Exercise 7.8-Additional Problems - Chapter 7 - Applications of Differential Calculus - 12th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

You can Download the Exercise 7.8-Additional Problems - Chapter 7 - Applications of Differential Calculus - 12th Maths Guide Samacheer Kalvi Solutions - Tamil Medium with expert answers for all chapters. Perfect for Tamil & English Medium students to revise the syllabus and score more marks in board exams. Download and share it with your friends

Share this to Friend on WhatsApp

Additional Problems
Question 1.

The top and bottom margins of a poster are each $6 \mathrm{cms}$ and the side margins are each $4 \mathrm{cms}$. If the area of the printed material on the poster is fixed at $384 \mathrm{cms}^2$, find the dimension of the poster with the smallest area.

Solution:
Let $\mathrm{x}$ and $\mathrm{y}$ be the length and breadth of printed area, then the area $\mathrm{xy}=384$
Dimensions of the poster area are $(x+8)$ and $(y+12)$ respectively.
Poster area $\mathrm{A}=(\mathrm{x}+8)(\mathrm{y}+12)$
$
\begin{aligned}
& =x y+12 x+8 y+96 \\
& =12 x+8 y+480
\end{aligned}
$

$
\begin{aligned}
= & 12 x+8\left(\frac{384}{x}\right)+480 \\
\mathrm{~A}^{\prime} & =12-8 \times 384 \times \frac{1}{x^2} \\
\mathrm{~A}^{\prime \prime} & =16 \times 384 \times \frac{1}{x^3} \\
\mathrm{~A}^{\prime} & =0 \Rightarrow x=\pm 16
\end{aligned}
$
But $\mathrm{x}>0$
$
\therefore \mathrm{x}=16
$
when $\mathrm{x}=16, \mathrm{~A}$ " $>0$
when $x=16$, the area is minimum
$
\begin{aligned}
& \mathrm{y}=24 \\
& \therefore \mathrm{x}+8=24, \\
& \mathrm{y}+12=36
\end{aligned}
$
Hence the dimensions are $24 \mathrm{~cm}$ and $36 \mathrm{~cm}$

Question 2.
Show that the volume of the largest right circular cone that can be inscribed in a sphere of radius a is $\frac{8}{27}$ (volume of the sphere).
Solution:
Given that a is the radius of the sphere and let $\mathrm{x}$ be the base radius of the cone. If $\mathrm{h}$ is the height of the cone, then its volume is

$
\begin{aligned}
\mathbf{V} & =\frac{1}{3} \pi x^2 h \\
& =\frac{1}{3} \pi x^2(a+y)
\end{aligned}
$
where $\mathrm{OC}=\mathrm{y}$ so that height $\mathrm{h}=\mathrm{a}+\mathrm{y}$

From the diagram $x^2+y^2=a^2$
Using (2) in (1) we have
$
\mathrm{V}=\frac{1}{3} \pi\left(a^2-y^2\right)(a+y)
$
For the volume to be maximum:
$
\begin{aligned}
& \mathrm{V}^{\prime}=0 \Rightarrow \frac{1}{3} \pi\left[a^2-2 a y-3 y^2\right)=0 \\
& \Rightarrow 3 \mathrm{y}=+\mathrm{a} \text { or } \mathrm{y}=-\mathrm{a} \\
& \Rightarrow y=\frac{a}{3} \text { and } y=-a \text { is not possible } \\
& \text { Now } \mathrm{V}^{\prime \prime}=-\pi \frac{2}{3}(a+3 y)<0 \text { at } y=\frac{a}{3}
\end{aligned}
$
The volume is maximum when $y=\frac{a}{3}$ and the maximum volume is
$
\frac{1}{3} \pi \times \frac{8 a^2}{9}\left(a+\frac{1}{3} a\right)=\frac{8}{27}\left(\frac{4}{3} \pi a^3\right)=\frac{8}{27} \text { (volume of the sphere) }
$
Question 3.
A closed (cuboid) box with a square base is to have a volume of 2000 c.c. The material for the top and bottom of the box is to cost Rs. 3 per square $\mathrm{cm}$ and the material for the sides is to cost Rs. $1.50$ per square $\mathrm{cm}$. If the cost of the materials is to be the least, find the dimensions of the box.
Solution:
Let $x, y$ respectively denote the length of the side of the square base and depth of the box. Let $\mathrm{C}$ be the cost of the material
Area of the bottom $=x^2$
Area of the top $=x^2$
Combined area of the top and bottom $=2 x^2$
Area of the four sides $=4 \mathrm{xy}$
Cost of the material for the top and bottom $=3(2 \mathrm{x})^2$

