NCERT Solutions for Miscellaneous Exercise Chapter 6 Class 12 - Application of Derivatives

Question:1(a) Using differentials, find the approximate value of each of the following:

$(17/81{)}^{1/4}$

Answer:

Let $y=x^{\frac{1}{4}}$ and $x=\frac{16}{81}and\mathrm{\Delta }x=\frac{1}{81}$
$\mathrm{\Delta }y=(x+\mathrm{\Delta }x{)}^{\frac{1}{4}}-x^{\frac{1}{4}}$
$=(\frac{16}{81}+\frac{1}{81}{)}^{\frac{1}{4}}-(\frac{16}{81}{)}^{\frac{1}{4}}$
$(\frac{17}{81}{)}^{\frac{1}{4}}=\mathrm{\Delta }y+\frac{2}{3}$
Now, we know that $\mathrm{\Delta }y$ is approximate equals to dy
So,
$$\begin{gathered} dy=\frac{dy}{dx}.\mathrm{\Delta }x \\ =\frac{1}{4x^{\frac{3}{4}}}.\frac{1}{81}(\because y=x^{\frac{1}{4}}and\mathrm{\Delta }x=\frac{1}{81}) \\ =\frac{1}{4(\frac{16}{81}{)}^{\frac{3}{4}}}.\frac{1}{81}=\frac{27}{4\times 8}.\frac{1}{81}=\frac{1}{96} \end{gathered}$$
Now,
$(\frac{17}{81}{)}^{\frac{1}{4}}=\mathrm{\Delta }y+\frac{2}{3}=\frac{1}{96}+\frac{2}{3}=\frac{65}{96}=0.677$
Hence, $(\frac{17}{81}{)}^{\frac{1}{4}}$ is approximately equal to 0.677

Question:1 Using differentials, find the approximate value of each of the following:
$(33{)}^{-1/5}$

Answer:

Let $y=x^{\frac{-1}{5}}$ and $x=32and\mathrm{\Delta }x=1$
$\mathrm{\Delta }y=(x+\mathrm{\Delta }x{)}^{\frac{-1}{5}}-x^{\frac{-1}{5}}$
$=(32+1{)}^{\frac{-1}{5}}-(32{)}^{\frac{-1}{5}}$
$(33{)}^{\frac{-1}{4}}=\mathrm{\Delta }y+\frac{1}{2}$
Now, we know that $\mathrm{\Delta }y$ is approximately equals to dy
So,
$$\begin{gathered} dy=\frac{dy}{dx}.\mathrm{\Delta }x \\ =\frac{-1}{5x^{\frac{6}{5}}}.1(\because y=x^{\frac{-1}{5}}and\mathrm{\Delta }x=1) \\ =\frac{-1}{5(32{)}^{\frac{6}{5}}}.1=\frac{-1}{5\times 64}.1=\frac{-1}{320} \end{gathered}$$
Now,
$(33{)}^{\frac{-1}{5}}=\mathrm{\Delta }y+\frac{1}{2}=\frac{-1}{320}+\frac{1}{2}=\frac{159}{320}=0.497$
Hence, $(33{)}^{\frac{-1}{5}}$ is approximately equals to 0.497

Question:2. Show that the function given by $f(x)=\frac{\log x}{x}$ has maximum at x = e.

Answer:

Given function is
$f(x)=\frac{\log x}{x}$
$f^{{}^{\prime }}(x)=\frac{1}{x}.\frac{1}{x}+logx\frac{-1}{x^2}=\frac{1}{x^2}(1-\log x)$
$$\begin{gathered} f^{{}^{\prime }}(x)=0 \\ \frac{1}{x^2}(1-\log x)=0 \\ \frac{1}{x^2}\ne 0Sologx=1\Rightarrow x=e \end{gathered}$$
Hence, x =e is the critical point
Now,
$$\begin{gathered} f^{{}^{″}}(x)=\frac{-2x}{x^3}(1-\log x)+\frac{1}{x^2}(-\frac{1}{x})=\frac{1}{x^3}(-2x+2xlogx-1) \\ {f}^{{}^{″}(e)}=\frac{-1}{e^3}<0 \end{gathered}$$
Hence, x = e is the point of maxima

