RD Sharma Class 12 Exercise 17.5 Maxima And Minima Solutions Maths - Download PDF Free Online

# RD Sharma Class 12 Exercise 17.5 Maxima And Minima Solutions Maths - Download PDF Free Online

Edited By Satyajeet Kumar | Updated on Jan 21, 2022 02:40 PM IST

RD Sharma class 12th exercise 17.5 is evidently one of the most trusted and beneficial books for students in high school. It covers comprehensive knowledge on the subject and imparts accurate information on every concept. This guidance aids students to gain clarity on complex concepts and prepares them to score high in their exams.
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## Maxima and Minima Excercise:17.5

Maxima and minima exercise 17.5 question 1

$x= \frac{15}{2}, y =\frac{15}{2}$
Hint: For maximum or minimum value of z must have $\frac{dz}{dx}=0$
Given:$x+y =15$
Solution:$x+y =15$ ..........(1)
Now,
$z= x^2+y^2$
$z= x^2+(15-x)^2$ (from equation 1)
$z= x^2+x^2+225 -30x$ $[(a-b)^2=a^2+b^2-2ab]$
$\frac{dz}{dx}=4x-30$
For maximum or minimum value of z
$\frac{dz}{dx}=0$
$4x-30=0$
$x=\frac{15}{2}$
$\frac{d^2z}{dx^2}=4>0$
Substituting $x= \frac{15}{2}$ in (1)
$y= \frac{15}{2}$
$\therefore$ z is minimum when x = y = $\frac{15}{2}$

### Maxima and minima exercise 17.5 question 2

x= 32,33
Hint: For maximum or minimum value of z must have $\frac{dz}{dx}=0$
Given: 64 divide into two parts, sum of cubes of two parts is minimum
Solution: Suppose 64 divide in two parts x and 64 - x then,
$z = x^3+(64-x)^3$
$\frac{dz}{dx}= 3x^2+3(64-x)^2$
For maximum and minimum value of z
$\frac{dz}{dx}=0$
$3x^2+3(64-x)^2=0$
$3x^2=3(64-x)^2$
$x^2=x^2+4096-128x$
$x=\frac{4096}{128}$
$x=32$
$\frac{d^2z}{dx^2}=6x+6(64-x)$
$=384>0$
$\therefore$ z is minimum when 64 is divided into two parts 32 and 32

## Maxima and minima exercise 17.5 question 3

$\left (\frac{1}{2}-\frac{1}{\sqrt{3}} \right )\text{ and} \; \frac{1}{\sqrt{3}}$
Hint: For maximum or minimum value of z must have $\frac{dz}{dx}=0$
Given:$x,y >-2$ and $x+y =\frac{1}{2}$
Solution: Let the numbers be x and y then
$x,y >-2$ and $x+y =\frac{1}{2}$ .....(1)
\begin{aligned} &\text { Now, }\\ &z=x+y^{3}\\ &z=x+\left(\frac{1}{2}-x\right)^{3} \quad[\text { from }(1)]\\ &\frac{d z}{d x}=1+3\left(\frac{1}{2}-x\right)^{2} \end{aligned}
For maximum or minimum value of z
$\frac{dz}{dx}=0$
\begin{aligned} &1+3\left(\frac{1}{2}-x\right)^{2}=0 \\ &\left(\frac{1}{2}-x\right)^{2}=\frac{1}{3} \\ &\left(\frac{1}{2}-x\right)=\pm \frac{1}{\sqrt{3}} \\ &x=\frac{1}{2} \pm \frac{1}{\sqrt{3}} \end{aligned}
\begin{aligned} &\frac{d^{2} z}{d x^{2}}=3-6\left(\frac{1}{2}+\frac{1}{\sqrt{3}}\right) \\ &=\frac{-6}{\sqrt{3}}<0 \end{aligned}
$\therefore$ z is minimum when $x=\frac{1}{2}+\frac{1}{\sqrt{3}}$
At $x=\frac{1}{2}-\frac{1}{\sqrt{3}}$
\begin{aligned} &\frac{d^{2} z}{d x^{2}}=3-6\left(\frac{1}{2}-\frac{1}{\sqrt{3}}\right) \\ &=\frac{-6}{\sqrt{3}}>0 \end{aligned}
$\therefore$ z is minimum when $x=\frac{1}{2}-\frac{1}{\sqrt{3}}$
$x+y = \frac{1}{2}$
Substituting the value in (1)
\begin{aligned} &y=\frac{-1}{2}+\frac{1}{\sqrt{3}}+\frac{1}{2} \\ &y=\frac{1}{\sqrt{3}} \end{aligned}
So the two required numbers are $\left (\frac{1}{2}-\frac{1}{\sqrt{3}} \right )\text{ and} \; \frac{1}{\sqrt{3}}$
x = 6 and y =9
Hint: For maximum or minimum value of z must have $\frac{dz}{dx}=0$
Given: x + y =15
Solution: Let the two numbers be x any y then,
x + y = 15 .......(1)
Now,
$z = x^2y^3$
$z = x^2(15-x)^3$ from (1)
$\frac{d z}{d x}=2 x(15-x)^{3}-3 x^{2}(15-x)^{2}$
For minimum and maximum value of z
\begin{aligned} &\frac{d z}{d x}=0 \\ &2 x(15-x)^{3}-3 x^{2}(15-x)^{2}=0 \\ &2 x(15-x)=3 x^{2} \\ &30 x-2 x^{2}=3 x^{2} \\ &30 x=5 x^{2} \\ &x=6 \text { and } y=9 \end{aligned}
$\frac{d^{2} z}{d x^{2}}=2(15-x)^{3}-6 x(15-x)^{2}-6 x(15-x)^{2}+6 x^{2}(15-x)$
At x =6
\begin{aligned} &\frac{d^{2} z}{d x^{2}}=2(9)^{3}-36(9)^{2}-36(9)^{2}+6(36)(9) \\ &\frac{d^{2} z}{d x^{2}}=-2430<1 \end{aligned}
Thus z is maximum when x =6 and y =9

## Maxima and minima exercise 17.5 question 4

x = 6 and y =9
Hint: For maximum or minimum value of z must have $\frac{dz}{dx}=0$
Given: x + y =15
Solution: Let the two numbers be x any y then,
x + y = 15 .......(1)
Now,
$z = x^2y^3$
$z = x^2(15-x)^3$ from (1)
$\frac{d z}{d x}=2 x(15-x)^{3}-3 x^{2}(15-x)^{2}$
For minimum and maximum value of z
\begin{aligned} &\frac{d z}{d x}=0 \\ &2 x(15-x)^{3}-3 x^{2}(15-x)^{2}=0 \\ &2 x(15-x)=3 x^{2} \\ &30 x-2 x^{2}=3 x^{2} \\ &30 x=5 x^{2} \\ &x=6 \text { and } y=9 \end{aligned}
$\frac{d^{2} z}{d x^{2}}=2(15-x)^{3}-6 x(15-x)^{2}-6 x(15-x)^{2}+6 x^{2}(15-x)$
At x =6
\begin{aligned} &\frac{d^{2} z}{d x^{2}}=2(9)^{3}-36(9)^{2}-36(9)^{2}+6(36)(9) \\ &\frac{d^{2} z}{d x^{2}}=-2430<1 \end{aligned}
Thus z is maximum when x =6 and y =9

## Maxima and minima exercise 17 5 question 5

$2\left ( \frac{50}{\pi} \right )^{\frac{1}{3}}$
Hint: For maximum or minimum value of we must have $\frac{ds}{dr} =0$
Given: Let r and h be in the radius and height of the cylinder respectively. Then ,
Volume (V) of cylinder = $\pi r^2h$
$100=\pi r^2h$
$h=\frac{100}{\pi r^2}$
Solution:$h=\frac{100}{\pi r^2}$
Surface area (S) of the cylinder = $2\pi r^2+2\pi rh$
\begin{aligned} &=2 \pi r^{2}+2 \pi r \times \frac{100}{\pi r^{2}} \\ &S=2 \pi r^{2}+\frac{200}{r} \\ &\frac{d s}{d r}=4 \pi r-\frac{200}{r^{2}} \end{aligned}
For maximum or minimum,
$\frac{ds}{dr}=0$
$4\pi r-\frac{200}{r^2}=0$
$4\pi r^3=200$
\begin{aligned} &r=\left(\frac{50}{\pi}\right)^{\frac{1}{3}} \\ &\frac{d^{2} s}{d r^{2}}=4 \pi+\frac{400}{r^{3}} \\ &\frac{d^{2} s}{d r^{2}}>0, \text { when } r=\left(\frac{50}{\pi}\right)^{\frac{1}{3}} \end{aligned}
Thus surface area is minimum when $r=\left (\frac{50}{\pi} \right )^{\frac{1}{3}}$
\begin{aligned} h &=\frac{100}{\pi\left(\frac{50}{\pi}\right)^{\frac{1}{3}}} \\ h &=2\left(\frac{50}{\pi}\right)^{\frac{1}{3}} \end{aligned}
Cylinder with radius $=\left (\frac{50}{\pi} \right )^{\frac{1}{3}}$ has min surface area.

