Exercise 7.8 - 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

$
\text { Ex } 7.8
$
Question 1.
Find two positive numbers whose sum is 12 and their product is maximum.

Solution:
Let the two numbers be $\mathrm{x}, 12-\mathrm{x}$.
Their product $p=x(12-x)=12 x-x^2$ To find the maximum product.
$
\begin{aligned}
& \mathrm{p}^{\prime}(\mathrm{x})=12-2 \mathrm{x} \\
& \mathrm{p}^{\prime \prime}(\mathrm{x})=-2 \\
& \mathrm{p}^{\prime}(\mathrm{x})=0 \Rightarrow 12-2 \mathrm{x}=0 \Rightarrow 2 \mathrm{x}=12 \\
& \Rightarrow \mathrm{x}=6 \\
& \text { at } \mathrm{x}=6, \mathrm{p}^{\prime \prime}(\mathrm{x})=-2=-\text { ve } \\
& \Rightarrow \mathrm{p} \text { is maximum at } \mathrm{x}=6 \\
& \text { when } \mathrm{x}=6,12-\mathrm{x}=12-6=6
\end{aligned}
$
So the two numbers are 6,6
Question 2.
Find two positive numbers whose product is 20 and their sum is minimum.

Solution:
Let the two numbers be $x, \frac{20}{x}$
Their sum $=x+\frac{20}{x}$
To get the minimum sum
$
\begin{aligned}
\mathrm{s} & =x+\frac{20}{x} \\
\mathrm{~s}^{\prime}(x) & =1-\frac{20}{x^2} \\
\mathrm{~s}^{\prime \prime}(x) & =-20\left(\frac{-2}{x^3}\right)=\frac{40}{x^3} \\
\mathrm{~s}^{\prime}(x) & =0 \Rightarrow 1=\frac{20}{x^2} \Rightarrow x^2=20 \\
x & =\sqrt{20}=\pm 2 \sqrt{5}
\end{aligned}
$

but $\mathrm{x}$ is positive (given)
$
\Rightarrow x=+2 \sqrt{5}
$
when $x=2 \sqrt{5}, \mathrm{~s}^{\prime \prime}(x)=\frac{40}{(2 \sqrt{5})^3}=+\mathrm{ve}$ $\Rightarrow x=+2 \sqrt{5}$ is a minimum point when
$
\begin{aligned}
x & =2 \sqrt{5}, y=\frac{20}{2 \sqrt{5}}=\frac{20}{2 \sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} \\
& =2 \sqrt{5}
\end{aligned}
$
So the two numbers are $2 \sqrt{5}, 2 \sqrt{5}$

Question 3.
Find the smallest possible value of $x^2+y^2$ given that $x+y=10$.
Solution:
Given $\mathrm{x}+\mathrm{y}=10 \Rightarrow 7=10-\mathrm{x}$
To find the smallest value of $x^2+y^2$
$
\begin{aligned}
f(x) & =x^2+y^2=x^2+(10-x)^2 \\
f^{\prime}(x) & =2 x+2(10-x)(-1) \\
& =2 x-20+2 x=4 x-20 \\
f^{\prime \prime}(x) & =4 \\
f^{\prime}(x) & =0 \Rightarrow 4 x-20=0 \\
& \Rightarrow 4 x=20 \Rightarrow x=5
\end{aligned}
$
at $x=5, f^{\prime}(x)=4$ at $x=5, y=10-5=5=+$ ve $\mathrm{x}=5$ is a minimum point.
So the minimum value of $x^2+y^2=5^2+5^2=50$

Question 4.
A garden is to be laid out in a rectangular area and protected by wire fence. What is the largest possible area of the fenced garden with 40 metres of wire.
Solution:
Perimeter $=40 \mathrm{~m}$
$2(1+b)=40 \Rightarrow 1+b=20$
Let $1=\mathrm{x} \mathrm{m}$
$\mathrm{b}=(20-\mathrm{x}) \mathrm{m}$
Area $=1 \times b=x(20-x)=20 x-x^2$
To find the maximum area
$\mathrm{A}(\mathrm{x})=20 \mathrm{x}-\mathrm{x}^2$
$\mathrm{A}^{\prime}(\mathrm{x})=20-2 \mathrm{x}$
$A^{\prime \prime}(\mathrm{x})=-2$
$\mathrm{A}^{\prime}(\mathrm{x})=0 \Rightarrow 20-2 \mathrm{x}=0$
$\Rightarrow \mathrm{x}=10$
$\mathrm{x}=10$ is a maximum point
$\therefore$ Maximum Area $=\mathrm{x}(20-\mathrm{x})$
$
\begin{aligned}
& =10(20-10) \\
& =10 \times 10=100 \text { sq.m. }
\end{aligned}
$

