathbf{R} .
$

Therefore, the minimum value of $f$ is 0 and the point of minimum value of $f$ is $x=0$. Further, it may be observed from the graph of the function that $f$ has no maximum value and hence no point of maximum value of $f$ in $\mathbf{R}$.

Example 15

Find the maximum and minimum values of $f$, if any, of the function given by $f(x)=|x|, x \in \mathbf{R}$.

Solution

From the graph of the given function (Fig 6.9), note that
$f(x) \geq 0$, for all $x \in \mathbf{R}$ and $f(x)=0$ if $x=0$.

Therefore, the function $f$ has a minimum value 0 and the point of minimum value of $f$ is $x=0$. Also, the graph clearly shows that $f$ has no maximum value in $\mathbf{R}$ and hence no point of maximum value in $\mathbf{R}$.

Example 16

Find the maximum and the minimum values, if any, of the function given by
$
f(x)=x, x \in(0,1) .
$

Solution

The given function is an increasing (strictly) function in the given interval $(0,1)$. From the graph (Fig 6.10) of the function $f$, it

seems that, it should have the minimum value at point closest to 0 on its right and the maximum value at a point closest to 1 on its left. Are such points available? Of course, not. It is not possible to locate such points. Infact, if a point $x_0$ is closest to 0 , ther we find $\frac{x_0}{2}x_1$ for all $x_1 \in(0,1)$.

Therefore, the given function has neither the maximum value nor the minimum value in the interval $(0,1)$.

Example 17 

Find all points of local maxima and local minima of the function $f$ given by
$
f(x)=x^3-3 x+3
$

Solution

We have
or
$
\begin{aligned}
f(x) & =x^3-3 x+3 \\
f^{\prime}(x) & =3 x^2-3=3(x-1)(x+1) \\
f^{\prime}(x) & =0 \text { at } x=1 \text { and } x=-1
\end{aligned}
$

Thus, $x= \pm 1$ are the only critical points which could possibly be the points of local maxima and/or local minima of $f$. Let us first examine the point $x=1$.

Note that for values close to 1 and to the right of $1, f^{\prime}(x)>0$ and for values close to 1 and to the left of $1, f^{\prime}(x)<0$. Therefore, by first derivative test, $x=1$ is a point of local minima and local minimum value is $f(1)=1$. In the case of $x=-1$, note that $f^{\prime}(x)>0$, for values close to and to the left of -1 and $f^{\prime}(x)<0$, for values close to and to the right of -1 . Therefore, by first derivative test, $x=-1$ is a point of local maxima and local maximum value is $f(-1)=5$.

Example 18

Find all the points of local maxima and local minima of the function $f$ given by
$
f(x)=2 x^3-6 x^2+6 x+5
$

Solution

We have
or
$
\begin{aligned}
f(x) & =2 x^3-6 x^2+6 x+5 \\
f^{\prime}(x) & =6 x^2-12 x+6=6(x-1)^2 \\
f^{\prime}(x) & =0 \text { at } x=1
\end{aligned}
$
or
Thus, $x=1$ is the only critical point of $f$. We shall now examine this point for local maxima and/or local minima of $f$. Observe that $f^{\prime}(x) \geq 0$, for all $x \in \mathbf{R}$ and in particular $f^{\prime}(x)>0$, for values close to 1 and to the left and to the right of 1 . Therefore, by first derivative test, the point $x=1$ is neither a point of local maxima nor a point of local minima. Hence $x=1$ is a point of inflexion.

Example 19

Find local minimum value of the function $f$ given by $f(x)=3+|x|, x \in \mathbf{R}$.

Solution

Note that the given function is not differentiable at $x=0$. So, second derivative test fails. Let us try first derivative test. Note that 0 is a critical point of $f$. Now to the left of $0, f(x)=3-x$ and so $f^{\prime}(x)=-1<0$. Also to the right of $0, f(x)=3+x$ and so $f^{\prime}(x)=1>0$. Therefore, by first derivative test, $x=$ 0 is a point of local minima of $f$ and local minimum value of $f$ is $f(0)=3$.