Question:3(i) Find the intervals in which the function f given by $f(x)=\frac{4\sin x-2x-x\cos x}{2+\cos x}$ is

increasing

Answer:

Given function is
$f(x)=\frac{4\sin x-2x-x\cos x}{2+\cos x}$
$f^{{}^{\prime }}(x)=\frac{(4\cos x-2-\cos x+x\sin x)(2+\cos x)-(4\sin x-2x-x\cos x)(-\sin x)}{(2+\cos x{)}^2}$
$=\frac{4\cos x-{\cos }^{2}x}{2+\cos x}$
$$\begin{gathered} f^{{}^{\prime }}(x)=0 \\ \frac{4\cos x-{\cos }^{2}x}{2+\cos x}=0 \\ \cos x(4-\cos x)=0 \\ \cos x=0and\cos x=4 \end{gathered}$$
But $\cos x\ne 4$
So,
$$\begin{gathered} \cos x=0 \\ x=\frac{\pi }{2}and\frac{3\pi }{2} \end{gathered}$$
Now three ranges are there $(0,\frac{\pi }{2}),(\frac{\pi }{2},\frac{3\pi }{2})and(\frac{3\pi }{2},2\pi )$
In interval $(0,\frac{\pi }{2})and(\frac{3\pi }{2},2\pi )$ , $f^{{}^{\prime }}(x)>0$

Hence, the given function $f(x)=\frac{4\sin x-2x-x\cos x}{2+\cos x}$ is increasing in the interval $(0,\frac{\pi }{2})and(\frac{3\pi }{2},2\pi )$
in interval $,(\frac{\pi }{2},\frac{3\pi }{2}),f^{{}^{\prime }}(x)<0$ so function is decreasing in this inter

Question:3 (ii) Find the intervals in which the function f given by f x is equal to

$f(x)=\frac{4\sin x-2x-x\cos x}{2+\cos x}$ is

decreasing

Answer:

Given function is
$f(x)=\frac{4\sin x-2x-x\cos x}{2+\cos x}$
$f^{{}^{\prime }}(x)=\frac{(4\cos x-2-\cos x+x\sin x)(2+\cos x)-(4\sin x-2x-x\cos x)(-\sin x)}{(2+\cos x{)}^2}$
$=\frac{4\cos x-{\cos }^{2}x}{2+\cos x}$
$$\begin{gathered} f^{{}^{\prime }}(x)=0 \\ \frac{4\cos x-{\cos }^{2}x}{2+\cos x}=0 \\ \cos x(4-\cos x)=0 \\ \cos x=0and\cos x=4 \end{gathered}$$
But $\cos x\ne 4$
So,
$$\begin{gathered} \cos x=0 \\ x=\frac{\pi }{2}and\frac{3\pi }{2} \end{gathered}$$
Now three ranges are there $(0,\frac{\pi }{2}),(\frac{\pi }{2},\frac{3\pi }{2})and(\frac{3\pi }{2},2\pi )$
In interval $(0,\frac{\pi }{2})and(\frac{3\pi }{2},2\pi )$ , $f^{{}^{\prime }}(x)>0$

Hence, given function $f(x)=\frac{4\sin x-2x-x\cos x}{2+\cos x}$ is increasing in interval $(0,\frac{\pi }{2})and(\frac{3\pi }{2},2\pi )$
in interval $,(\frac{\pi }{2},\frac{3\pi }{2}),f^{{}^{\prime }}(x)<0$
Hence, given function $f(x)=\frac{4\sin x-2x-x\cos x}{2+\cos x}$ is decreasing in interval $,(\frac{\pi }{2},\frac{3\pi }{2})$

Question:4(i) Find the intervals in which the function f given by $f(x)=x^3+\frac{1}{x^3},x\ne 0$

Increasing

Answer:

