Exercise 7.8-Additional Problems - Chapter 7 Applications of Differential Calculus 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

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 .