Example 20

Find local maximum and local minimum values of the function $f$ given by
$
f(x)=3 x^4+4 x^3-12 x^2+12
$

Solution

We have
or
$
\begin{aligned}
f(x) & =3 x^4+4 x^3-12 x^2+12 \\
f^{\prime}(x) & =12 x^3+12 x^2-24 x=12 x(x-1)(x+2)
\end{aligned}
$
or
$
f^{\prime}(x)=0 \text { at } x=0, x=1 \text { and } x=-2 \text {. }
$

Now
$
f^{\prime \prime}(x)=36 x^2+24 x-24=12\left(3 x^2+2 x-2\right)
$
or
$
\left\{\begin{array}{l}
f^{\prime \prime}(0)=-24<0 \\
f^{\prime \prime}(1)=36>0 \\
f^{\prime \prime}(-2)=72>0
\end{array}\right.
$

Therefore, by second derivative test, $x=0$ is a point of local maxima and local maximum value of $f$ at $x=0$ is $f(0)=12$ while $x=1$ and $x=-2$ are the points of local minima and local minimum values of $f$ at $x=-1$ and -2 are $f(1)=7$ and $f(-2)=-20$, respectively.

Example 21

Find all the points of local maxima and local minima of the function $f$ given by
$
f(x)=2 x^3-6 x^2+6 x+5
$

Solution

We have
or
$
\begin{gathered}
f(x)=2 x^3-6 x^2+6 x+5 \\
\left\{\begin{array}{l}
f^{\prime}(x)=6 x^2-12 x+6=6(x-1)^2 \\
f^{\prime \prime}(x)=12(x-1)
\end{array}\right.
\end{gathered}
$

Now $f^{\prime}(x)=0$ gives $x=1$. Also $f^{\prime \prime}(1)=0$. Therefore, the second derivative test fails in this case. So, we shall go back to the first derivative test.

We have already seen (Example 18) that, using first derivative test, $x=1$ is neither a point of local maxima nor a point of local minima and so it is a point of inflexion.

Example 22

Find two positive numbers whose sum is 15 and the sum of whose squares is minimum.

Solution

Let one of the numbers be $x$. Then the other number is $(15-x)$. Let $\mathrm{S}(x)$ denote the sum of the squares of these numbers. Then

$
\begin{gathered}
\mathrm{S}(x)=x^2+(15-x)^2=2 x^2-30 x+225 \\
\left\{\begin{array}{l}
\mathrm{S}^{\prime}(x)=4 x-30 \\
\mathrm{~S}^{\prime \prime}(x)=4
\end{array}\right.
\end{gathered}
$

Now $\mathrm{S}^{\prime}(x)=0$ gives $x=\frac{15}{2}$. Also $\mathrm{S}^{\prime \prime}\left(\frac{15}{2}\right)=4>0$. Therefore, by second derivative test, $x=\frac{15}{2}$ is the point of local minima of S. Hence the sum of squares of numbers is minimum when the numbers are $\frac{15}{2}$ and $15-\frac{15}{2}=\frac{15}{2}$.

Example 23

Find the shortest distance of the point $(0, c)$ from the parabola $y=x^2$, where $\frac{1}{2} \leq c \leq 5$.

Solution

Let $(h, k)$ be any point on the parabola $y=x^2$. Let $\mathrm{D}$ be the required distance between $(h, k)$ and $(0, c)$. Then
$
\mathrm{D}=\sqrt{(h-0)^2+(k-c)^2}=\sqrt{h^2+(k-c)^2}
$

Since $(h, k)$ lies on the parabola $y=x^2$, we have $k=h^2$. So (1) gives
$
\begin{aligned}
\mathrm{D} & \equiv \mathrm{D}(k)=\sqrt{k+(k-c)^2} \\
\mathrm{D}^{\prime}(k) & =\frac{1+2(k-c)}{2 \sqrt{k+(k-c)^2}}
\end{aligned}
$
$
\mathrm{D}^{\prime}(k)=0 \text { gives } k=\frac{2 c-1}{2}
$

Now
or
$
\begin{aligned}
\mathrm{D} & \equiv \mathrm{D}(k)=\sqrt{k+(k-c)^2} \\
\mathrm{D}^{\prime}(k) & =\frac{1+2(k-c)}{2 \sqrt{k+(k-c)^2}}
\end{aligned}
$
$
\mathrm{D}^{\prime}(k)=0 \text { gives } k=\frac{2 c-1}{2}
$

Observe that when $k<\frac{2 c-1}{2}$, then $2(k-c)+1<0$, i.e., $\mathrm{D}^{\prime}(k)<0$. Also when $k>\frac{2 c-1}{2}$, then $\mathrm{D}^{\prime}(k)>0$. So, by first derivative test, $\mathrm{D}(k)$ is minimum at $k=\frac{2 c-1}{2}$.