Given function is
$f(x)=x^3+\frac{1}{x^3}$
$$\begin{gathered} f^{{}^{\prime }}(x)=3x^2+\frac{-3x^2}{x^4} \\ {f}^{{}^{\prime }}(x)=0 \\ 3x^2+\frac{-3x^2}{x^4}=0 \\ {x}^4=1 \\ x=\pm 1 \end{gathered}$$
Hence, three intervals are their $(-\mathrm{\infty },-1),(-1,1)and(1,\mathrm{\infty })$
In interval $(-\mathrm{\infty },-1)and(1,\mathrm{\infty }),f^{{}^{\prime }})x>0$
Hence, given function $f(x)=x^3+\frac{1}{x^3}$ is increasing in interval $(-\mathrm{\infty },-1)and(1,\mathrm{\infty })$
In interval (-1,1) , $f^{{}^{\prime }}(x)<0$
Hence, given function $f(x)=x^3+\frac{1}{x^3}$ is decreasing in interval (-1,1)

Question:4(ii) Find the intervals in which the function f given by $f(x)=x^3+\frac{1}{x^3},x\ne 0$

decreasing

Answer:

Given function is
$f(x)=x^3+\frac{1}{x^3}$
$$\begin{gathered} f^{{}^{\prime }}(x)=3x^2+\frac{-3x^2}{x^4} \\ {f}^{{}^{\prime }}(x)=0 \\ 3x^2+\frac{-3x^2}{x^4}=0 \\ {x}^4=1 \\ x=\pm 1 \end{gathered}$$
Hence, three intervals are their $(-\mathrm{\infty },-1),(-1,1)and(1,\mathrm{\infty })$
In interval $(-\mathrm{\infty },-1)and(1,\mathrm{\infty }),f^{{}^{\prime }})x>0$
Hence, given function $f(x)=x^3+\frac{1}{x^3}$ is increasing in interval $(-\mathrm{\infty },-1)and(1,\mathrm{\infty })$
In interval (-1,1) , $f^{{}^{\prime }}(x)<0$
Hence, given function $f(x)=x^3+\frac{1}{x^3}$ is decreasing in interval (-1,1)

Question:5 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.

Answer:

Given the equation of the ellipse
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Now, we know that ellipse is symmetrical about x and y-axis. Therefore, let's assume coordinates of A = (-n,m) then,
Now,
Put(-n,m) in equation of ellipse
we will get
$m=\pm \frac{b}{a}.\sqrt{a^2-n^2}$
Therefore, Now
Coordinates of A = $(-n,\frac{b}{a}.\sqrt{a^2-n^2})$
Coordinates of B = $(-n,-\frac{b}{a}.\sqrt{a^2-n^2})$
Now,
Length AB(base) = $2\frac{b}{a}.\sqrt{a^2-n^2}$
And height of triangle ABC = (a+n)
Now,
Area of triangle = $\frac{1}{2}bh$
$A=\frac{1}{2}.\frac{2b}{a}.\sqrt{a^2-n^2}.(a+n)=ab\sqrt{a^2-n^2}+bn\sqrt{a^2-n^2}$
Now,
$\frac{dA}{dn}=\frac{-abn}{\sqrt{a^2-n^2}}+n\sqrt{a^2-n^2}-\frac{b{n}^2}{\sqrt{a^2-n^2}}$
Now,
$$\begin{gathered} \frac{dA}{dn}=0 \\ \frac{-abn}{\sqrt{a^2-n^2}}+n\sqrt{a^2-n^2}-\frac{b{n}^2}{\sqrt{a^2-n^2}}=0 \\ -abn+n(a^2-n^2)-b{n}^2=0 \\ \Rightarrow n=-a,\frac{a}{2} \end{gathered}$$
but n cannot be zero
therefore, $n=\frac{a}{2}$
Now, at $n=\frac{a}{2}$
$\frac{d^{2}A}{d{n}^2}<0$
Therefore, $n=\frac{a}{2}$ is the point of maxima
Now,
$b=2\frac{b}{a}.\sqrt{a^2-(\frac{a}{2}{)}^2}=\sqrt{3}b$
$h=(a+n)=a+\frac{a}{2}=\frac{3a}{2}$
Now,
Therefore, Area (A) $=\frac{1}{2}bh=\frac{1}{2}\sqrt{3}b\frac{3a}{2}=\frac{3\sqrt{3}ab}{4}$

Question:6 A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2 m and volume is 8 m3. If building of tank costs Rs 70 per sq metres for the base and Rs 45 per square metre for sides. What is the cost of least expensive tank?

