Case study (4 Marks)

Question 14 Marks

A magazine company in a town has $5000$ subscribers on its list and collects fix charges of $₹ 3000$ per year from each subscriber. 'The company proposes to increase the annual charges, and it is believed that for every increase of $₹ 1$ one subscriber will discontinue service.

Based on the above information, answer the following questions.

If $x$ denote the amount of increase in annual charges, then revenue $R,$ as a function of $x$ can be represented as.

$R(x) = 3000 × 5000 × x$
$R(x) = (3000 - 2x)(5000 + 2x)$
$R(x) = (5000 + x)(3000 - x)$
$R(x) = (3000 + x)(5000 - x)$

If magazine company increases $₹\ 500$ as annual charges, then R is equal to.

$₹\ 15750000$
$₹\ 16750000$
$₹\ 17500000$
$₹\ 15000000$

If revenue collected by the magazine company is $₹\ 15640000$, then value of amount increased as annual charges for each subscriber, is.

$400$
$1600$
Both $(a)$ and $(b)$
None of these

What amount of increase in annual charges will bring maximum revenue?

$₹\ 1000$
$₹\ 2000$
$₹\ 3000$
$₹\ 4000$

Maximum revenue is equal to.

$₹\ 15000000$
$₹\ 16000000$
$₹\ 20500000$
$₹\ 25000000$

Answer

(d) $R(x) = (3000 + x)(5000 - x)$

Solution:

If $x$ be the amount of increase in annual charges, then number of subscriber reduces to $5000 -x $

$\therefore$ Revenue, $R(x) = (3000 + x)(5000 - x)$

$= 15000000 + 2000x - x^2, 0 < x < 5000$

(a) $₹\ 15750000$

Solution:

Clearly, at $x = 500$

$R(500) = 15000000 + 2000(500) - (500)^2$

$= 15000000 + 1000000 - 250000 = ₹\ 15750000$

Solution:

Since, $15000000 + 2000x - x^2 = 15640000 ($Given$)$

$\Rightarrow x^2 - 2000x + 640000 = 0$

$\Rightarrow x^2 - 1600x - 400x + 640000 = 0$

$\Rightarrow x(x - 1600) - 400(x - 1600) = 0$

$\Rightarrow x = 400, 1600$

(a) $₹\ 1000$

Solution:

$\frac{\text{dR}}{\text{dx}}=2000-2\text{x}\text{ and } \frac{\text{d}^2\text{R}}{\text{dx}^2}=-2<0$

For maximum revenue $\frac{\text{dR}}{\text{dx}}=0$

$\Rightarrow\text{x}=0=1000$

$\therefore $ Required amount $=₹\ 1000$

(b) $₹\ 16000000$

Solution:

Maximum revenue $= R(1000)$

$= (3000 + 1000)(5000 - 1000)$

$= 4000 × 4000 = ₹\ 16000000$

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Question 24 Marks

Shreya got a rectangular parallelepiped shaped box and spherical ball inside it as return gift. Sides of the box are x, 2x, and $\frac{\text{x}}{3},$ while radius of the ball is r.

Based on the above information, answer the following questions.

If S represents the sum of volume of parallelepiped and sphere, then Scan be written as.

$\frac{4\text{x}^3}{3}+\frac{2}{2}\pi\text{r}^2$
$\frac{2\text{x}^2}{3}+\frac{4}{3}\pi\text{r}^2$
$\frac{2\text{x}^3}{3}+\frac{4}{3}\pi\text{r}^3$
$\frac{2}{3}\text{x}+\frac{4}{3}\pi\text{r}$

If sum of the surface areas of box and ball are given to be constant $k^2$ then x is equal to.

$\sqrt{\frac{\text{k}^2-4\pi\text{r}^2}{6}}$
$\sqrt{\frac{\text{k}^2-4\pi\text{r}}{6}}$
$\sqrt{\frac{\text{k}^2-4\pi}{6}}$
$\text{None of these}$

The radius of the ball, when Sis minimum, is.

$\sqrt{\frac{\text{k}^2}{54+\pi}}$
$\sqrt{\frac{\text{k}^2}{54+4}}$
$\sqrt{\frac{\text{k}^2}{64+3\pi}}$
$\sqrt{\frac{\text{k}^2}{4\pi+3}}$

Relation between length of the box and radius of the ball can be represented as.