## Maxima and minima exercise 17.5 question 6 sub question (i)

$\frac{L}{2}$
Hint: For maximum or minimum value of we must have $\frac{dm}{dx}=0$
Given:$m=\frac{WL}{2}x-\frac{W}{2}x^2$
Solution:$m=\frac{WL}{2}x-\frac{W}{2}x^2$
\begin{aligned} &\frac{d m}{d x}=\frac{W L}{2}-2 \frac{W}{2} x \\ &=\frac{W L}{2}-W x \end{aligned}
For maximum or minimum ,
$\frac{dm}{dx}=0$
$\frac{WL}{2}-Wx=0$
$\frac{WL}{2}=Wx$
$x=\frac{L}{2}$
$\frac{d^2m}{dx^2}=-W<0$
So m is maximum at $x=\frac{L}{2}$

## Maxima and minima exercise 17.5 question 6 sub question (ii)

$\frac{L}{\sqrt{3}}$
Hint: For maximum or minimum value of we must have $\frac{dm}{dx}=0$
Given:$m=\frac{Wx}{3}-\frac{Wx^3}{3L^2}$
Solution:$m=\frac{Wx}{3}-\frac{Wx^3}{3L^2}$
\begin{aligned} &\frac{d m}{d x}=\frac{W}{3}-3 \times \frac{W x^{2}}{3 L^{2}} \\ &=\frac{W}{3}-\frac{W x^{2}}{L^{2}} \end{aligned}
For maximum or minimum,
$\frac{dm}{dx}=0$
\begin{aligned} &\frac{W}{3}-\frac{W x^{2}}{L^{2}}=0 \\ &\frac{W}{3}=\frac{W x^{2}}{L^{2}} \\ &x=\frac{L}{\sqrt{3}} \end{aligned}
$\frac{d^{2} m}{d x^{2}}=\frac{-2 W x}{L^{2}}<0$
So m is minimum at $x=\frac{L}{\sqrt{3}}$

## Maxima and Minima exercise 17.5 question 7.

$\frac{112}{\pi +4}\; \text{and}\; \frac{28 \pi }{\pi +4}$
Hint: For maximum or minimum value of z must have $\frac{dz}{dx}=0$
Given: Suppose the wire which is to be made into a square and circle is cut into two pieces of length x,m and ym respectively.
Solution: x +y = 28 ......(1)
Perimeter of square 4a=x
$a=\frac{x}{4}$
Area of square $a^2$ $=\left ( \frac{x}{4} \right )^4=\frac{x^2}{16}$
Circumference of circle $2\pi r=y$
$r=\frac{y}{2\pi}$
Area of circle $=\pi r^2$
$=\pi \left ( \frac{y}{2\pi } \right )^2=\frac{y^2}{4\pi }$
z = Area of square + Area of circle
\begin{aligned} &=\frac{x^{2}}{16}+\frac{y^{2}}{4 \pi} \\ &=\frac{x^{2}}{16}+\frac{(28-x)^{2}}{4 \pi} \\ &\frac{d z}{d x}=\frac{2 x}{16}-\frac{2(28-x)}{4 \pi} \end{aligned}
For maximum or minimum value of z
$\frac{dz}{dx}=0$
\begin{aligned} &\frac{2 x}{16}-\frac{2(28-x)}{4 \pi}=0 \\ &\frac{x}{4}=\frac{(28-x)}{\pi} \\ &\frac{x \pi}{4}+x=28, x\left(\frac{\pi}{4}+1\right)=28 \end{aligned}
\begin{aligned} x &=\frac{28}{\left(\frac{\pi}{4}+1\right)} \\ x &=\frac{112}{\pi+4} \end{aligned}
$y=28-\frac{112}{\pi +4}$ from equation 1
$\frac{d^{2} z}{d x^{2}}=\frac{1}{8}+\frac{1}{2 \pi}>0$
Thus z is maximum when $\frac{112}{\pi +4}$ and $\frac{28 \pi }{\pi +4}$

### Maxima and Minima exercise 17.5 question 8.

$\frac{80\sqrt{3}}{9+4\sqrt{3}}\; \text{and}\; \frac{180}{9+4\sqrt{3}}$
Hint: For maximum or minimum value of z must have $\frac{dz}{dx}=0$
Given: Suppose wire which has to made into squre and triangle is cut into two pieces x and y respectively.
Solution: x + y = 20 ......(1)
Perimeter of square = 4a = x
$a=\frac{x}{4}$
Area of square $a^2=\left (\frac{x}{4} \right )^2=\frac{x^2}{16}$
Perimeter of triangle 3a=y
$a=\frac{y}{3}$
Area of triangle $=\frac{\sqrt{3}}{4}a^2$
$=\frac{\sqrt{3}}{4}\left (\frac{y}{3} \right )^2=\frac{\sqrt{3}y^2}{36}$
Now,
z =Area of square + area of triangle
\begin{aligned} &z=\frac{x^{2}}{16}+\frac{\sqrt{3} y^{2}}{36} \\ &=\frac{x^{2}}{16}+\frac{\sqrt{3}(20-x)^{2}}{36} \\ &\frac{d z}{d x}=\frac{2 x}{16}+\frac{2 \sqrt{3}(20-x)}{36} \end{aligned}
For minimum or maximum of z
$\frac{dz}{dx}=0$
\begin{aligned} &\frac{2 x}{16}+\frac{\sqrt{3}(20-x)}{18}=0 \\ &\frac{9 x}{4}=\sqrt{3}(20-x) \\ &\frac{9 x}{4}+x \sqrt{3}=20 \sqrt{3} \end{aligned}
\begin{aligned} &x\left(\frac{9}{4}+\sqrt{3}\right)=20 \sqrt{3} \\ &x=\frac{20 \sqrt{3}}{\left(\frac{9}{4}+\sqrt{3}\right)} \\ &x=\frac{80 \sqrt{3}}{(9+4 \sqrt{3})} \end{aligned}
$y=20-\frac{80 \sqrt{3}}{(9+4 \sqrt{3})}$
$y=20-\frac{180}{(9+4 \sqrt{3})}$ from equation (1)
$\frac{d^{2} z}{d x^{2}}=\frac{1}{8}+\frac{\sqrt{3}}{18}>0$
Thus z is maximum when x $\frac{80 \sqrt{3}}{(9+4 \sqrt{3})}$

## Maxima and Minima exercise 17.5 question 9.

x =2r
Hint: For maximum or minimum value of z must have $\frac{dz}{dx}=0$
Given: Let the length of the side of square and radius of the circle x and y respectively. It's given that the sum of the perimeters of square and circle is constant
Solution: $4x + 2\pi r=k$ (k is constant)
$x= \frac{k-2\pi r}{4}$ .......(1)
Now,
$A=x^2+4\pi r^2$
$A=\frac{(k-2\pi r)^2}{16}+\pi r ^2$ .....rom equation (1)
\begin{aligned} &\frac{d A}{d x}=\frac{(k-2 \pi r)^{2}}{16}+\pi r^{2} \\ &=\frac{2(k-2 \pi r)-2 \pi}{16}+2 \pi r \\ &=\frac{(k-2 \pi r)-\pi}{4}+2 \pi r \end{aligned}
\begin{aligned} &=\frac{(k-2 \pi r)-\pi}{4}+2 \pi r=0 \\ &\frac{(k-2 \pi r) \pi}{4}=2 \pi r \\ &k-2 \pi r=8 r \\ &\frac{d^{2} A}{d x^{2}}=\frac{\pi^{2}}{2}+2 \pi>0 \end{aligned}
So the sum of the areas, A is least when $k-2\pi r=8r$
From (1) and (2) we get
$x=\frac{k-2\pi r}{4}$
$x=\frac{8r}{4}=2r$
$\therefore$ Side of square = Diameter of circle
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.

Maxima and Minima exercise 17.5 question 10

$\frac{25}{4}$ square units
Hint: For maxima and minima we must have $f'(x)=0$
Given:$x^2 +y^2=5^2$
Solution:
Let the base of the triangle be x, and height be y
$x^2 +y^2=5^2$
$y^2=25-x^2$
$y=\sqrt{25-x^2}$
As area of triangle A, $A=\frac{1}{2}xy$
\begin{aligned} &A=\frac{1}{2} x \times \sqrt{25-x^{2}} \\ &A(x)=\frac{x \sqrt{25-x^{2}}}{2} \\ &A^{\prime}(x)=\frac{\sqrt{25-x^{2}}}{2}+\frac{x(-2 x)}{4 \sqrt{25-x^{2}}} \end{aligned}
\begin{aligned} &=\frac{\sqrt{25-x^{2}}}{2}+\frac{x^{2}}{2 \sqrt{25-x^{2}}} \\ &=\frac{25-x^{2}-x^{2}}{2 \sqrt{25-x^{2}}} \\ &A^{\prime}(x)=\frac{25-2 x^{2}}{2 \sqrt{25-x^{2}}} \\ &A^{\prime}(x)=0 \end{aligned}
\begin{aligned} &\frac{25-2 x^{2}}{2 \sqrt{25-x^{2}}}=0 \\ &25-2 x^{2}=0 \\ &x=\frac{5}{\sqrt{2}}, y=\sqrt{25-\frac{25}{2}} \\ &x=\frac{5}{\sqrt{2}}, y=\frac{5}{\sqrt{2}} \end{aligned}
Also $A''(x)$ $=\frac{\left[-4 x \sqrt{25-x^{2}}-\frac{\left(25-2 x^{2}\right)(-2 x)}{2 \sqrt{25-x^{2}}}\right]}{25-x^{2}}$
\begin{aligned} &=\frac{\left[\frac{-4 x\left(25-x^{2}\right)+\left(25-2 x^{3}\right)}{2 \sqrt{25-x^{2}}}\right]}{25-x^{2}} \\ &=\frac{-100 x+4 x^{3}+25 x-2 x^{3}}{\left(25-x^{2}\right) \sqrt{25-x^{2}}} \\ &=\frac{-75 x+2 x^{3}}{\left(25-x^{2}\right) \sqrt{25-x^{2}}} \end{aligned}
$A^{\prime \prime}\left(\frac{5}{\sqrt{2}}\right)=\frac{-75\left(\frac{5}{\sqrt{2}}\right)+2\left(\frac{5}{\sqrt{2}}\right)^{3}}{\left(25-\left(\frac{5}{\sqrt{2}}\right)^{2}\right)^{\frac{3}{2}}}<0$
So, $x=\left (\frac{5}{\sqrt{2}} \right )$ is point of maxima
$\therefore$ Largest possible area of triangle,
$=\frac{1}{2}\left(\frac{5}{\sqrt{2}}\right)\left(\frac{5}{\sqrt{2}}\right)=\frac{25}{4}$ square units

Maxima and Minima exercise 17.5 question 11.