Question 5.
A rectangular page is to contain $24 \mathrm{~cm}^2$ of print. The margins at the top and bottom of the page are $1.5 \mathrm{~cm}$ and the margins at other sides of the page is $1 \mathrm{~cm}$. What should be the dimensions of the page so that the area of the paper used is minimum.
Solution:
Let the length of the printed page be $=\mathrm{x} \mathrm{cm}$ and breadth $=y \mathrm{~cm}$
Now xy $=24$
$
\Rightarrow \mathrm{y}=\frac{24}{x}
$
The length of the paper $=y+3$
Area $A=(x+2)(y+3)$
$=x y+3 x+2 y+6$
$=24+3 \mathrm{x}+2 \mathrm{y}+6$
$=3 \mathrm{x}+2 \mathrm{y}+30$
Substituting (1) in (2) we get

$
\begin{aligned}
& \mathrm{A}=3 x+2\left(\frac{24}{x}\right)+30 \\
& \mathrm{~A}^{\prime}(x)=3+48\left(\frac{-1}{x^2}\right)=3-\frac{48}{x^2} \\
& \mathrm{~A}^{\prime \prime}(x)=-48\left(\frac{-2}{x^3}\right)=\frac{96}{x^3} \\
& \mathrm{~A}^{\prime}(x)= 0 \Rightarrow 3 x^2=48 \\
& x^2=16 \Rightarrow x=4
\end{aligned}
$
at $x=4, \mathrm{~A}^{\prime \prime}(x)=\frac{96}{4^3}=+$ ve is a minimum point
So at $x=4, y=\frac{24}{x}=\frac{24}{4}=6$
$\therefore$ Dimensions of the paper are
$
\mathrm{x}+2=4+2=6 \mathrm{~cm}
$
and $\mathrm{y}+3=6+3=9 \mathrm{~cm}$

Question 6.
A farmer plans to fence a rectangular pasture adjacent to a river. The pasture must contain $1,80,000$ sq. mtrs in order to provide enough grass for herds. No fencing is needed along the river. What is the length of the minimum needed fencing material?
Solution:
Given Area $=180000$ sq. mtrs
Let length be $=x$
and breadth be $=y$
$
\Rightarrow y=\frac{180000}{x}
$
Now perimeter $=2 x+y(\because$ one side is along the river $)$
Now $p=2 x+\frac{180000}{x}$ from (1)
$
\begin{aligned}
& \frac{d p}{d x}=2-\frac{180000}{x^2} \\
& \frac{d^2 p}{d x^2}=-180000\left(\frac{-2}{x^3}\right)=\frac{360000}{x^3}=+\mathrm{ve} \\
& \frac{d p}{d x}=0 \Rightarrow 2=\frac{180000}{x^2} \\
& x^2=90000 \\
& x=\sqrt{90000}=300 \mathrm{~m}
\end{aligned}
$
at $x=300 \mathrm{~m}, p^{\prime \prime}$ is positive $\Rightarrow x=300$ is a minimum point
when $x=300 \mathrm{~m}, y=\frac{180000}{300}=600 \mathrm{~m}$
So minimum perimeter $=2 x+y$
$
=2(300)+600=1200 \mathrm{~m}
$

Question 7.
Find the dimensions of the rectangle with maximum area that can be inscribed in a circle of radius $10 \mathrm{~cm}$.
Solution:
$\mathrm{P}$ is a point on the circumference of a circle of radius $10 \mathrm{~cm} \mathrm{P}=(10 \cos \alpha, 10 \sin \alpha)$
$\therefore \mathrm{PQ}=20 \sin \alpha$ and
$\mathrm{PS}=20 \cos \alpha$
$\mathrm{A}=$ area of $\mathrm{PQRS}=(20 \sin \alpha)(20 \cos \alpha)$
$=400 \sin \alpha \operatorname{cosc} \alpha$
$=(200)(2 \sin \alpha \cos \alpha)$