Hence, the required shortest distance is given by
$
\mathrm{D}\left(\frac{2 c-1}{2}\right)=\sqrt{\frac{2 c-1}{2}+\left(\frac{2 c-1}{2}-c\right)^2}=\frac{\sqrt{4 c-1}}{2}
$

Example 24

Let AP and BQ be two vertical poles at points $A$ and $B$, respectively. If $A P=16 \mathrm{~m}, B Q=22 \mathrm{~m}$ and $A B=20 \mathrm{~m}$, then find the distance of a point $R$ on $A B$ from the point $A$ such that $R P^2+R Q^2$ is minimum.

Solution

Let $\mathrm{R}$ be a point on $\mathrm{AB}$ such that $\mathrm{AR}=x \mathrm{~m}$. Then $R B=(20-x) \mathrm{m}($ as $A B=20 \mathrm{~m})$. From Fig 6.16, we have

$\begin{array}{ll} 
& \mathrm{RP}^2=\mathrm{AR}^2+\mathrm{AP}^2 \\
\text { and } & \mathrm{RQ}^2=\mathrm{RB}^2+\mathrm{BQ}^2
\end{array}$

Therefore
$
\begin{aligned}
\mathrm{RP}^2+\mathrm{RQ}^2 & =\mathrm{AR}^2+\mathrm{AP}^2+\mathrm{RB}^2+\mathrm{BQ}^2 \\
& =x^2+(16)^2+(20-x)^2+(22)^2 \\
& =2 x^2-40 x+1140
\end{aligned}
$

Let
$
\mathrm{S} \equiv \mathrm{S}(x)=\mathrm{RP}^2+\mathrm{RQ}^2=2 x^2-40 x+1140 .
$

Therefore
$
\mathrm{S}^{\prime}(x)=4 x-40 \text {. }
$

Now $\mathrm{S}^{\prime}(x)=0$ gives $x=10$. Also $\mathrm{S}^{\prime \prime}(x)=4>0$, for all $x$ and so $\mathrm{S}^{\prime \prime}(10)>0$.
Therefore, by second derivative test, $x=10$ is the point of local minima of S. Thus, the distance of $\mathrm{R}$ from $\mathrm{A}$ on $\mathrm{AB}$ is $\mathrm{AR}=x=10 \mathrm{~m}$.
Example 25

If length of three sides of a trapezium other than base are equal to $10 \mathrm{~cm}$, then find the area of the trapezium when it is maximum.
Solution

The required trapezium is as given in Fig 6.17. Draw perpendiculars DP and

$\mathrm{CQ}$ on $\mathrm{AB}$. Let $\mathrm{AP}=x \mathrm{~cm}$. Note that $\triangle \mathrm{APD} \sim \Delta \mathrm{BQC}$. Therefore, $\mathrm{QB}=x \mathrm{~cm}$. Also, by Pythagoras theorem, $\mathrm{DP}=\mathrm{QC}=\sqrt{100-x^2}$. Let $\mathrm{A}$ be the area of the trapezium. Then
$
\begin{aligned}
\mathrm{A} \equiv \mathrm{A}(x) & =\frac{1}{2} \text { (sum of parallel sides) (height) } \\
& =\frac{1}{2}(2 x+10+10)\left(\sqrt{100-x^2}\right) \\
& =(x+10)\left(\sqrt{100-x^2}\right) \\
\mathrm{A}^{\prime}(x) & =(x+10) \frac{(-2 x)}{2 \sqrt{100-x^2}}+\left(\sqrt{100-x^2}\right) \\
& =\frac{-2 x^2-10 x+100}{\sqrt{100-x^2}}
\end{aligned}
$
or