$\frac{ab}{2}$
Hint: For maxima or minima we must have $f'(x)=0$
Given: From the question, the area of triangle,
$A= \frac{1}{2}ab\sin \theta$
Solution:$A(\theta)= \frac{1}{2}ab\sin \theta$
$A'(\theta)= \frac{1}{2}ab\cos \theta$
For maxima and minima $A'(\theta)= 0$
$\frac{1}{2}ab \cos \theta =0$
$\cos \theta =0$
$\theta =\frac{\pi}{2}$
Also $A''(\theta)=\frac{1}{2}ab \sin \theta$ or $A''\left ( \frac{\pi}{2} \right )=-\frac{1}{2}ab \sin \frac{\pi}{2}$
$A^{\prime \prime}\left(\frac{\pi}{2}\right)=-\frac{1}{2} a b<0$
i.e,
$\theta =\frac{\pi}{2}$ is teh point of maxima
The maximum area of triangle $\frac{1}{2}ab \sin \frac{\pi}{2}$
$= \frac{ab}{2}$

## Maxima and Minima exercise 17.5 question 12.

432 cm2
Hint: For maxima and minima we must have f'(x) =0
Given: From the question
$V(x)=x(18-2x)^2$
Solution:
Let the square side be cutoff be xcm. Then, the length and breadth of the box will be (18-2x)cm each and height of the box will be xcm
Volume of the box $V(x)=x(18-2x)^2$
\begin{aligned} &V^{\prime}(x)=(18-2 x)^{2}-4 x(18-2 x) \\ &=(18-2 x)^{2}(18-2 x-4 x) \\ &=(18-2 x)^{2}(18-6 x) \end{aligned}
\begin{aligned} &=12(9-x)(3-x) \\ &V^{\prime \prime}(x)=12(-(9-x)-(3-x)) \\ &=-12(9-x+3-x)=-24(6-x) \end{aligned}
For maxima or minima V'(x) =0
$12(9-x)(3-x)=0$
x=9 or x =3
if x=9, then l and b will become 0
So, $x\neq 9$
if $x =3$
Now, $V''(3)= -24(6-3)=-72<0$
x= 3 is the point of maximum
$V(3)= 3(18-6)^3=3 \times 144$
$=432 \; cm^2$

## Maxima and Minima exercise 17.5 question 13.

5cm
Hint: For maxima and minima we must have f'(x) =0
Given: For the question,
$V= (45-2x)(24-2x)x$
Solution: Let the side of the square to be xcm is cutoff.
Then, the l and b, h of the loon will be (45-x), (25-x),xcm
Volume of the box V = (45-x) (25-x)x
$\frac{d V}{d x}=(45-2 x)(24-2 x)-2 x(45-2 x)-2 x(24-2 x)$
$\frac{d V}{d x}=0$
\begin{aligned} &(45-2 x)(24-2 x)-2 x(45-2 x)-2 x(24-2 x)=0 \\ &4 x^{2}+1080-138 x-48 x+4 x^{2}+4 x^{2}-90 x=0 \\ &12 x^{2}-276 x+1080=0 \end{aligned}
Divide by 12
\begin{aligned} &x^{2}-23 x+90=0 \\ &x^{2}-18 x-5 x+90=0 \\ &x(x-15)-5(x-18)=0 \\ &x-18=0 \text { or } x-5=0 \\ &x=18 \text { or } x=5 \end{aligned}
\begin{aligned} &\frac{d^{2} V}{d x^{2}}=24 x-276 \\ &\frac{d^{2} V}{d x^{2}}=120-276=-156<0 \\ &\frac{d^{2} V}{d x^{2}}_{x=18}=432-276=156>0 \end{aligned}
Thus volume of the box is max when x = 5cm.

## Maxima and Minima exercise 17.5 question 14

Rs 1000
Hint: For maxima and minima value of A, we must have $\frac{dA}{dl}=0$
Given: From the question
$l \times b\times 2 =8$
Solution:
Let l,b and h be the length, breadth and height of the tank
h = 2cm
Volume of tank = 8m3
Volume of the tank = $l \times b\times h$
$l \times b\times 2=8$
$lb =4$
$b =\frac{4}{l}$
Area of base $=lb = 4m^3$
Area of 4 walls, $A=2h (l+b)$
\begin{aligned} &\therefore A=4\left(l+\frac{4}{l}\right) \\ &\frac{d A}{d l}=4\left(l-\frac{4}{l^{2}}\right) \\ &\frac{d A}{d l}=0 \\ &4\left(l-\frac{4}{l^{2}}\right)=0 \\ &l=\pm 2 \end{aligned}
The length cannot be negative.
now
\begin{aligned} &\frac{d^{2} A}{d l^{2}}=\frac{32}{l^{3}} \\ &\text { At } l=2 \\ &\frac{d^{2} A}{d l^{2}}=\frac{32}{8}=4>0 \end{aligned}
Thus the area is minimum when l=2,
Cost of the building the base = $Rs 70 \times lb$
$=Rs 70 \times 4$
$=Rs 280$
Cost of the building walls $=Rs\; 2h(l+b)\times 45$
$=Rs\; 90(2)(2+2)$
$=Rs\; 720$
Total cost = Rs (280 +720) = 1000

## Maxima and Minima exercise 17.5 question 15.

$\frac{20}{\pi +4},\frac{10 }{\pi +4}$
Hint: For maxima and minima value of A, we must have $\frac{dA}{dx}=0$
Given:$(x+2y)+\pi \left (\frac{x}{2} \right )=10$
Solution:
Let the dimensions of the rectangular part be x,y
Radius of the semicircle $=\frac{x}{2}$
Total perimeter = 10
$(x+2y)+\pi \left (\frac{x}{2} \right )=10$
$2 y=\left[10-x-\pi\left(\frac{x}{2}\right)\right]$
$y=\frac{1}{2}\left[10-x\left(1+\frac{\pi}{2}\right)\right]$ (1)
Area $A=\frac{\pi}{2}\left(\frac{x}{2}\right)^{2}+x y$
$=\frac{\pi x^{2}}{8}+\frac{x}{2}\left[10-x\left(1+\frac{\pi}{2}\right)\right]$ ..... from (1)
\begin{aligned} &A=\frac{\pi x^{2}}{8}+\frac{10 x}{2}-\frac{x^{2}}{2}\left(1+\frac{\pi}{2}\right) \\ &\frac{d A}{d x}=\frac{\pi x}{4}+\frac{10}{2}-\frac{2 x}{2}\left(1+\frac{\pi}{2}\right) \\ &\frac{d A}{d x}=0 \\ &\frac{\pi x}{4}+\frac{10}{2}-\frac{2 x}{2}\left(1+\frac{\pi}{2}\right)=0 \end{aligned}
\begin{aligned} &x\left[\frac{\pi}{4}-1-\frac{\pi}{2}\right]=-5 \\ &x=\frac{-5}{\left(\frac{-4-\pi}{4}\right)}=\frac{20}{\pi+4} \end{aligned}
Substitute value of x in eqn (1)
\begin{aligned} &y=\frac{1}{2}\left[10-\left(\frac{20}{\pi+4}\right)\left(1+\frac{\pi}{2}\right)\right] \\ &=5-\frac{10(\pi+2)}{\pi+4}=\frac{10}{\pi+4} \end{aligned}
\begin{aligned} &\frac{d^{2} A}{d x^{2}}=\frac{\pi}{4}-\frac{\pi}{2}-1=\frac{\pi-2 \pi-4}{4} \\ &=\frac{-\pi-4}{4}<0 \end{aligned}
Thus the area is max when x$x=\frac{20}{\pi +4}, y=\frac{10}{\pi +4}$
So, $l=\frac{20}{\pi +4}m, b=\frac{10}{\pi +4}m$

## Maxima and Minima exercise 17.5 question 16.

$x = \frac{12}{6-\sqrt{3}} \; \text{and}\; \: y = \frac{18-6\sqrt{3}}{6-\sqrt{3}}$