$
\begin{aligned}
& =200 \sin 2 \alpha \\
\frac{d \mathrm{~A}}{d \alpha} & =(200)(\cos 2 \alpha)(2)=400 \cos 2 \alpha \\
\frac{d^2 \mathrm{~A}}{d \alpha^2} & =400(-\sin 2 \alpha) 2=-800 \sin 2 \alpha \\
\frac{d \mathrm{~A}}{d \alpha} & =0 \Rightarrow \cos 2 \alpha=0 \Rightarrow 2 \alpha=\frac{\pi}{2} \Rightarrow \alpha=\frac{\pi}{4} \\
\text { at } \alpha & =\frac{\pi}{4} \Rightarrow \frac{d^2 \mathrm{~A}}{d \alpha^2}=(-800)(1)=-\mathrm{ve}
\end{aligned}
$
$\Rightarrow$ Area is maximum when $\alpha=\frac{\pi}{4}$
So $\mathrm{PQ}=20\left(\frac{1}{\sqrt{2}}\right)=\frac{20 \times \sqrt{2}}{\sqrt{2} \sqrt{2}}=10 \sqrt{2}$ and $\mathrm{PS}=20\left(\frac{1}{\sqrt{2}}\right)=10 \sqrt{2}$
So the dimensions of the rectangle are $10 \sqrt{2} \mathrm{~cm}, 10 \sqrt{2} \mathrm{~cm}$,
Question 8.
Prove that among all the rectangles of the given perimeter, the square has the maximum area.
Solution:
Let the length and breadth of the rectangle be $\mathrm{x}$ and $\mathrm{y}$ respectively.
$
\begin{aligned}
& \mathrm{P}=2(\mathrm{x}+\mathrm{y}) \text { [given }] \\
& \Rightarrow x+y=\frac{\mathrm{P}}{2} \text { (or) } y=\frac{\mathrm{P}}{2}-x \\
& \text { Area }=\mathrm{A}=x y=x\left[\frac{\mathrm{P}}{2}-x\right]
\end{aligned}
$
Substitute (3) in (2) we get

$
\begin{aligned}
\mathrm{V} & =x^2\left(\frac{108-x^2}{4 x}\right)=27 x-\frac{x^3}{4} \\
\mathrm{~V} & =27 x-\frac{x^3}{4} \\
\frac{d \mathrm{~V}}{d x} & =27-\frac{3 x^2}{4} \\
\frac{d^2 \mathrm{~V}}{d x^2} & =\frac{-6 x}{4}=\frac{-3 x}{2} \\
\frac{d \mathrm{~V}}{d x} & =0 \Rightarrow 27-\frac{3 x^2}{4}=0 \\
\Rightarrow 3 x^2 & =108 \\
\Rightarrow x^2 & =36 \Rightarrow x=6
\end{aligned}
$
at $x=6, \frac{d^2 \mathrm{~V}}{d x^2}$ is $-\mathrm{ve}$
$\Rightarrow x=6$ is a maximum point.
at $x=6, y=\frac{108-36}{4 \times 6}=\frac{72}{24}=3$
so $\mathrm{x}=6 \mathrm{~cm}$ and $\mathrm{y}=3 \mathrm{~cm}$

Question 9.
Find the dimensions of the largest rectangle that can be inscribed in a semi circle of radius $\mathrm{r} \mathrm{cm}$.

Solution:
Let $\theta$ be the angle made by $O P$ with the positive direction of $x$-axis. Then the area of the rectangle $A$ is

$
\begin{aligned}
& \mathrm{A}(\theta)=(2 \mathrm{r} \cos \theta)(\mathrm{r} \sin \theta) \\
& =\mathrm{r}^2 2 \sin \theta \cos \theta=\mathrm{r}^2 \sin 2 \theta
\end{aligned}
$
Now $\mathrm{A}(\theta)$ is maximum when $\sin 2 \theta$ is maximum. The maximum value of $\sin 2 \theta=1 \Rightarrow 2 \theta=\frac{\pi}{2}$ or $\theta=\frac{\pi}{4}$. (Note that $A^{\prime}(\theta)=0$ when $\theta=\frac{\pi}{4}$ ) Therefore the critical number is $\frac{\pi}{4}$. The Area A $\left(\frac{\pi}{4}\right)=r^2$.