Now $\quad \mathrm{A}^{\prime}(x)=0$ gives $2 x^2+10 x-100=0$, i.e., $x=5$ and $x=-10$.
Since $x$ represents distance, it can not be negative.
So, $x=5$. Now
$
\mathrm{A}^{\prime \prime}(x)=\frac{\sqrt{100-x^2}(-4 x-10)-\left(-2 x^2-10 x+100\right) \frac{(-2 x)}{2 \sqrt{100-x^2}}}{100-x^2}
$

$
\begin{aligned}
& =\frac{2 x^3-300 x-1000}{\left(100-x^2\right)^{\frac{3}{2}}} \text { (on simplification) } \\
\text { or } \quad \mathrm{A}^{\prime \prime}(5) & =\frac{2(5)^3-300(5)-1000}{\left(100-(5)^2\right)^{\frac{3}{2}}}=\frac{-2250}{75 \sqrt{75}}=\frac{-30}{\sqrt{75}}<0
\end{aligned}
$

Thus, area of trapezium is maximum at $x=5$ and the area is given by
$
\mathrm{A}(5)=(5+10) \sqrt{100-(5)^2}=15 \sqrt{75}=75 \sqrt{3} \mathrm{~cm}^2
$

Example 26

Prove that the radius of the right circular cylinder of greatest curved surface area which can be inscribed in a given cone is half of that of the cone.
Solution

Let $\mathrm{OC}=r$ be the radius of the cone and $\mathrm{OA}=h$ be its height. Let a cylinder with radius $\mathrm{OE}=x$ inscribed in the given cone (Fig 6.18). The height $\mathrm{QE}$ of the cylinder is given by

$
\begin{aligned}
& \frac{\mathrm{QE}}{\mathrm{OA}}=\frac{\mathrm{EC}}{\mathrm{OC}} \quad(\text { since } \Delta \mathrm{QEC} \sim \Delta \mathrm{AOC}) \\
& \text { or } \quad \frac{\mathrm{QE}}{h}=\frac{r-x}{r} \\
& \text { or } \quad \mathrm{QE}=\frac{h(r-x)}{r} \\
&
\end{aligned}
$

Let $S$ be the curved surface area of the given cylinder. Then

$\begin{aligned}
& \mathrm{S} \equiv \mathrm{S}(x)=\frac{2 \pi x h(r-x)}{r}=\frac{2 \pi h}{r}\left(r x-x^2\right) \\
&\left\{\begin{array}{l}
\mathrm{S}^{\prime}(x)=\frac{2 \pi h}{r}(r-2 x) \\
\mathrm{S}^{\prime \prime}(x)=\frac{-4 \pi h}{r}
\end{array}\right.
\end{aligned}$

Now $\mathrm{S}^{\prime}(x)=0$ gives $x=\frac{r}{2}$. Since $\mathrm{S}^{\prime \prime}(x)<0$ for all $x, \mathrm{~S}^{\prime \prime}\left(\frac{r}{2}\right)<0$. So $x=\frac{r}{2}$ is a point of maxima of S. Hence, the radius of the cylinder of greatest curved surface area which can be inscribed in a given cone is half of that of the cone.

Example 27

Find the absolute maximum and minimum values of a function $f$ given by $f(x)=2 x^3-15 x^2+36 x+1$ on the interval $[1,5]$.

Solution

We have
or
$
\begin{aligned}
f(x) & =2 x^3-15 x^2+36 x+1 \\
f^{\prime}(x) & =6 x^2-30 x+36=6(x-3)(x-2)
\end{aligned}
$

Note that $f^{\prime}(x)=0$ gives $x=2$ and $x=3$.
We shall now evaluate the value of $f$ at these points and at the end points of the interval $[1,5]$, i.e., at $x=1, x=2, x=3$ and at $x=5$. So
$
\begin{aligned}
& f(1)=2\left(1^3\right)-15\left(1^2\right)+36(1)+1=24 \\
& f(2)=2\left(2^3\right)-15\left(2^2\right)+36(2)+1=29 \\
& f(3)=2\left(3^3\right)-15\left(3^2\right)+36(3)+1=28 \\
& f(5)=2\left(5^3\right)-15\left(5^2\right)+36(5)+1=56
\end{aligned}
$

Thus, we conclude that absolute maximum value of $f$ on $[1,5]$ is 56 , occurring at $x=5$, and absolute minimum value of $f$ on $[1,5]$ is 24 which occurs at $x=1$.