Answer:

Let l , b and h are length , breath and height of tank
Then, volume of tank = l X b X h = 8 $m^3$
h = 2m (given)
lb = 4 = $l=\frac{4}{b}$
Now,
area of base of tank = l X b = 4
area of 4 side walls of tank = hl + hl + hb + hb = 2h(l + b)
Total area of tank (A) = 4 + 2h(l + b)
$A(b)=4+2h(\frac{4}{b}+b)$
$$\begin{gathered} A^{{}^{\prime }}(b)=2h(\frac{-4}{b^2}+1) \\ {A}^{{}^{\prime }}(b)=0 \\ 2h(\frac{-4}{b^2}+1)=0 \\ {b}^2=4 \\ b=2 \end{gathered}$$
Now,
$$\begin{gathered} A^{{}^{″}}(b)=2h(\frac{-4\times -2b}{b^3}) \\ {A}^{{}^{″}}(2)=8>0 \end{gathered}$$
Hence, b = 2 is the point of minima
$l=\frac{4}{b}=\frac{4}{2}=2$
So, l = 2 , b = 2 and h = 2 m
Area of base = l X B = 2 X 2 = $4m^2$
building of tank costs Rs 70 per sq metres for the base
Therefore, for $4m^2$ Rs = 4 X 70 = 280 Rs
Area of 4 side walls = 2h(l + b)
= 2 X 2(2 + 2) = $16m^2$
building of tank costs Rs 45 per square metre for sides
Therefore, for $16m^2$ Rs = 16 X 45 = 720 Rs
Therefore, total cost for making the tank is = 720 + 280 = 1000 Rs

Question:7 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.

Answer:

It is given that
the sum of the perimeter of a circle and square is k = $2\pi r+4a=k\Rightarrow a=\frac{k-2\pi r}{4}$
Let the sum of the area of a circle and square(A) = $\pi r^2+a^2$
$A=\pi r^2+(\frac{k-2\pi r}{4}{)}^2$
$$\begin{gathered} A^{{}^{\prime }}(r)=2\pi r+2(\frac{k-2\pi r}{16})(-2\pi ) \\ {A}^{{}^{\prime }}(r)=0 \\ 2\pi (\frac{8r-k-2\pi r}{8})=0 \\ r=\frac{k}{8-2\pi } \end{gathered}$$
Now,
$$\begin{gathered} A^{{}^{″}}(r)=2\pi (\frac{8-2\pi }{8})=0 \\ {A}^{{}^{″}}(\frac{k}{8-2\pi })>0 \end{gathered}$$
Hence, $r=\frac{k}{8-2\pi }$ is the point of minima
$a=\frac{k-2\pi r}{4}=\frac{k-2\pi \frac{k}{8-2\pi }}{4}=2\frac{k}{8-2\pi }=2r$
Hence proved that the sum of their areas is least when the side of the square is double the radius of the circle

Question:8 A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window to admit maximum light through the whole opening.

Answer:

Let l and bare the length and breadth of rectangle respectively and r will be the radius of circle $(r=\frac{l}{2})$
The total perimeter of window = perimeter of rectangle + perimeter of the semicircle

= $l+2b+\pi \frac{l}{2}$

$$\begin{gathered} l+2b+\pi \frac{l}{2}=10 \\ l=\frac{2(10-2b)}{2+\pi } \end{gathered}$$
Area of window id given by (A) = $lb+\frac{\pi }{2}{\left(\frac{l}{2}\right)}^2$
$=\frac{2(10-2b)}{2+\pi }b+\frac{\pi }{2}{\left(\frac{10-2b}{2+\pi }\right)}^2$
$A^{{}^{\prime }}(b)=\frac{20-8b}{2+\pi }+\frac{\pi }{2}.2(\frac{10-2b}{2+\pi }).\frac{(-2)}{2+\pi }$
$$\begin{gathered} =\frac{20-8b}{2+\pi }-2\pi (\frac{10-2b}{(2+\pi {)}^2}) \\ {A}^{{}^{\prime }}(b)=0 \\ \frac{20-8b}{2+\pi }=2\pi (\frac{10-2b}{(2+\pi {)}^2}) \\ 40+20\pi -16b-8\pi b=20\pi -4\pi b \\ 40=4b(\pi +4) \\ b=\frac{10}{\pi +4} \end{gathered}$$
Now,
$$\begin{gathered} A^{{}^{″}}(b)=\frac{-8}{2+\pi }+\frac{4\pi }{(2+\pi {)}^2}=\frac{-16-8\pi +4\pi }{(2+\pi {)}^2}=\frac{-16-4\pi }{(2+\pi {)}^2} \\ {A}^{{}^{″}}(\frac{10}{\pi +4})<0 \end{gathered}$$
Hence, b = 5/2 is the point of maxima
$l=\frac{2(10-2b)}{2+\pi }=\frac{2(10-2.\frac{10}{4+\pi })}{2+\pi }=\frac{20}{4+\pi }$
$r=\frac{l}{2}=\frac{20}{2(4+\pi )}=\frac{10}{4+\pi }$
Hence, these are the dimensions of the window to admit maximum light through the whole opening