Cost of the material for the sides $=(1.5)(4 x y)=6 x y$
Total cost $C=6 x^2+6 x y$
Volume of the box $V=($ area $)($ depth $)=x^2 y=2000$
Eliminating $y$ from (1) \& (2) we get $\mathrm{C}(x)=6 x^2+\frac{12000}{x}$
where $x>0$, ie., $x \in(0,+\infty)$ and $\mathrm{C}(x)$ is continuous on $(0,+\infty)$.
$
\begin{aligned}
C^{\prime}(x) & =12 x-\frac{12000}{x^2} \\
C^{\prime}(x)=0 \Rightarrow 12 x^3-12000=0 \Rightarrow 12\left(x^3-10^3\right) & =0 \\
x=10 \text { or } x^2+10 x+100 & =0 \\
x^2+10 x+100 & =0 \text { is not possible } \\
\text { The critical numbers is } x & =10 \\
\text { Now C " }(x)=12+\frac{24000}{x^3} ; C^{\prime \prime}(10) & =12+\frac{24000}{1000}=36>0 \\
C \text { is a minimum at }(10, C(10)) & =(10,1800) \\
\therefore \text { The base length is } 10 \mathrm{~cm} \text { and depth is } y & =\frac{2000}{100}=20 \mathrm{~cm}
\end{aligned}
$

Question 4.
Find two numbers whose sum is 100 and whose product is a maximum.
Solution:
Let the two numbers be $\mathrm{x}$ and $\mathrm{y}$.
$
\begin{aligned}
& x+y=100 \\
& \Rightarrow y=100-x
\end{aligned}
$
Product $=x y=x(100-x)$
$
f=x(100-x)=100 x-x^2
$
We have to find $x$ at which $f$ is maximum.
$
\begin{aligned}
f & =100 x-x^2 \\
\frac{d f}{d x} & =100-2 x \\
\frac{d^2 f}{d x^2} & =-2 \\
\frac{d f}{d x} & =0 \Rightarrow 100-2 x=0 \\
100 & =2 x \Rightarrow x=\frac{100}{2}=50
\end{aligned}
$
At $x=50, \frac{d^2 f}{d x^2}=-2<0 \Rightarrow x=50$ is a maximum point.
$\therefore \mathrm{f}$ is maximum at $\mathrm{x}=50$
So, $y=100-x=100-50=50$
So, the two numbers are 50,50 .

Question 5.
Find two positive numbers whose product is 100 and whose sum is minimum.
Solution:
Let the two numbers be $\mathrm{x}$ and $\mathrm{y}$.
Given $x y=100 \Rightarrow y=\frac{100}{x}$
$
\text { Sum }=x+y=x+\frac{100}{x}=f
$
To find $\mathrm{x}$ at which $\mathrm{f}$ is maximum
$
\begin{aligned}
\text { Here, } f & =x+\frac{100}{x} \\
\frac{d f}{d x} & =1+100\left(\frac{-1}{x^2}\right)=1-\frac{100}{x^2} \\
\frac{d^2 f}{d x^2} & =-100\left(\frac{-2}{x^3}\right)=\frac{200}{x^3} \\
\frac{d f}{d x} & =0 \Rightarrow 1-\frac{100}{x^2}=0 \Rightarrow 1=\frac{100}{x^2} \\
\text { i.e., } x^2 & =100 \Rightarrow x=\pm 10 \\
\text { At } x & =+10, \frac{d^2 f}{d x^2}=+\mathrm{ve} \Rightarrow x=10 \text { is a minimum point. } \\
\text { At } x & =-10, \frac{d^2 f}{d x^2}=-\mathrm{ve} \Rightarrow x=-10 \text { is a maximum point. } \\
\therefore \text { at } & x=10, \text { the sum is minimum } \\
\therefore \text { at } & x=10, y=\frac{100}{x}=\frac{100}{10}=10
\end{aligned}
$
So the two numbers are 10,10 .