Example 28

Find absolute maximum and minimum values of a function $f$ given by
$
f(x)=12 x^{\frac{4}{3}}-6 x^{\frac{1}{3}}, x \in[-1,1]
$

Solution

We have
$
\begin{aligned}
f(x) & =12 x^{\frac{4}{3}}-6 x^{\frac{1}{3}} \\
f^{\prime}(x) & =16 x^{\frac{1}{3}}-\frac{2}{x^{\frac{2}{3}}}=\frac{2(8 x-1)}{x^{\frac{2}{3}}}
\end{aligned}
$

Thus, $f^{\prime}(x)=0$ gives $x=\frac{1}{8}$. Further note that $f^{\prime}(x)$ is not defined at $x=0$. So the critical points are $x=0$ and $x=\frac{1}{8}$. Now evaluating the value of $f$ at critical points $x=0, \frac{1}{8}$ and at end points of the interval $x=-1$ and $x=1$, we have
$
\begin{aligned}
f(-1) & =12(-1)^{\frac{4}{3}}-6(-1)^{\frac{1}{3}}=18 \\
f(0) & =12(0)-6(0)=0
\end{aligned}
$

$
\begin{aligned}
f\left(\frac{1}{8}\right) & =12\left(\frac{1}{8}\right)^{\frac{4}{3}}-6\left(\frac{1}{8}\right)^{\frac{1}{3}}=\frac{-9}{4} \\
f(1) & =12(1)^{\frac{4}{3}}-6(1)^{\frac{1}{3}}=6
\end{aligned}
$

Hence, we conclude that absolute maximum value of $f$ is 18 that occurs at $x=-1$ and absolute minimum value of $f$ is $\frac{-9}{4}$ that occurs at $x=\frac{1}{8}$.

Example 29

An Apache helicopter of enemy is flying along the curve given by $y=x^2+7$. A soldier, placed at $(3,7)$, wants to shoot down the helicopter when it is nearest to him. Find the nearest distance.

Solution

For each value of $x$, the helicopter's position is at point $\left(x, x^2+7\right)$. Therefore, the distance between the helicopter and the soldier placed at $(3,7)$ is
$
\sqrt{(x-3)^2+\left(x^2+7-7\right)^2} \text {, i.e., } \sqrt{(x-3)^2+x^4} .
$

Let
$
\begin{aligned}
f(x) & =(x-3)^2+x^4 \\
f^{\prime}(x) & =2(x-3)+4 x^3=2(x-1)\left(2 x^2+2 x+3\right)
\end{aligned}
$
or
Thus, $f^{\prime}(x)=0$ gives $x=1$ or $2 x^2+2 x+3=0$ for which there are no real roots. Also, there are no end points of the interval to be added to the set for which $f^{\prime}$ is zero, i.e., there is only one point, namely, $x=1$. The value of $f$ at this point is given by $f(1)=(1-3)^2+(1)^4=5$. Thus, the distance between the solider and the helicopter is $\sqrt{f(1)}=\sqrt{5}$
Note that $\sqrt{5}$ is either a maximum value or a minimum value. Since
$
\sqrt{f(0)}=\sqrt{(0-3)^2+(0)^4}=3>\sqrt{5}
$
it follows that $\sqrt{5}$ is the minimum value of $\sqrt{f(x)}$. Hence, $\sqrt{5}$ is the minimum distance between the soldier and the helicopter.

Example 30

$\mathrm{~A}$ car starts from a point $\mathrm{P}$ at time $t=0$ seconds and stops at point $\mathrm{Q}$. The distance $x$, in metres, covered by it, in $t$ seconds is given by
$
x=t^2\left(2-\frac{t}{3}\right)
$

Find the time taken by it to reach $Q$ and also find distance between $P$ and $Q$.
Solution

Let $v$ be the velocity of the car at $t$ seconds.