Question:9 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 $(a^{\frac{2}{3}}+b^{\frac{2}{3}}{)}^{\frac{3}{2}}$

Answer:

It is given that
A point on the hypotenuse of a triangle is at a distance a and b from the sides of the triangle

Let the angle between AC and BC is $\theta$
So, the angle between AD and ED is also $\theta$
Now,
CD = $bcosec\theta$
And
AD = $a\sec \theta$
AC = H = AD + CD
= $a\sec \theta$ + $bcosec\theta$
$$\begin{gathered} \frac{dH}{d\theta }=a\sec \theta \tan \theta -b\cot \theta cosec\theta \\ \frac{dH}{d\theta }=0 \\ a\sec \theta \tan \theta -b\cot \theta cosec\theta =0 \\ a\sec \theta \tan \theta =b\cot \theta cosec\theta \\ a{\sin }^3\theta =b{\cos }^3\theta \\ {\tan }^3\theta =\frac{b}{a} \\ \tan \theta ={\left(\frac{b}{a}\right)}^{\frac{1}{3}} \end{gathered}$$
Now,
$\frac{d^{2}H}{d{\theta }^2}>0$
When $\tan \theta ={\left(\frac{b}{a}\right)}^{\frac{1}{3}}$
Hence, $\tan \theta ={\left(\frac{b}{a}\right)}^{\frac{1}{3}}$ is the point of minima
$\sec \theta =\frac{a\sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}}{a^{\frac{1}{3}}}$ and $cosec\theta =\frac{b\sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}}{b^{\frac{1}{3}}}$

AC = $\frac{a\sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}}{a^{\frac{1}{3}}}+$ $\frac{b\sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}}{b^{\frac{1}{3}}}$ = $(a^{\frac{2}{3}}+b^{\frac{2}{3}}{)}^{\frac{3}{2}}$
Hence proved

Question:10 Find the points at which the function f given by $f(x)=(x-2{)}^4(x+1{)}^3$ has (i) local maxima (ii) local minima (iii) point of inflexion

Answer:

Given function is
$f(x)=(x-2{)}^4(x+1{)}^3$
$$\begin{gathered} f^{{}^{\prime }}(x)=4(x-2{)}^3(x+1{)}^3+3(x+1{)}^2(x-2{)}^4 \\ {f}^{{}^{\prime }}(x)=0 \\ 4(x-2{)}^3(x+1{)}^3+3(x+1{)}^2(x-2{)}^4=0 \\ (x-2{)}^3(x+1{)}^2(4(x+1)+3(x-2)) \\ x=2,x=-1andx=\frac{2}{7} \end{gathered}$$
Now, for value x close to $\frac{2}{7}$ and to the left of $\frac{2}{7}$ , $f^{{}^{\prime }}(x)>0$ ,and for value close to $\frac{2}{7}$ and to the right of $\frac{2}{7}$ $f^{{}^{\prime }}(x)<0$
Thus, point x = $\frac{2}{7}$ is the point of maxima
Now, for value x close to 2 and to the Right of 2 , $f^{{}^{\prime }}(x)>0$ ,and for value close to 2 and to the left of 2 $f^{{}^{\prime }}(x)<0$
Thus, point x = 2 is the point of minima
There is no change in the sign when the value of x is -1
Thus x = -1 is the point of inflexion