Hint: For maxima and minima value of A, we must have $\frac{dA}{dx}=0$
Given:$3x + 2y = 12$
Solution:
Let the dimensions of the rectangular part x, y
Perimeter of the window ⇒ $x + y + x + x + y = 12$
$3x + 2y = 12$
$y = \frac{12-3x}{2}$ .....(1)
Area = $xy +\frac{ \sqrt{3}}{4}x^2$
\begin{aligned} &A=x\left(\frac{12-3 x}{2}\right)+\frac{\sqrt{3}}{4} x^{2} \\ &=6 x-\frac{3 x^{2}}{2}+\frac{\sqrt{3}}{4} x^{2} \\ &\frac{d A}{d x}=6-\frac{6 x}{2}+\frac{2 \sqrt{3}}{4} x \\ &\frac{d A}{d x}=6-3 x+\frac{\sqrt{3}}{2} x \\ &=6-x\left(3-\frac{\sqrt{3}}{2}\right) \end{aligned}
$\frac{dA}{dx}=0$
\begin{aligned} &6=x\left(3-\frac{\sqrt{3}}{2}\right) \\ &x=\frac{12}{6-\sqrt{3}} \end{aligned}
Substitute x value in eqn (1)
\begin{aligned} &y=\frac{12-3\left(\frac{12}{6-\sqrt{3}}\right)}{2}=\frac{18-6 \sqrt{3}}{6-\sqrt{3}} \\ &\frac{d^{2} A}{d x^{2}}=-3+\frac{\sqrt{3}}{2}<0 \end{aligned}
Thus the area is max when $x = \frac{12}{6-\sqrt{3}} \; \text{and}\; \: y = \frac{18-6\sqrt{3}}{6-\sqrt{3}}$
$h=\frac{2R}{\sqrt{3}}$
Hint: For maxima and minima value of A, we must have $\frac{dA}{dx}=0$
Given:$h = 2\sqrt{R^2-r^2}$
Solution: Let the height and radius of the base of the cylinder be h and r
$\frac{h^2}{4}+r^2=R^2$
$h = 2\sqrt{R^2-r^2}$ .....(1)
Volume of the cylinder V $=\pi r^2 h$
Square on both the sides
$V^2=\pi ^2 r^4h^2$
$V^2=4\pi ^2 r^4 (R^2-r^2)$ ..........from (1)
Now,
\begin{aligned} &Z=4 \pi^{2}\left(r^{4} R^{2}-r^{6}\right) \\ &\frac{d Z}{d x}=4 \pi^{2}\left(4 r^{3} R^{2}-6 r^{5}\right) \\ &\frac{d Z}{d x}=0 \\ &4 \pi^{2}\left(4 r^{3} R^{2}-6 r^{5}\right)=0 \end{aligned}
\begin{aligned} &4 r^{3} R^{2}=6 r^{5} \\ &6 r^{2}=4 R^{2} \\ &r^{2}=\frac{4 R^{2}}{6} \\ &r=\frac{2 R}{\sqrt{6}} \end{aligned}
Subsitute r value in eqn (1)
\begin{aligned} &h=2 \sqrt{R^{2}-\left(\frac{2 R}{\sqrt{6}}\right)^{2}} \\ &=2 \sqrt{\frac{6 R^{2}-4 R^{2}}{6}}=2 \sqrt{\frac{R^{2}}{3}}=\frac{2 R}{\sqrt{3}} \\ &\frac{d^{2} Z}{d x^{2}}=4 \pi^{2}\left(12 r^{2} R^{2}-30 r^{4}\right) \end{aligned}
\begin{aligned} &=4 \pi^{2}\left(12\left(\frac{2 R}{\sqrt{6}}\right)^{2} R^{2}-30\left(\frac{2 R}{\sqrt{6}}\right)^{4}\right) \\ &=4 \pi^{2}\left(8 R^{4}-\frac{80 R^{4}}{6}\right) \end{aligned}
\begin{aligned} &=4 \pi^{2}\left(\frac{48 R^{4}-80 R^{4}}{6}\right) \\ &=4 \pi^{2}\left(\frac{-16 R^{4}}{3}\right)<0 \end{aligned}
Volume of the cylinder is max when, $h=\frac{2R}{\sqrt{3}}$
$A=r^2$
Hint: For maxima or minima f'(x) must be zero f'(x) = 0
Given: $\frac{x^2}{4}+y^2=r^2$
Solution:$\frac{x^2}{4}+y^2=r^2$
$x^2 +4y^2=4r^2$
$x^2 =4(r^2 -y^2)$ ......(1)
Area = xy
Square on both sides
$A^2=x^2 y^2$
$Z= 4y^2(r^2-y^2)$ ......from (1)
Now,
\begin{aligned} &\frac{d Z}{d y}=8 y r^{2}-16 y^{3} \\ &\frac{d Z}{d y}=0 \\ &8 y r^{2}-16 y^{3}=0 \\ &8 r^{2}=16 y^{2} \\ &y^{2}=\frac{r^{2}}{2}, y=\frac{r}{\sqrt{2}} \end{aligned}
Substitute y value in eqn (1)
\begin{aligned} &x^{2}=4\left(r^{2}-\left(\frac{r}{\sqrt{2}}\right)^{2}\right) \\ &=4\left(r^{2}-\frac{r^{2}}{2}\right) \\ &=4\left(\frac{r^{2}}{2}\right) \end{aligned}
$x^2 = 2r^2$
$x= r\sqrt{2}$
\begin{aligned} &\frac{d^{2} Z}{d y^{2}}=8 r^{2}-48 y^{2} \\ &\frac{d^{2} Z}{d y^{2}}=8 r^{2}-48\left(\frac{r^{2}}{2}\right) \\ &=-16 r^{2}<0 \end{aligned}
Thus the area is maximum when $x=r\sqrt{2}$ and $y= \frac{r}{\sqrt{2}}$
Area = xy
$=r \sqrt{2} \times \frac{r}{\sqrt{2}}$
$A=r ^2$

## Maxima and Minima exercise 17.5 question 19.

$h = \sqrt{2}r$
Hint: For maxima or minima f'(x) must be zero f'(x) = 0
Given: Surface area of conical tank $S=\pi r \sqrt{r^2+h^2}$
$V= \frac{1}{3}\pi r^2 h$
Solution:$V= \frac{1}{3}\pi r^2 h$
\begin{aligned} &h=\frac{3 V}{\pi r^{2}} \\ &\therefore S=\pi r \sqrt{r^{2}+\left(\frac{3 V}{\pi r^{2}}\right)^{2}} \\ &=\frac{1}{r} \sqrt{\pi^{2} r^{6}+9 V^{6}} \end{aligned}
Now,
\begin{aligned} &\frac{d S}{d r}=\frac{d}{d r}\left[\frac{1}{r} \sqrt{\pi^{2} r^{6}+9 V^{6}}\right] \\ &=\frac{1}{r} \frac{6 \pi^{2} r^{5}}{\sqrt{\pi^{2} r^{6}+9 V^{6}}}-\frac{\sqrt{\pi^{2} r^{6}+9 V^{6}}}{r^{2}} \end{aligned}
For minima $\frac{dS}{dr}=0$
\begin{aligned} &\frac{3 \pi^{2} r^{4}}{\sqrt{\pi^{2} r^{6}+9 V^{6}}}=\frac{\sqrt{\pi^{2} r^{6}+9 V^{6}}}{r^{2}} \\ &3 \pi^{2} r^{6}=\pi^{2} r^{6}+9 V^{6} \\ &2 \pi^{2} r^{6}=9 V^{6} \end{aligned}
Substitute V value in equation (1)
\begin{aligned} &2 \pi^{2} r^{6}=9\left(\frac{1}{3} \pi r^{2} h\right)^{6} \\ &2 \pi^{2} r^{6}=\pi^{2} r^{4} h^{2} \\ &2 r^{2}=h^{2} \\ &h=\sqrt{2} r \end{aligned}

Maxima and Minima exercise 17.5 question 20.

$h =\frac{2}{3}$
Hint: For maxima or minima f'(x) must be zero f'(x) = 0
Given: $h = R+\sqrt{R^2-r^2}$
Solution: Let h, r and R be in the height and radius and base of the cone
\begin{aligned} &h=R+\sqrt{R^{2}-r^{2}} \\ &(h-R)^{2}=R^{2}-r^{2} \\ &h^{2}+R^{2}-2 h r=R^{2}-r^{2} \end{aligned}
$r^2=2hR-h^2$ .......(1)
Volume of cone $= \frac{1}{3}\pi r^2h$
$V= \frac{1}{3}\pi h \left (2hR -h^2 \right )$ ..... from (1)
$V= \frac{1}{3}\pi \left (2h^2R -h^3 \right )$
Now,
$\frac{d V}{d h}=\frac{\pi}{3}\left(4 h R-3 h^{2}\right)$
$\frac{d V}{d h}=0$
$\frac{\pi }{3}(4hR-3h^2)=0$
$4hR=3h^2$
$h=\frac{4R}{3}$
Substituting y value in eqn (1)
\begin{aligned} &r^{2}=4\left(r^{2}-\left(\frac{r}{\sqrt{2}}\right)^{2}\right) \\ &=4\left(r^{2}-\frac{r^{2}}{2}\right)=4\left(\frac{r^{2}}{2}\right)=2 r^{2} \end{aligned}
$x=r\sqrt{2}$
\begin{aligned} &\frac{d^{2} V}{d h^{2}}=\frac{\pi}{3}(4 R-6 h) \\ &=\frac{\pi}{3}\left(4 R-6 \times \frac{4 R}{3}\right)=-\frac{4 \pi R}{3}<0 \end{aligned}
Volume is maximum when $h =\frac{4R}{3}$
$h =\frac{2}{3}$ (Diameter of sphere)
Hence proved.

## Maxima and Minima exercise 17.5 question 21 maths.

$r =\left ( \frac{3V}{\pi \sqrt{2}} \right )^{\frac{1}{3}}$ or $V=\frac{\pi r^2 \sqrt{2}}{3}$
Hint: For maxima or minima f'(x) must be zero f'(x) = 0
Given: Volume $=\frac{1}{3}\pi r^2 h$
Solution: $V=\frac{1}{3}\pi r^2 h,h =\frac{3V}{\pi r^2}$
Slant height, $l = \sqrt{h^2+r^2}$
\begin{aligned} &=\sqrt{\left(\frac{3 V}{\pi r^{2}}\right)^{2}+r^{2}} \\ &=\sqrt{\left(\frac{9 V^{2}}{\pi^{2} r^{4}}\right)+r^{2}} \\ &I=\sqrt{\frac{9 V^{2}+\pi^{2} r^{6}}{\pi^{2} r^{4}}}=\frac{\sqrt{9 V^{2}+\pi^{2} r^{6}}}{\pi r^{2}} \end{aligned}
Curved surface area, $e = \pi r l$
\begin{aligned} &C(r)=\frac{\pi r \sqrt{9 V^{2}+\pi^{2} r^{6}}}{\pi r^{2}} \\ &=\frac{\sqrt{9 V^{2}+\pi^{2} r^{6}}}{r} \end{aligned}
\begin{aligned} &C^{\prime}(r)=\frac{\frac{r \times 6 \pi^{2} r^{5}}{2 \sqrt{9 V^{2}+\pi^{2} r^{6}}}-\sqrt{9 V^{2}+\pi^{2} r^{6}}}{r^{2}} \\ &=\frac{\left[\frac{3 \pi^{2} r^{6}-\left(9 V^{2}+\pi^{2} r^{6}\right)}{\sqrt{9 V^{2}+\pi^{2} r^{6}}}\right]}{r^{2}} \end{aligned}
\begin{aligned} &=\frac{3 \pi^{2} r^{6}-9 V^{2}-\pi^{2} r^{6}}{r^{2} \sqrt{9 V^{2}+\pi^{2} r^{6}}} \\ &=\frac{2 \pi^{2} r^{6}-9 V^{2}}{r^{2} \sqrt{9 V^{2}+\pi^{2} r^{6}}} \end{aligned}
$C'(r)=0$
\begin{aligned} &\frac{2 \pi^{2} r^{6}-9 V^{2}}{r^{2} \sqrt{9 V^{2}+\pi^{2} r^{6}}}=0 \\ &2 \pi^{2} r^{6}=9 V^{2} \\ &V^{2}=\frac{2 \pi^{2} r^{6}}{9}, V=\sqrt{\frac{2 \pi^{2} r^{6}}{9}} \end{aligned}
$V=\frac{\pi r^{3} \sqrt{2}}{3} \text { or } r=\left(\frac{3 V}{\pi \sqrt{2}}\right)^{\frac{1}{3}}$
So,
$h =\frac{3}{\pi r^2}\times \frac{\pi r^3 \sqrt{2}}{3}$
$h =r \sqrt{2}$
$\frac{h }{r}= \sqrt{2}$
$\cot \theta = \sqrt{2}, \theta = \cot^{-1}\sqrt{2}$
Since, $r <\left ( \frac{3V}{\pi \sqrt{2}} \right )^{\frac{1}{3}}, C'(r)<0$ and for $r >\left ( \frac{3V}{\pi \sqrt{2}} \right )^{\frac{1}{3}}, C'(r)>0$
Curved surface area for $r =\left ( \frac{3V}{\pi \sqrt{2}} \right )^{\frac{1}{3}}$ or $V=\frac{\pi r^2 \sqrt{2}}{3}$
$\theta = \frac{\pi}{6}$
Hint: For maxima or minima f'(x) must be zero f'(x) = 0
Given:

Solution:
Let ABC be an isoceles triangle inscribed in the circle with the radius a such that
AB=AC
$AD =AO+OD =a + a\cos \theta = a (1+2\cos \theta)$
$BC =2BD =2a \sin 2\theta$
Area of triangle ACA $=\frac{1}{2}BC\times AD$
\begin{aligned} &A(\theta)=\frac{1}{2} \times 2 a \sin 2 \theta \times a(1+\cos 2 \theta) \\ &=a^{2} \sin 2 \theta(1+\cos 2 \theta) \\ &A(\theta)=a^{2} \sin 2 \theta+a^{2} \sin 2 \theta \cos 2 \theta \\ &A(\theta)=a^{2} \sin 2 \theta+\frac{a^{2} \sin 4 \theta}{2} \end{aligned}
\begin{aligned} &A^{\prime}(\theta)=2 a^{2} \cos 2 \theta+\frac{4 a^{2} \cos 4 \theta}{2} \\ &A^{\prime}(\theta)=2 a^{2} \cos 2 \theta+2 a^{2} \cos 4 \theta \\ &=2 a^{2}(\cos 2 \theta+\cos 4 \theta) \end{aligned}
\begin{aligned} &A^{\prime}(\theta)=0 \\ &2 a^{2}(\cos 2 \theta+\cos 4 \theta)=0 \\ &\cos 2 \theta+\cos 4 \theta=0 \\ &\cos 2 \theta=-\cos 4 \theta \\ &=\cos (\pi-4 \theta) \\ &2 \theta=\pi-4 \theta \\ &6 \theta=\pi \\ &\theta=\frac{\pi}{6} \\ &A^{\prime \prime}(\theta)=2 a^{2}(-\sin 2 \theta-\sin 4 \theta) \\ &=-2 a^{2}(\sin 2 \theta+\sin 4 \theta)<0 \\ &at \ \ \theta=\frac{\pi}{6} \end{aligned}

## Maxima and Minima exercise 17.5 question 23

$6\sqrt{3}r$
Hint: Using Pythagoras theorem solve the problem easily
Given: To prove the least perimeter of an isosceles triangle in which a circle of radius r
Can be inscribed in $6\sqrt{3}r$
Solution:
Let ABC an isoceles triangle with AB = AC =x and BC =y and a circle with centre O and radius r inscribed in triangle ABC.
AO= 2r, OF =r

Using phythagoras theorem in triabgle ABF
$AF^2+BF^2 =AB^2$
$(3r)^2+\left ( \frac{y}{2} \right )^2=x^2$ ..........(1)
Again for $\Delta ADO$,
$(2r)^2=r^2+AD^2$
$3r^2=AD^2$
$AD=\sqrt{3}r$
BD=BF, EC=FC
Now, AD + DB = x
$\sqrt{3}r+ \left (\frac{y}{2} \right )=x$
$\frac{y}{2} =x =\sqrt{3}r$ ........(2)
\begin{aligned} &\therefore(3 r)^{2}+(x-\sqrt{3} r)^{2}=x^{2} \\ &9 r^{2}+x^{2}-2 \sqrt{3} r x+3 r^{2}=x^{2} \\ &12 r^{2}=2 \sqrt{3} r x \\ &6 r=\sqrt{3} x \\ &x=\frac{6 r}{\sqrt{3}} \end{aligned}
From (2)
\begin{aligned} &\frac{y}{2}=\frac{6 r}{\sqrt{3}}-\sqrt{3} r \\ &=\frac{(6 \sqrt{3}-3 \sqrt{3}) r}{3}=\frac{3 \sqrt{3}}{3} r \\ &y=2 \sqrt{3} r \end{aligned}
Perimeter = 2x +y
\begin{aligned} &=2\left(\frac{6 r}{\sqrt{3}}\right)+2 \sqrt{3} r \\ &=\frac{12 r}{\sqrt{3}}+2 \sqrt{3} r \\ &=\frac{12 r+6 r}{\sqrt{3}}=\frac{18 r}{\sqrt{3}} \\ &=\frac{18 \times \sqrt{3} r}{\sqrt{3} \times \sqrt{3}} \end{aligned}
Perimeter $=6\sqrt{3}$

Maxima and Minima exercise 17.5 question 24

l =12
Hint: For maxima or minima f'(x) must be zero f'(x) = 0
Given: Let l,b,v be length, breadth and volume
Solution:
$2(l+b)=36$
$l =18-b$ .....(1)
$V= \pi l^2 b$
$V= \pi (18-b)^2 b$ ..... from (1)
\begin{aligned} &=\pi\left(324 b+b^{3}-36 b^{2}\right) \\ &\frac{d V}{d b}=\pi\left(324+3 b^{2}-72 b\right) \\ &\frac{d V}{d b}=0 \end{aligned}
\begin{aligned} &\pi\left(324+3 b^{2}-72 b\right)=0 \\ &324+3 b^{2}-72 b=0 \\ &b^{2}-24 b+108=0 \\ &b^{2}-6 b-18 b+108=0 \\ &(b-6)(b-18)=0 \\ &b=6,18 \end{aligned}
Now,
\begin{aligned} &\frac{d^{2} V}{d b^{2}}=\pi(6 b-72) \\ &\text { At } b=6 \\ &\frac{d^{2} V}{d b^{2}}=\pi(36-72)=-36 \pi<0 \\ &\text { At } b=18 \\ &\frac{d^{2} V}{d b^{2}}=\pi(18 \times 6-72)=36 \pi>0 \end{aligned}
Substitute the value of b in (1)
$l=18-6=12$
So, the volume is maximum when a = 12cm and b = 6cm

## Maxima and Minima exercise 17.5 question 25

h = 16cm
Hint: For maxima or minima f'(x) must be zero f'(x) = 0
Given: Let r,l,v be radius, height and volume respectively.
Solution:
$h = R + \sqrt{R^2-r^2}$
Squaring on both the sides
$h^2+R^2-2hR=R^2-r^2$
$r^2=2hR-h^2$ ........(1)
Now,
$V=\frac{1}{3}\pi r^2h$
$V=\frac{\pi }{3}(2h^2R-h^3)$ ....... from (1)
\begin{aligned} &\frac{d V}{d h}=\frac{\pi}{3}\left(4 h R-3 h^{2}\right) \\ &\frac{d V}{d h}=0 \\ &\frac{\pi}{3}\left(4 h R-3 h^{2}\right)=0 \\ &4 h R=3 h^{2} \\ &h=\frac{4 R}{3} \end{aligned}
Now,
\begin{aligned} &\frac{d^{2} V}{d h^{2}}=\frac{\pi}{3}(4 R-6 h) \\ &\frac{\pi}{3}(4 R-8 R)=0 \\ &\frac{-4 \pi R}{3}<0 \end{aligned}
So, the volume is maximum when $h = \frac{4R}{3}$
$h = \frac{4\times 12}{3}=16cm$

## Maxima and Minima exercise 17.5 question 26

r = 7cm
Hint: For maxima or minima value of s we must have $\frac{ds}{dr}=0$
Given: Let the height, radius of base and surface area of cylinder be h,r,v.
Solution:
Volume $=\pi r^2 h$
$2156=\pi r^2 h$
$2156=\frac{22}{7} r^2 h$
$h = \frac{2156 \times 7}{22r^2},h = \frac{686}{r^2}$ .........(1)
Surface area = $2\pi rh+\pi r^2h$
$=\frac{4132}{r}+\frac{44}{r^2}$ ......from (1)
\begin{aligned} &\frac{d s}{d r}=\frac{4312}{-r^{2}}+\frac{88 r}{7} \\ &\frac{d s}{d r}=0 \\ &-\frac{4312}{r}+\frac{88 r}{7}=0 \\ &\frac{4312}{r}=\frac{88 r}{7} \Rightarrow r^{3}=\frac{4321 \times 7}{88} \\ &r^{3}=343, r=7 \mathrm{~cm} \end{aligned}
Now,
\begin{aligned} &\frac{d^{2} s}{d r^{2}}=\frac{8624}{r^{3}}+\frac{88}{7} \\ &=\frac{8624}{343}+\frac{88}{7}=\frac{176}{7}>0 \end{aligned}
So, the surface area is maximum when r = 7