Now
$
x=t^2\left(2-\frac{t}{3}\right)
$

Therefore
$
v=\frac{d x}{d t}=4 t-t^2=t(4-t)
$

Thus, $v=0$ gives $t=0$ and/or $t=4$.
Now $v=0$ at $\mathrm{P}$ as well as at $\mathrm{Q}$ and at $\mathrm{P}, t=0$. So, at $\mathrm{Q}, t=4$. Thus, the car will reach the point $Q$ after 4 seconds. Also the distance travelled in 4 seconds is given by
$
x]_{t=4}=4^2\left(2-\frac{4}{3}\right)=16\left(\frac{2}{3}\right)=\frac{32}{3} \mathrm{~m}
$

Example 31

A water tank has the shape of an inverted right circular cone with its axis vertical and vertex lowermost. Its semi-vertical angle is $\tan ^{-1}(0.5)$. Water is poured into it at a constant rate of 5 cubic metre per hour. Find the rate at which the level of the water is rising at the instant when the depth of water in the tank is $4 \mathrm{~m}$.

Solution

Let $r, h$ and $\alpha$ be as in Fig 6.20. Then $\tan \alpha=\frac{r}{h}$.

So
$
\alpha=\tan ^{-1}\left(\frac{r}{h}\right) .
$

But
$
\alpha=\tan ^{-1}(0.5)
$
(given)
or
$
\frac{r}{h}=0.5
$
or
$
r=\frac{h}{2}
$

Let $\mathrm{V}$ be the volume of the cone. Then
$
\mathrm{V}=\frac{1}{3} \pi r^2 h=\frac{1}{3} \pi\left(\frac{h}{2}\right)^2 h=\frac{\pi h^3}{12}
$

Therefore
$
\begin{aligned}
\frac{d V}{d t} & =\frac{d}{d h}\left(\frac{\pi h^3}{12}\right) \cdot \frac{d h}{d t} \\
& =\frac{\pi}{4} h^2 \frac{d h}{d t}
\end{aligned}
$
(by Chain Rule)

Now rate of change of volume, i.e., $\frac{d \mathrm{~V}}{d t}=5 \mathrm{~m}^3 / \mathrm{h}$ and $h=4 \mathrm{~m}$.

Therefore
$
5=\frac{\pi}{4}(4)^2 \cdot \frac{d h}{d t}
$
or
$
\frac{d h}{d t}=\frac{5}{4 \pi}=\frac{35}{88} \mathrm{~m} / \mathrm{h}\left(\pi=\frac{22}{7}\right)
$

Thus, the rate of change of water level is $\frac{35}{88} \mathrm{~m} / \mathrm{h}$.
Example 32

A man of height 2 metres walks at a uniform speed of $5 \mathrm{~km} / \mathrm{h}$ away from a lamp post which is 6 metres high. Find the rate at which the length of his shadow increases.

Solution

In Fig 6.21, Let AB be the lamp-post, the lamp being at the position $B$ and let $M N$ be the man at a particular time $t$ and let $\mathrm{AM}=l$ metres. Then, $\mathrm{MS}$ is the shadow of the man. Let MS $=s$ metres.

Note that
$
\triangle \mathrm{MSN} \sim \Delta \mathrm{ASB}
$
or
$
\frac{\mathrm{MS}}{\mathrm{AS}}=\frac{\mathrm{MN}}{\mathrm{AB}}
$

or
$
\mathrm{AS}=3 s(\text { as } \mathrm{MN}=
$
2 and $A B=6$ (given) $)$
Thus
So
$
\begin{aligned}
\mathrm{AM} & =3 s-s=2 s . \text { But } \mathrm{AM}=l \\
l & =2 s
\end{aligned}
$

Therefore
$
\frac{d l}{d t}=2 \frac{d s}{d t}
$

$\text { Since } \frac{d l}{d t}=5 \mathrm{~km} / \mathrm{h} \text {. Hence, the length of the shadow increases at the rate } \frac{5}{2} \mathrm{~km} / \mathrm{h} \text {. }$