Question 10.
A manufacturer wants to design an open box having a square base and a surface area of $108 \mathrm{sq} . \mathrm{cm}$. Determine the dimensions of the box for the maximum volume.
Solution:
Let the side of the square base be $=x \mathrm{~cm}$ and the height be $=\mathrm{y} \mathrm{cm}$
Surface area $=108 \mathrm{sq} \mathrm{cm}$
$
\begin{aligned}
& \Rightarrow x^2+4 x y=108 \mathrm{sq} \mathrm{cm} \\
& \begin{array}{l}
\text { Volume }=x^2 y \\
\text { from (1) } \Rightarrow 4 x y=108-x^2
\end{array} \\
& \Rightarrow \quad y=\frac{108-x^2}{4 x} \ldots . \text { (3) }
\end{aligned}
$
Substituting (3) in (2) we get
$
\begin{aligned}
\mathrm{V} & =x^2\left(\frac{108-x^2}{4 x}\right)=27 x-\frac{x^3}{4} \\
\mathrm{~V} & =27 x-\frac{x^3}{4} \\
\frac{d \mathrm{~V}}{d x} & =27-\frac{3 x^2}{4} \\
\frac{d^2 \mathrm{~V}}{d x^2} & =\frac{-6 x}{4}=\frac{-3 x}{2} \\
\frac{d \mathrm{~V}}{d x} & =0 \Rightarrow 27-\frac{3 x^2}{4}=0 \\
\Rightarrow 3 x^2 & =108 \\
\Rightarrow x^2 & =36 \Rightarrow x=6
\end{aligned}
$
at $x=6, \frac{d^2 \mathrm{~V}}{d x^2}$ is $-\mathrm{ve}$
$\Rightarrow x=6$ is a maximum point.
at $x=6, y=\frac{108-36}{4 \times 6}=\frac{72}{24}=3$
so $\mathrm{x}=6 \mathrm{~cm}$ and $\mathrm{y}=3 \mathrm{~cm}$

Question 11.
The volume of a cylinder is given by the formula $V=\pi r^2 h$. Find the greatest and least values of $V$ if $r+h=6$.

Solution:
$
\mathrm{V}=\pi \mathrm{r}^2 \mathrm{~h}
$
Given $r+h=6 \Rightarrow h=6-r$
$
\begin{aligned}
\mathrm{V}=\pi \mathrm{r}^2(6-\mathrm{r}) & =6 \pi r^2-\pi \mathrm{r}^3 \\
\frac{d \mathrm{~V}}{d r} & =6 \pi(2 r)-3 \pi r^2 \\
& =12 \pi r-3 \pi r^2 \\
\frac{d^2 \mathrm{~V}}{d r^2} & =12 \pi-6 \pi r \\
\frac{d \mathrm{~V}}{d r} & =0 \Rightarrow 12 \pi r-3 \pi r^2=0 \\
3 \pi \mathrm{r}(4-\mathrm{r})=0 & \Rightarrow \mathrm{r}=0 \text { or } 4
\end{aligned}
$
when $r=4, h=2$
So $\mathrm{v}=\pi(16)(2)=32 \pi$
when $r=0, \mathrm{~V}=0$
So the maximum volume $=32 \pi$ and the minimum volume $=0$

Question 12.
A hollow cone with base radius a $\mathrm{cm}$ and height $\mathrm{b} \mathrm{cm}$ is placed on a table. Show that the volume of the largest cylinder that can be hidden underneath is - times volume of the cone.

Solution:
The height of cone $=\mathrm{h}=\mathrm{b}$
The base radius $=r=a$
The base radius of cylinder $=r$
The height of cylinder $=\mathrm{h}$

From the diagram $\Rightarrow \frac{h}{a-r}=\frac{b}{a}$ (using similar triangles)
$
\Rightarrow h=\frac{b}{a}(a-r)=b-\frac{b}{a} r
$
Volume of cylinder $\mathrm{V}=\pi r^2 h$
(i.e.,)
$
=\pi r^2\left[b-\frac{b}{a} r\right]
$
$
\begin{aligned}
\frac{d \mathrm{~V}}{d r} & =2 \pi b r-\frac{\pi b}{a}\left(3 r^2\right) \\
& =\frac{2 \pi a b r-3 \pi b r^2}{a}
\end{aligned}
$
$
\frac{d \mathrm{~V}}{d r}=0 \Rightarrow \pi b r(2 a-3 r)=0 \Rightarrow 2 a=3 r \Rightarrow r=\frac{2 a}{3}
$
$
\frac{d^2 \mathrm{~V}}{d r^2}=2 \pi b-\frac{6 \pi b r}{a}
$
at $r=\frac{2 a}{3}, \frac{d^2 \mathrm{~V}}{d r^2}=2 \pi b-\frac{6 \pi b}{a}\left(\frac{2 a}{3}\right)=-2 \pi b$
$r=\frac{2 a}{3}$ is a maximum point
So volume is maximum at $r=\frac{2 a}{3}$

So $h=b-\frac{b}{a}\left(\frac{2 a}{3}\right)=\frac{b}{3}$
$
\begin{aligned}
\text { Volume of cylinder } & =\pi r^2 h=\pi\left(\frac{2 a}{3}\right)^2\left(\frac{b}{3}\right) \\
& =\pi\left(\frac{4 a}{a}\right)^2\left(\frac{b}{3}\right)=\frac{4}{9}\left(\frac{1}{3} \pi a^2 b\right) \\
& =\frac{4}{9} \text { (volume of cone) }
\end{aligned}
$