### Maxima and Minima exercise 17.5 question 27

$500 \pi cm^3$
Hint: For the maximum value of v, we must have $\frac{dv}{dh}=0$
Given: Let height, radius of base and volume of cylinder be h,r and v respectively
Solution:$\frac{h^2}{4}+r^2=R^2$
$h^2=4(R^2-r^2)$
$r^2=R^2-\frac{h^4}{4}$ .......(1)
Now, $v=\pi r^2h$
$v=\pi \left [ hR^2-\frac{h^3}{4} \right ]$ ......... from (1)
\begin{aligned} &\frac{d v}{d h}=\pi\left(R^{2}-\frac{3 h^{2}}{4}\right) \\ &\frac{d v}{d h}=0 \\ &\pi\left(R^{2}-\frac{3 h^{2}}{4}\right)=0 \\ &R^{2}=\frac{3 h^{2}}{4}, h=\frac{2 R}{\sqrt{3}} \end{aligned}
\begin{aligned} &\frac{d^{2} v}{d h^{2}}=\frac{-3 \pi h}{2}=\frac{-3 \pi}{2} \times \frac{2 R}{\sqrt{3}} \\ &\frac{d^{2} v}{d h^{2}}=\frac{-3 \pi R}{\sqrt{3}}<0 \end{aligned}
Volume of maximum $h =\frac{2R}{\sqrt{3}}$
Maximum volume $=\pi h =\left ( R^2-\frac{h^2}{4} \right )$
\begin{aligned} &=\pi \times \frac{2 R}{\sqrt{3}}\left(R^{2}-\frac{4 R^{2}}{12}\right) \\ &=\frac{2 \pi R}{\sqrt{3}}\left(\frac{8 R^{2}}{12}\right)=\frac{4 \pi R^{3}}{3 \sqrt{3}}=\frac{4 \pi(5 \sqrt{3})^{3}}{3 \sqrt{3}} \\ &=500 \pi \mathrm{cm}^{3} \end{aligned}

## Maxima and Minima exercise 17.5 question 28

$x=y=\frac{r}{\sqrt{2}}$
Hint: For maximum or minimum values of z, we must have $\frac{dz}{dx}=0$
Given:
$x^2+y^2=r^2$
$y = \sqrt{r^2-x^2}$ .........(1)
Solution:
Now,
$z=x+y$
$z=x+\sqrt{r^2-x^2}$ ......from (1)
\begin{aligned} &\frac{d z}{d x}=1+\frac{(-2 x)}{2 \sqrt{r^{2}-x^{2}}} \\ &\frac{d z}{d x}=0 \\ &1+\frac{(-2 x)}{2 \sqrt{r^{2}-x^{2}}}=0 \\ &2 x=2 \sqrt{r^{2}-x^{2}} \\ &x=\sqrt{r^{2}-x^{2}} \end{aligned}
Squaring on the both sides
$x^2=r^2-x^2$
$x=\frac{r}{\sqrt{2}}$
Substituting the value of x in equation (1)
$y = \sqrt{r^2-x^2}$
\begin{aligned} &y=\sqrt{r^{2}-\left(\frac{r}{\sqrt{2}}\right)^{2}} \\ &\frac{d^{2} z}{d x^{2}}=\frac{-\sqrt{r^{2}-x^{2}}+x(-x)}{\sqrt{r^{2}-x^{2}}}=\frac{-r^{2}+x^{2}-x^{2}}{r^{2}-x^{2}} \end{aligned}
$=\frac{-r^2}{r^3}\times 2 \sqrt{2}$
$=-\frac{ 2 \sqrt{2}}{r}<0$
So z = x + y is maximum when $x=y=\frac{r}{\sqrt{2}}$

## maxima and minima exercise 17.5 question 29

$(\pm 2\sqrt{3},3)$
Hint: For maximum or minimum value of z must have $\frac{dz}{dy}=0$
Given: Let the point (x,y) on the curve $x^2=4y$ nearest to (0,5)
$x^2=4y$
$y=\frac{x^2}{4}$ .........(1)
Also,
$d^2=(x)^2+(y-5)^2$ using the distance formula
Solution:
Now,
$z=d^2=(x)^2+(y-5)^2$
$z=(x)^2+\left ( \frac{x^2}{4} -5\right )^2$ ....... from (1)
\begin{aligned} &=x^{2}+\frac{x^{4}}{16}+25-\frac{5 x^{2}}{2} \\ &\frac{d z}{d y}=2 x+\frac{4 x^{3}}{16}-5 x \\ &\frac{d z}{d y}=0 \\ &2 x+\frac{4 x^{3}}{16}-5 x=0 \end{aligned}
$\frac{4x^3}{16}=3x$
$x^3=12x$
$x^2=12$
$x=\pm 2\sqrt{3}$
Substituting the value of x is equal (1)
y =3
$\frac{d^{2} z}{d y^{2}}=2+\frac{12 x^{2}}{16}-5=9-3=6>0$
The required nearest point is $(\pm 2\sqrt{3},3)$

## maxima and minima exercise 17.5 question 30

(4,-4)
Hint: For maximum or minimum value of z must have $\frac{dz}{dy}=0$
Given:$y^2=4x$
$x=\frac{y^2}{4}$ ........(1)
Also,
$d^2=(x-2)^2+(y+8)^2$ using distance formula
Solution:
Now,
$z=d^2=(x-2)^2+(y+8)^2$
$z=\left ( \frac{y^2}{4}-2 \right )^2+(y+8)^2$ ...... from (1)
\begin{aligned} &=\frac{y^{4}}{16}+4-y^{2}+y^{2}+54+16 y \\ &\frac{d z}{d y}=\frac{4 y^{3}}{16}+16 \\ &\frac{d z}{d y}=0 \\ &\frac{4 y^{3}}{16}+16=0 \end{aligned}
$4y^3=-64, y=-4$
Substituting the value of y in equation (1)
$x =4$
$\frac{d^{2} z}{d y^{2}}=\frac{12 y^{2}}{16}=12>0$
The required nearest point is (4,-4)
(4,2)
Hint: For maximum or minimum value of z must have $\frac{dz}{dy}=0$
Given: Let the point (x,y) on the curves $x^2=8y$ nearest to (2,4)
$x^2=8y$
$y = \frac{x^2}{8}$ ....... (1)
Also,
$d^2=(x-2)^2+(y-4)^2$ using distance formula
Solution:
Now,
$z=d^2=(x-2)^2+(y-4)^2$
$z=(x-2)^2+\left ( \frac{x^2}{8}-4 \right )$ ..... from (1)\
\begin{aligned} &=x^{2}+4-4 x+\frac{x^{4}}{64}+16-x^{2} \\ &\frac{d z}{d y}=-4+\frac{4 x^{3}}{64} \\ &\frac{d z}{d y}=0 \\ &\frac{4 x^{3}}{64}=-4 \\ &x^{3}=64, x=4 \end{aligned}
Substituting the value of y in eqn (1)
y =2
$\frac{d^2z}{dy^2}=\frac{12x^2}{64}=3>0$
The required nearest point is (4,2)

$(\pm 2\sqrt{2},4)$
Hint: For maximum or minimum value of z must have $\frac{dz}{dy}=0$
Given: Let the point (x,y) on the curve $x^2=2y$ nearest (0,5)
$x^2=2y$
$y=\frac{x^2}{2}$ .........(1)
Also,
$d^2=(x)^2+(y-5)^2$ using the distance formula

Solution:
Now,
$z=d^2=(x)^2+(y-5)^2$
$z=(x)^2+\left ( \frac{x^2}{2}-5 \right )^2$ ....... from (1)
\begin{aligned} &=x^{2}+\frac{x^{4}}{4}+25-5 x^{2} \\ &\frac{d z}{d y}=2 x+x^{3}-10 x \\ &\frac{d z}{d y}=0 \\ &x^{3}-8 x=0 \\ &x^{2}=8, x=\pm 2 \sqrt{2} \end{aligned}
Substituting the value of x in eqn (1)
y =4
$\frac{d^2z}{dy^2}=3x^2-8$
$=24-8=16>0$
The required nearest point $(\pm 2\sqrt{2},4)$.

(-2, -8)
Hint: For maximum or minimum value of S, we must have $\frac{dS}{dt}=0$
Given: Let coordinate of the point on the parabola be (x,y). Then,
$y= x^2+7x +2$ ......(1)
Solution: Let the distance of the point $(x,(x^2+7x+2))$ from the line $y=3x-3$ be S.
\begin{aligned} &\mathrm{S}=\left|\frac{-3 x+\left(x^{2}+7 x+2\right)+3}{\sqrt{10}}\right| \\ &\frac{d S}{d t}=\frac{-3+2 x+7}{\sqrt{10}} \\ &\frac{d S}{d t}=0 \\ &\frac{-3+2 x+7}{\sqrt{10}}=0 \\ &2 x=-4, x=-2 \\ &\frac{d^{2} S}{d t^{2}}=\frac{2}{\sqrt{10}}>0 \end{aligned}
The required nearest point is $(x,(x^2+7x+2))$
$=(-2,4-14+2)$
$=(-2,-8)$

maxima and minima exercise 17.5 question 34

(2, 2)
Hint: For maximum or minimum value of f, we must be have f'(x) =0
Given:
$y^2=2x$
$x=\frac{y^2}{2}$ ........(1)
$d^2= (x-1)^2+(y-4)^2$ using the distance formula
Solution:
Now,
$z=d^2= (x-1)^2+(y-4)^2$
$z=\left ( \frac{y^2}{2}-1 \right )^2+(y-4)^2$ ........ from (1)
$=\frac{y^4}{4}+1-y^2+y^2+16-8y$
$\frac{dz}{dy}=y^3-8$
$\frac{dz}{dy}=0$
$y^3-8=0$
$y^3=8$
$y=2$
Substituting the value of y in eqn (1)
x = 2
$\frac{d^2z}{dy^2}=3y^2$
$=12>0$
The required nearest point is (2,2)
5
Hint: For maximum or minimum value of S, we must have $\frac{dS}{dx}=0$
Given:
$y= -x^3+3x^2+2x-27$ ......(1)
Slope = $\frac{dy}{dx}=-3x^2+6x+2$
Solution:
Now,
$m=-3x^2+6x+2$
$\frac{dm}{dx}=-6x+6$
$\frac{dm}{dx}=0$
$-6x+6=0$
x = 1
Substituting the value of x in eqn(1)
$y =(1)^3+3(1)^2+2(1)-27$
$=-23$
$\frac{d^2m}{dx^2}=-6<0$
So the slope is maximum x= 1, y = -23 at (1,-23)
maximum slope $= -3(1)^2+6(1)+2$
$= 5$
10 items
Hint: For maximum or minimum value of P, we must have $\frac{dP}{dx}=0$
Given: Profit = S.P - C.P
$P= x\left ( 50-\frac{x}{2} \right )-\left ( \frac{x^2}{4}+35x+25 \right )$
Solution: $P= x\left ( 50-\frac{x}{2} \right )-\left ( \frac{x^2}{4}+35x+25 \right )$
\begin{aligned} &P=50 x-\frac{x^{2}}{2}-\frac{x^{2}}{4}-35 x-25 \\ &\frac{d P}{d x}=50-x-\frac{x}{2}-35 \\ &\frac{d P}{d x}=0 \\ &15-\frac{3 x}{2}=0 \\ &15=\frac{3 x}{2}, x=\frac{30}{3}=10 \end{aligned}
Now,
$\frac{d^2P}{dx^2}==\frac{-3}{2}<0$
Profit is the maximum if daily output is 10 units.