Example 33

Find intervals in which the function given by
$
f(x)=\frac{3}{10} x^4-\frac{4}{5} x^3-3 x^2+\frac{36}{5} x+11
$
is (a) increasing (b) decreasing.
Solution

We have
$
\begin{aligned}
f(x) & =\frac{3}{10} x^4-\frac{4}{5} x^3-3 x^2+\frac{36}{5} x+11 \\
f^{\prime}(x) & =\frac{3}{10}\left(4 x^3\right)-\frac{4}{5}\left(3 x^2\right)-3(2 x)+\frac{36}{5} \\
& =\frac{6}{5}(x-1)(x+2)(x-3) \quad \text { (on simplification) }
\end{aligned}
$

Now $f^{\prime}(x)=0$ gives $x=1, x=-2$, or $x=3$. The points $x=1,-2$, and 3 divide the real line into four disjoint intervals namely, $(-\infty,-2),(-2,1),(1,3)$ and $(3, \infty)$ (Fig 6.22).

Consider the interval $(-\infty,-2)$, i.e., when $-\infty<x<-2$.
In this case, we have $x-1<0, x+2<0$ and $x-3<0$.
(In particular, observe that for $x=-3, f^{\prime}(x)=(x-1)(x+2)(x-3)=(-4)(-1)$ $(-6)<0)$
Therefore, $\quad f^{\prime}(x)<0$ when $-\infty<x<-2$.
Thus, the function $f$ is decreasing in $(-\infty,-2)$.
Consider the interval $(-2,1)$, i.e., when $-2<x<1$.
In this case, we have $x-1<0, x+2>0$ and $x-3<0$
(In particular, observe that for $x=0, f^{\prime}(x)=(x-1)(x+2)(x-3)=(-1)(2)(-3)$ $=6>0)$
So $f^{\prime}(x)>0$ when $-2<x<1$.

Thus, $f$ is increasing in $(-2,1)$.
Now consider the interval $(1,3)$, i.e., when $1<x<3$. In this case, we have $x-1>0, x+2>0$ and $x-3<0$.

So, $f^{\prime}(x)<0$ when $1<x<3$.

Thus, $f$ is decreasing in $(1,3)$.
Finally, consider the interval $(3, \infty)$, i.e., when $x>3$. In this case, we have $x-1>0$, $x+2>0$ and $x-3>0$. So $f^{\prime}(x)>0$ when $x>3$.
Thus, $f$ is increasing in the interval $(3, \infty)$.

Example 34

Show that the function $f$ given by
$
f(x)=\tan ^{-1}(\sin x+\cos x), x>0
$

is always an increasing function in $\left(0, \frac{\pi}{4}\right)$.
Solution

We have
$
\begin{aligned}
f(x) & =\tan ^{-1}(\sin x+\cos x), x>0 \\
f^{\prime}(x) & =\frac{1}{1+(\sin x+\cos x)^2}(\cos x-\sin x)
\end{aligned}
$

$
=\frac{\cos x-\sin x}{2+\sin 2 x}
$
(on simplification)

Note that $2+\sin 2 x>0$ for all $x$ in $0, \frac{\pi}{4}$.
Therefore
$
f^{\prime}(x)>0 \text { if } \cos x-\sin x>0
$
or
$
f^{\prime}(x)>0 \text { if } \cos x>\sin x \text { or } \cot x>1
$

Now $\cot x>1$ if $\tan x<1$, i.e., if $0

Thus
$
f^{\prime}(x)>0 \text { in }\left(0, \frac{\pi}{4}\right)
$

Hence $f$ is increasing function in $\left(0, \frac{\pi}{4}\right)$.

Example 35

$\mathrm{~A}$ circular disc of radius $3 \mathrm{~cm}$ is being heated. Due to expansion, its radius increases at the rate of $0.05 \mathrm{~cm} / \mathrm{s}$. Find the rate at which its area is increasing when radius is $3.2 \mathrm{~cm}$.