Question:12 Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is 4r/3

Answer:

The volume of a cone (V) = $\frac{1}{3}\pi R^{2}h$
The volume of the sphere with radius r = $\frac{4}{3}\pi r^3$
By Pythagoras theorem in $\mathrm{\Delta }ADC$ we ca say that
$$\begin{gathered} O{D}^2=r^2-R^2 \\ OD=\sqrt{r^2-R^2} \\ h=AD=r+OD=r+\sqrt{r^2-R^2} \end{gathered}$$
V = $$\begin{gathered} \frac{1}{3}\pi R^2(r+\sqrt{r^2+R^2}) \\ =\frac{1}{3}\pi R^{2}r+\frac{1}{3}\pi R^2\sqrt{r^2+R^2} \end{gathered}$$
$$\begin{gathered} \frac{1}{3}\pi R^2(r+\sqrt{r^2-R^2}) \\ {V}^{{}^{\prime }}(R)=\frac{2}{3}\pi Rr+\frac{2}{3}\pi R\sqrt{r^2-R^2}+\frac{1}{3}\pi R^2.\frac{-2R}{2\sqrt{r^2-R^2}} \\ {V}^{{}^{\prime }}(R)=0 \\ \frac{1}{3}\pi R(2r+2\sqrt{r^2-R^2}-\frac{R^2}{\sqrt{r^2-R^2}})=0 \\ \frac{1}{3}\pi R\left(\frac{2r\sqrt{r^2-R^2}+2r^2-2R^2-R^2}{\sqrt{r^2-R^2}}\right)=0 \\ R\ne 0So, \\ 2r\sqrt{r^2-R^2}=3R^2-2r^2 \\ Squarebothsides \\ 4r^4-4r^2{R}^2=9R^4+4r^4-12R^2{r}^2 \\ 9R^4-8R^2{r}^2=0 \\ {R}^2(9R^2-8r^2)=0 \\ R\ne 0So,9R^2=8r^2 \\ R=\frac{2\sqrt{2}r}{3} \end{gathered}$$
Now,
$$\begin{gathered} V^{{}^{″}}(R)=\frac{2}{3}\pi r+\frac{2}{3}\pi \sqrt{r^2-R^2}+\frac{2}{3}\pi R.\frac{-2R}{2\sqrt{r^2-R^2}}-\frac{3\pi R^2}{\sqrt{r^2-R^2}}-\frac{(-1)(-2R)}{(r^2+R^2)\frac{3}{2}} \\ {V}^{{}^{″}}(\frac{2\sqrt{2}r}{3})<0 \end{gathered}$$
Hence, the point $R=\frac{2\sqrt{2}r}{3}$ is the point of maxima
$h=r+\sqrt{r^2-R^2}=r+\sqrt{r^2-\frac{8r^2}{9}}=r+\frac{r}{3}=\frac{4r}{3}$
Hence, the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is $\frac{4r}{3}$

Question:13 Let f be a function defined on [a, b] such that $f(x)>0$ , for all $xϵ(a,b)$ . Then prove that f is an increasing function on (a, b).

Answer:

Let's do this question by taking an example
suppose
$f(x)=x^3>0,(a.b)$
Now, also
$f{}^{\prime }(x)=3x^2>0,(a,b)$
Hence by this, we can say that f is an increasing function on (a, b)

Question:14 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.

Answer:

The volume of the cylinder (V) = $\pi r^{2}h$
By Pythagoras theorem in $\mathrm{\Delta }OAB$
$OA=\sqrt{R^2-r^2}$
h = 2OA
$h=2\sqrt{R^2-r^2}$
$V=2\pi r^2\sqrt{R^2-r^2}$
$$\begin{gathered} V^{{}^{\prime }}(r)=4\pi r\sqrt{R^2-r^2}+2\pi r^2.\frac{-2r}{2\sqrt{R^2-r^2}} \\ {V}^{{}^{\prime }}(r)=0 \\ 4\pi r\sqrt{R^2-r^2}-\frac{2\pi r^3}{\sqrt{R^2-r^2}}=0 \\ 4\pi r(R^2-r^2)-2\pi r^3=0 \\ 6\pi r^3=4\pi r{R}^2 \\ r=\frac{\sqrt{6}R}{3} \end{gathered}$$
Now,
$$\begin{gathered} V^{{}^{″}}(r)=4\pi \sqrt{R^2-r^2}+4\pi r.\frac{-2r}{2\sqrt{R^2-r^2}}-\frac{6\pi r^2}{\sqrt{R^2-r^2}}.\frac{(-1)-2r}{2(R^2-r^2)\frac{3}{2}} \\ {V}^{{}^{″}}(\frac{\sqrt{6}R}{3})<0 \end{gathered}$$
Hence, the point $r=\frac{\sqrt{6}R}{3}$ is the point of maxima
$h=2\sqrt{R^2-r^2}==2\sqrt{R^2-\frac{2R^2}{3}}=\frac{2R}{\sqrt{3}}$
Hence, the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is $\frac{2R}{\sqrt{3}}$
and maximum volume is
$V=\pi r^{2}h=\pi \frac{2R^2}{3}.\frac{2R}{\sqrt{3}}=\frac{4\pi R^3}{3\sqrt{3}}$

Question:15 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 a is one-third that of the cone and the greatest volume of cylinder is

$\frac{4}{27}\pi h^3{\tan }^2\alpha$

Answer:

Let's take radius and height of cylinder = r and h ' respectively
Let's take radius and height of cone = R and h respectively

Volume of cylinder = $\pi r^2{h}^{\prime }$
Volume of cone = $\frac{1}{3}\pi R^{2}h$
Now, we have
$R=h\tan a$
Now, since $\mathrm{\Delta }AOG\text{and}\mathrm{\Delta }CEG$ are similar
$\frac{OA}{OG}=\frac{CE}{EG}$
$\frac{h}{R}=\frac{h^{\prime }}{R-r}$
$h^{\prime }=\frac{h(R-r)}{R}$
$h^{\prime }=\frac{h(h\tan a-r)}{h\tan a}=\frac{h\tan a-r}{\tan a}$
Now,
$V=\pi r^2{h}^{\prime }=\pi r^2.\frac{h\tan a-r}{\tan a}=\pi r^{2}h-\frac{\pi r^3}{\tan a}$
Now,
$$\begin{gathered} \frac{dV}{dr}=2\pi rh-\frac{3\pi r^2}{\tan a} \\ \frac{dV}{dr}=0 \\ 2\pi rh-\frac{3\pi r^2}{\tan a}=0 \\ 2\pi rh=\frac{3\pi r^2}{\tan a} \\ r=\frac{2h\tan a}{3} \end{gathered}$$
Now,
$\frac{d^{2}V}{d{r}^2}=2\pi h-\frac{6\pi r}{\tan a}$
at $r=\frac{2h\tan a}{3}$
$\frac{d^{2}V}{d{r}^2}=2\pi h-4\pi h<0$
Hence, $r=\frac{2h\tan a}{3}$ is the point of maxima
$h^{\prime }=\frac{h\tan a-r}{\tan a}=\frac{h\tan a-\frac{2h\tan a}{3}}{\tan a}=\frac{1}{3}h$
Hence proved
Now, Volume (V) at $h^{\prime }=\frac{1}{3}h$ and $r=\frac{2h\tan a}{3}$ is
$V=\pi r^2{h}^{\prime }=\pi {\left(\frac{2h\tan a}{3}\right)}^2.\frac{h}{3}=\frac{4}{27}.\pi h^3{\tan }^{2}a$
hence proved

Question:16 A cylindrical tank of radius 10 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) 1 m/h

(B) 0.1 m/h

(C) 1.1 m/h

(D) 0.5 m/h

Answer:

It is given that
$\frac{dV}{dt}=314m^3/h$
Volume of cylinder (V) = $\pi r^{2}h=100\pi h(\because r=10m)$
$$\begin{gathered} \frac{dV}{dt}=100\pi \frac{dh}{dt} \\ 314=100\pi \frac{dh}{dt} \\ \frac{dh}{dt}=\frac{3.14}{\pi }=1m/h \end{gathered}$$
Hence, (A) is correct answer