Maxima and Minima exercise 17.5 question 37.

240 items
Hint: For maximum or minimum value of P, we must have $\frac{dP}{dx}=0$
Given: Profit = S.P - C.P
$P= x\left ( 5-\frac{x}{100} \right )-\left ( 500+\frac{x}{5} \right )$
Solution:$P= x\left ( 5-\frac{x}{100} \right )-\left ( 500+\frac{x}{5} \right )$
\begin{aligned} &P=5 x-\frac{x^{2}}{100}-500-\frac{x}{5} \\ &\frac{d P}{d x}=5-\frac{x}{50}-\frac{1}{5} \\ &\frac{d P}{d x}=0 \\ &5-\frac{x}{50}-\frac{1}{5}=0 \\ &\frac{24}{5}=\frac{x}{50}, x=\frac{24 \times 50}{5}=240 \end{aligned}
Now,
$\frac{d^2P}{dx^2}=\frac{-1}{50}<0$
Profit is maximum if 240 items are sold.
$h =\frac{l}{2}$
Hint: For maxima and minima value of S, we must have $\frac{dS}{dl}=0$
Given: Let l,h,v and S be the length, height, volume and surface area of the tank constructed
Since volume, v is constant $l^2h=v\Rightarrow h=\frac{v}{l^2 }$ →i
Solution:
Let S denote the total surface area, then we have $S=l^2+4lh$
$\Rightarrow S=l^2+4l\times \frac{v}{l^2 }$ →fromi
$\Rightarrow S=l^2+\frac{4v}{l}$ …(ii)
Differentiating S w.r.t 'l', we get
$\frac{dS}{dl}= \frac{d}{dl}\left (l^2+\frac{4v}{l} \right )$
\begin{aligned} &\Rightarrow \frac{d S}{d l}=\frac{d}{d l}\left(l^{2}\right)+\frac{d}{d l}\left(\frac{4 v}{l}\right) \quad\left[\because \frac{d}{d x}\left\{(f(x)+g(x)\}=\frac{d}{d x}(f(x))+\frac{d}{d x}(g(x))\right]\right. \\ &\begin{aligned} \Rightarrow \frac{d S}{d l}=\frac{d}{d l}\left(l^{2}\right)+4 v \frac{d}{d l}\left(\frac{1}{2}\right) \quad\left[\because \frac{d}{d x}\left(a x^{n}\right)=a \frac{d}{d x}\left(x^{n}\right)\right] \\ \Rightarrow \frac{d S}{d l}=\frac{d}{d l}\left(l^{2}\right)+4 v \frac{d}{d l}\left(l^{-1}\right) \end{aligned} \\ &\Rightarrow \frac{d S}{d l}=2 l^{2-1}+4 v\left(-1 l^{-1-1}\right) \quad\left[\because \frac{d}{d x}\left(x^{n}\right)=n x^{n-1}\right] \end{aligned}
$\Rightarrow \frac{dS}{dl}=2l-\frac{4v}{l^2 }$ …(iii)
For maximum or minimum of S, we have
$\Rightarrow \frac{dS}{dl}=0$
$\Rightarrow 2l-\frac{4v}{l^2 }=0$ [using (iii)]⇒$\frac{2l^3-4v}{l^2}=0$]
$\Rightarrow 2l^3-4v=0$
$\Rightarrow 2l^3=4v\Rightarrow l^3=2v$ …(iv)
Again, differentiating S w.r.t 'l', then $\frac{d^{2} S}{d l^{2}}=\frac{d}{d l}\left(\frac{d S}{d l}\right)=\frac{d}{d l}\left(2 l-\frac{4 v}{l^{2}}\right)$ [using (iii)]
\begin{aligned} &\Rightarrow \frac{d^{2} S}{d l^{2}}=\frac{d}{d l}(2 l)+\frac{d}{d l}\left(\frac{-4 v}{l^{2}}\right) \quad\left[\because \frac{d}{d x}\left\{(f(x)+g(x)\}=\frac{d}{d x}(f(x))+\frac{d}{d x}(g(x))\right]\right. \\ &\begin{aligned} \Rightarrow \frac{d^{2} S}{d l^{2}}=2 \frac{d}{d l}(l)-4 v \frac{d}{d l}\left(\frac{1}{l^{2}}\right) &\left[\because \frac{d}{d x}\left(a x^{n}\right)=a \frac{d}{d x}\left(x^{n}\right)\right] \\ & \Rightarrow \frac{d^{2} S}{d l^{2}}=2 \frac{d}{d l}\left(l^{1}\right)-4 v \frac{d}{d l}\left(l^{-2}\right) \end{aligned} \end{aligned}
\begin{aligned} &\begin{aligned} \Rightarrow \frac{d^{2} S}{d l^{2}}=2\left(1 l^{1-1}\right)-4 v\left(-2 l^{-2-1}\right) &\left[\because \frac{d}{d x}\left(x^{n}\right)=n x^{n-1}\right] \\ \Rightarrow \frac{d^{2} S}{d l^{2}}=2(1)+8 v\left(l^{-3}\right) \end{aligned} \\ &\begin{aligned} \Rightarrow \frac{d^{2} S}{d l^{2}} &=2+\frac{8 v}{l^{3}} \\ &=2+\frac{8}{2} \\ &=2+4 \end{aligned} \end{aligned}
$=6>0$