Solution

Let $r$ be the radius of the given disc and $\mathrm{A}$ be its area. Then
$
\mathrm{A}=\pi r^2
$
or
$
\frac{d \mathrm{~A}}{d t}=2 \pi r \frac{d r}{d t}
$
(by Chain Rule)
Now approximate rate of increase of radius $=d r=\frac{d r}{d t} \Delta t=0.05 \mathrm{~cm} / \mathrm{s}$.
Therefore, the approximate rate of increase in area is given by
$
\begin{aligned}
d \mathrm{~A} & =\frac{d \mathrm{~A}}{d t}(\Delta t)=2 \pi r\left(\frac{d r}{d t} \Delta t\right) \\
& =2 \pi(3.2)(0.05)=0.320 \pi \mathrm{cm}^2 / \mathrm{s} \quad(r=3.2 \mathrm{~cm})
\end{aligned}
$

Example 36

An open topped box is to be constructed by removing equal squares from each corner of a 3 metre by 8 metre rectangular sheet of aluminium and folding up the sides. Find the volume of the largest such box.

Solution

Let $x$ metre be the length of a side of the removed squares. Then, the height of the box is $x$, length is $8-2 x$ and breadth is $3-2 x$ (Fig 6.23). If $\mathrm{V}(x)$ is the volume of the box, then

$
\begin{gathered}
\mathrm{V}(x)=x(3-2 x)(8-2 x) \\
=4 x^3-22 x^2+24 x \\
\left\{\begin{array}{l}
\mathrm{V}^{\prime}(x)=12 x^2-44 x+24=4(x-3)(3 x-2) \\
\mathrm{V}^{\prime \prime}(x)=24 x-44
\end{array}\right.
\end{gathered}
$

Therefore $\quad\left\{\begin{array}{l}\mathrm{V}^{\prime}(x)=12 x^2-44 x+24=4(x-3)(3 x-2) \\ \mathrm{V}^{\prime \prime}(x)=24 x-44\end{array}\right.$
Now $\quad \mathrm{V}^{\prime}(x)=0$ gives $x=3, \frac{2}{3}$. But $x \neq 3$ (Why?)
Thus, we have $x=\frac{2}{3}$. Now $\mathrm{V}^{\prime \prime}\left(\frac{2}{3}\right)=24\left(\frac{2}{3}\right)-44=-28<0$.
Therefore, $x=\frac{2}{3}$ is the point of maxima, i.e., if we remove a square of side $\frac{2}{3}$ metre from each corner of the sheet and make a box from the remaining sheet, then the volume of the box such obtained will be the largest and it is given by
$
\begin{aligned}
\mathrm{V}\left(\frac{2}{3}\right) & =4\left(\frac{2}{3}\right)^3-22\left(\frac{2}{3}\right)^2+24\left(\frac{2}{3}\right) \\
& =\frac{200}{27} \mathrm{~m}^3
\end{aligned}
$

Example 37

Manufacturer can sell $x$ items at a price of rupees $\left(5-\frac{x}{100}\right)$ each. The cost price of $x$ items is $\operatorname{Rs}\left(\frac{x}{5}+500\right)$. Find the number of items he should sell to earn maximum profit.

Solution

Let $\mathrm{S}(x)$ be the selling price of $x$ items and let $\mathrm{C}(x)$ be the cost price of $x$ items. Then, we have
$
\mathrm{S}(x)=\left(5-\frac{x}{100}\right) x=5 x-\frac{x^2}{100}
$
and
$
\mathrm{C}(x)=\frac{x}{5}+500
$

Thus, the profit function $\mathrm{P}(x)$ is given by
$
\begin{aligned}
& \mathrm{P}(x)=\mathrm{S}(x)-\mathrm{C}(x)=5 x-\frac{x^2}{100}-\frac{x}{5}-500 \\
& \mathrm{P}(x)=\frac{24}{5} x-\frac{x^2}{100}-500 \\
& \mathrm{P}^{\prime}(x)=\frac{24}{5}-\frac{x}{50}
\end{aligned}
$

Now $\mathrm{P}^{\prime}(x)=0$ gives $x=240$. Also $\mathrm{P}^{\prime \prime}(x)=\frac{-1}{50}$. So $\mathrm{P}^{\prime \prime}(240)=\frac{-1}{50}<0$
Thus, $x=240$ is a point of maxima. Hence, the manufacturer can earn maximum profit, if he sells 240 items.