$\therefore \frac{d^2S}{dl^2}>0$ Hence the surface area is minimum, $h = \frac{v}{l^2}$ Substitute the value of $v=\frac{l^3}{2}$ from (iv) in eqni, we get $\frac{l^{3}}{2 l^{2}} \Rightarrow h=\frac{l}{2}$
Hence proved.
$h=\left(\frac{32 c}{81}\right)^{\frac{1}{3}}, b=\left(\frac{9 c}{16}\right)^{\frac{1}{3}}, a=\left(\frac{9 c}{16}\right)^{\frac{1}{2}}$
Hint: For maximum or minimum value of T, we must have $\frac{dT}{db}=0$
Given:
Let l,b,h be the length , breadth and height of the box
Volume of the box = c
Solution:
l = 2b ....(1)
c = lbh
$c= 2b^2h$
$h = \frac{c}{2b^2}$ .......(2)
Let the cost of material required to the bottom be $k/m^2$.
Cost of material required for 4 wall stand top = $Rs 3k/m^2$
Total cost , $T=k(lb)+3k(2lh +2bh + lb)$
$T=2kb^2+3k\left ( \frac{4bc}{2b^2}+\frac{2bc}{2b^2}+b^2 \right )$ [From (1) and (2)}
\begin{aligned} &\frac{d T}{d b}=4 k b+3 k\left(\frac{-3 c}{b^{2}}+4 b\right) \\ &\frac{d T}{d b}=0 \\ &4 k b+3 k\left(\frac{-3 c}{b^{2}}+4 b\right)=0 \\ &4 b=3\left(\frac{3 c}{b^{2}}+4 b\right) \end{aligned}
\begin{aligned} &4 b=\frac{9 c-12 b^{3}}{b^{2}} \\ &4 b^{3}=9 c-12 b^{3} \\ &16 b^{3}=9 c \\ &b=\left(\frac{9 c}{16}\right)^{\frac{1}{3}} \end{aligned}
\begin{aligned} &\frac{d^{2} T}{d b^{2}}=4 k+3 k\left(\frac{6 c}{b^{3}}+4\right) \\ &=4 k+3 k\left(\frac{6 c}{9 c} \times 16+4\right) \\ &=k\left(\frac{44}{3} \times 3+4\right) \\ &=48 k>0 \end{aligned}
$\therefore$ cost of the minimum when $b=\left(\frac{9 c}{16}\right)^{\frac{1}{3}}$
Sub b in eqn(1) , eqn (2)
\begin{aligned} &l=2\left(\frac{9 c}{16}\right)^{\frac{1}{3}} \\ &h=\frac{c}{2 b c^{2}}, h=\frac{c}{2\left(\frac{9 c}{16}\right)^{\frac{2}{3}}}=\left(\frac{32 c}{81}\right)^{\frac{1}{3}} \\ &\therefore l=2\left(\frac{9 c}{16}\right)^{\frac{1}{3}}, b=\left(\frac{9 c}{16}\right)^{\frac{1}{3}}, h=\left(\frac{32 c}{81}\right)^{\frac{1}{3}} \end{aligned}
x = 2r
Hint: For maximum or minimum value of V, we must have $\frac{dV}{dr}=0$
Given:
Let r be the radius of the sphere, x is side of the cube and S be the sum of the surface area of both of them
$S= 4 \pi r^2 + 6x^2$
$x= \left ( \frac{S-4\pi r^2}{6} \right )$ ......(1)
Solution:
Sum of Volumes, $V=\frac{4}{3}\pi r^3+x^3$
$V=\frac{4}{3} \pi r^{3}+\left[\left(\frac{S-4 \pi r^{2}}{6}\right)\right]^{\frac{3}{2}}$
From eqn (1)
$\frac{dV}{dr}=4 \pi r^{2}-2 \pi r\left[\left(\frac{S-4 \pi r^{2}}{6}\right)\right]^{\frac{1}{2}}$
$\frac{dV}{dr}=0$ .........(2)
\begin{aligned} &4 \pi r^{2}-2 \pi r\left[\frac{\left(S-4 \pi r^{2}\right)}{6}\right]^{\frac{1}{2}}=0 \\ &4 \pi r^{2}=2 \pi r\left[\frac{\left(S-4 \pi r^{2}\right)}{6}\right]^{\frac{1}{2}} \end{aligned}
$4 \pi r^2 -2\pi rx$ from eqn (1),
\begin{aligned} &\frac{d^{2} V}{d r^{2}}=8 \pi r-2 \pi\left[\frac{\left(S-4 \pi r^{2}\right)}{6}\right]^{\frac{1}{2}}-\frac{2 \pi r}{2}\left[\frac{\left(S-4 \pi r^{2}\right)}{6}\right]^{-\frac{1}{2}}\left(\frac{-8 \pi r}{b}\right) \\ &=8 \pi r-2 \pi\left[\frac{\left(S-4 \pi r^{2}\right)}{6}\right]^{\frac{1}{2}}+\frac{4 \pi^{2} r^{2}}{3}\left[\frac{6}{\left(S-4 \pi r^{2}\right)}\right]^{\frac{1}{2}} \\ &=8 \pi r-2 \pi x+\frac{4 \pi^{2} r^{2}}{3} \frac{1}{x}=8 \pi r-4 \pi r+\frac{2}{3} \pi^{2} r \\ &\frac{d^{2} V}{d r^{2}}=4 \pi r+\frac{2}{3} \pi^{2} r>0 \end{aligned}
Volume is minimum when x = 2r
$\frac{l}{D}=\frac{\pi}{\pi +2}$
Hint: For maximum or minimum value of S, we must have $\frac{dS}{dD}=0$
Given: Volume, $V =\frac{1}{2}\pi l \left ( \frac{D}{2} \right )^2$
$V =\frac{\pi lD^2}{8}$
$I=\frac{8V}{\pi D^2}$ .........(1)
Solution:
Total surface area = $\frac{\pi D^2}{4}+lD+\frac{\pi Dl}{2}$
$S=\frac{\pi D^2}{4}+\frac{8V}{\pi D}+\frac{8V}{2D}$ .....from (1)
\begin{aligned} &\frac{d S}{d D}=\frac{\pi D}{2}-\frac{8 V}{\pi D^{2}}-\frac{8 V}{2 D^{2}} \\ &\frac{d S}{d D}=0 \\ &\frac{\pi D}{2}-\frac{8 V}{\pi D^{2}}-\frac{8 V}{2 D^{2}}=0 \end{aligned}
\begin{aligned} &\frac{\pi D}{2}=\frac{8 V}{D^{2}}\left(\frac{1}{\pi}+\frac{1}{2}\right) \\ &D^{3}=\frac{16 V}{\pi}\left(\frac{1}{\pi}+\frac{1}{2}\right) \\ &\frac{d^{2} S}{d D^{2}}=\frac{\pi}{2}+\frac{16 V}{D^{3}}\left(\frac{1}{\pi}+\frac{1}{2}\right) \\ &=\frac{\pi}{2}+\pi>0 \end{aligned}
\begin{aligned} &l=\frac{8 V}{\pi D^{2}} \\ &l=\frac{8}{\pi D^{2}}\left[\frac{\pi D^{3}}{16}\left[\frac{2 \pi}{\pi+2}\right]\right] \\ &l=D\left(\frac{\pi}{\pi+2}\right), \frac{l}{D}=\frac{\pi}{\pi+2} \end{aligned}
Hence proved
$b= \frac{2a}{\sqrt{3}}, h=\frac{2\sqrt{2}}{\sqrt{3}}a$
Hint: For maximum or minimum value of S, we must have $\frac{dS}{dD}=0$
Given:
Let the breadth, height, strength of the beam be b,h,S
$a^2=\frac{h^2+b^2}{4}$
$4a^2-b^2=h^2$ .......(1)
Solution:
Stength beam, $S=kbh^2$
$S=kb(4a^2-b^2)$ ......from (1)
$S=k(b4a^2-b^3)$
$\frac{dS}{db}=k(4a^2-3b^2)$
$\frac{dS}{db}=0$
\begin{aligned} &k\left(4 a^{2}-3 b^{2}\right)=0 \\ &4 a^{2}-3 b^{2}=0 \\ &4 a^{2}=3 b^{2} \\ &b=\frac{2 a}{\sqrt{3}} \end{aligned}
Sub b value in (1)
\begin{aligned} &4 a^{2}-\left(\frac{2 a}{\sqrt{3}}\right)^{2}=h^{2} \\ &\frac{12 a^{2}-4 a^{2}}{3}=h^{2} \\ &h=\frac{2 \sqrt{2}}{\sqrt{3}} a \end{aligned}
\begin{aligned} &\frac{d^{2} S}{d b^{2}}=-6 k b \\ &=-6 k \frac{2 a}{\sqrt{3}} \\ &=\frac{-12 k a}{\sqrt{3}}<0 \end{aligned}
So the strength of the beam is maximum when $b= \frac{2a}{\sqrt{3}}, h=\frac{2\sqrt{2}}{\sqrt{3}}a$
S =9
Hint: For maximum or minimum value of S, we must have $\frac{dS}{dD}=0$
Given:
The equation of line passing through (1,4) with slope m is given by
$y-4=m(x-1)$ ........(1)
Solution:
Sub (y=0) value in eqn (1)
$0-4=m(x-1)$
$\frac{-4}{m}=x-1$
$x=\frac{m-4}{m}$
Sub (x=0) value in eqn (1)
$y-4=m (0-1)$
$y=-m+4$
$x=-(m-4)$
So the intercepts coordinates axes are $\frac{m-4}{m},-(m-4)$
\begin{aligned} &S=\frac{m-4}{m}-(m-4) \\ &\frac{d S}{d m}=\frac{4}{m^{2}}-1 \\ &\frac{d S}{d m}=0 \\ &\frac{4}{m^{2}}-1=0 \\ &\frac{4}{m^{2}}=1 \\ &m^{2}=4, m=\pm 2 \end{aligned}
\begin{aligned} &\frac{d^{2} S}{d m^{2}}=\frac{-8}{m^{3}} \\ &\left(\frac{d^{2} S}{d m^{2}}\right)_{m=2}=\frac{-8}{2^{3}}=-1<0 \end{aligned}
Sum is minimum at m = 2
$\left(\frac{d^{2} S}{d m^{2}}\right)_{m=-2}=\frac{-8}{(-2)^{3}}=1>0$
So the sum is maximum at m = -2
Thus,
$\mathrm{S}=\left[\frac{-2-4}{-2}-(-2-4)\right]=3+6=9$
x =15, y = 10
Hint: For maxima and minima value of S, we must have $\frac{dS}{dD}=0$
Given:
Let x and y be the length and breadth of the rectangle
Area of page = 150
xy = 150
$y = \frac{150}{x}$ ......(1)
Solution:
Area of printed matter = $(x-3)(y-2)$
$A=xy-2x-3y+6$
$A=150-2x-\frac{450}{x}+6$
$\frac{dA}{dx}=-2+\frac{450}{x^2}$
$\frac{dA}{dx}=0$
$-2+\frac{450}{x^2}=0$
$2x^2=450$
$x=15$
Sub x=15 value in eqn (1)
y = 10
Now,
$\frac{d^2A}{dx^2}=\frac{-900}{x^3}=\frac{-900}{(15)^3}$
$=\frac{-900}{3375}<0$
Thus area of printed matter is max when x = 15 and y = 10
a = -260
Hint: For maxima and minima value of S, we must have $\frac{dS}{dD}=0$
Given: $S= t^5-40t^3-30t^2-80t-250$
Solution:$\frac{dS}{dt}= 5t^5-120t^2-60t-80$
Acceleration, $a=\frac{d^2S}{dt^2}= 20t^3-240t-60$
$\frac{da}{dt}= 60t^2-240$
$\frac{da}{dt}=0$
$60t^2-240=0$
$60t^2=240, t=2$
Now,
$\frac{d^2a}{dt^2}=120t$
$\frac{d^2a}{dt^2}=240>0$
So, acceleration is minimum at t =2
$a_{min}=20(2)^3-240(2)+60$
$a_{min}=160-480+60$
$a_{min}=-260$
$\therefore$ At =2, a = -260

t = 2
Hint: For maxima and minima value of V, we must have $\frac{dV}{dt}=0$
Given:
\begin{aligned} &S=\frac{t^{4}}{4}-2 t^{3}+4 t^{2}-7 \\ &V=\frac{d S}{d t}=t^{3}-6 t^{2}+8 t \\ &a=\frac{d V}{d t}=3 t^{2}-12 t+8 \end{aligned}
Solution:
$\frac{dV}{dt}=0$
$3t^2-12t+8 =0$
On solving the equation we get
$t=2\pm \frac{2}{\sqrt{3}}$
Now,
\begin{aligned} &\frac{d^{2} V}{d t^{2}}=6 t-12 \\ &\text { At } t=2 \pm \frac{2}{\sqrt{3}} \\ &\frac{d^{2} V}{d t^{2}}=6\left(2 \pm \frac{2}{\sqrt{3}}\right)-12,=\frac{-12}{3}<0 \end{aligned}
So, the velocity is maximum at $t = 2-\frac{2}{\sqrt{3}}$
Again,
$\frac{da}{dt}=6t-12$
For the maximum or minimum value of a
$\frac{da}{dt}=0$
$6t-12=0$
$t=2$
Now, $\frac{d^2a}{dt^2}=6>0$
So acceleration is minimum at t = 2

RD Sharma Class 12th exercise 17.5 comprises 37 level 1 questions and 9 level 2 questions including its subparts, and it incorporates elaborate solutions for all our queries and doubts. It includes the following concepts

• Maximum and minimum values of a function in its domain
• Local maxima and Local minima
• Point of inflection
• Maximum and minimum values in a closed interval
• Applied problems on maxima and minima

## RD Sharma Chapter-wise Solutions

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