$\text{x} = \frac{2}{\text{r}}$
$\text{x}=\frac{\text{r}}{2}$
$\text{x}=\frac{2}{\text{r}}$
$\text{x}=3\text{r}$

Minimum value of S is.

$\frac{\text{k}^2}{2(3\pi+54)^\frac{2}{3}}$
$\frac{\text{k}}{2(3\pi+54)^\frac{3}{2}}$
$\frac{\text{k}^3}{2(4\pi+54)^\frac{1}{2}}$
$\text{None of these}$

Answer

Solution:
Let S be the sum of volume of parallelepiped and sphere, then
$\text{S}=\text{x}(2\text{x})\Big(\frac{\text{x}}{3}\Big)+\frac{4}{3}\pi\text{r}^3=\frac{2\text{x}^3}{3}+\frac{4}{3}\pi\text{r}^3$

(a) $\sqrt{\frac{\text{k}^2-4\pi\text{r}^2}{6}}$

Solution:
Since, sum of surface area of box and sphere is given to be constant $k^2.$
$\therefore2\Big(\text{x}\times\text{2x}+\text{2x}\times\frac{\text{x}}{3}+\frac{\text{x}}{3}\times\text{x}\Big)+4\pi\text{r}^2=\text{k}^2$
$\Rightarrow6\text{x}^2+4\pi\text{r}^2=\text{k}^2$
$\Rightarrow\text{x}^2=\frac{\text{k}^2-4\pi\text{r}}{6}\Rightarrow\text{x}=\sqrt{\frac{\text{k}^2-4\pi\text{r}}{6}}$

(b) $\sqrt{\frac{\text{k}^2}{54+4}}$

Solution:
From (1) and (2), we get
$\text{s}=\frac{2}{3}\bigg(\frac{\text{k}^2-4\pi\text{r}^2}{6}\bigg)^\frac{3}{2}+\frac{4}{3}\pi\text{r}^3$
$=\frac{2}{3\times6\sqrt{6}}(\text{k}^2-4\pi\text{r}^2)^\frac{3}{2}+\frac{4}{3}\pi\text{r}^3$
$\Rightarrow\frac{\text{ds}}{\text{dr}}=\frac{1}{9\sqrt{6}}\frac{3}{2}(\text{k}^2-4\pi\text{r}^2)^\frac{1}{2}-8\pi\text{r}+4\pi\text{r}^2$
$=4\pi\text{r}\bigg[\text{r}\frac{1}{3\sqrt{6}}\sqrt{\text{k}^2-4\pi\text{r}^2}\bigg]$
For maximum/minimum, $\frac{\text{ds}}{\text{dr}}=0$
$\Rightarrow\frac{4\pi\text{r}}{3\sqrt{6}}\sqrt{\text{k}-4\pi\text{r}^2}=-4\pi\text{r}^2$
$\text{k}^2-4\pi\text{r}^2=54\text{r}^2$
$\Rightarrow\text{r}^2=\frac{\text{k}^2}{54+4\pi}\Rightarrow\text{r}=\sqrt{\frac{\text{k}^2}{54+4\pi}}$

(d) $\text{x}=3\text{r}$

Solution:
$\text{Since},\text{x}^2=\frac{\text{k}-4\pi\text{r}^2}{6}=\frac{1}{6}\bigg[\text{k}^2-4\pi\Big(\frac{\text{k}^2}{54+4\pi}\Big)\bigg]$
[From (2) and (3)
$=\frac{9\text{k}^2}{54+4\pi}=9\Big(\frac{\text{k}^2}{54+4\pi}\Big)=9\text{r}^2=(3\text{r})^2$
⇒ x = 3r

Solution:
Minimum value of S is given by
$\frac{2}{3}(3\text{r})^3+\frac{4}{3}\pi\text{r}^3$
$=18\text{r}^3+\frac{4}{3}\pi\text{r}^3=\bigg(18+\frac{4}{3}\pi\bigg)\text{r}^3$
$\bigg(18+\frac{4}{3}\pi\bigg)\text{r}^3\bigg(\frac{\text{k}^2}{54+4\pi}\bigg)^\frac{3}{2}$
$\frac{1}{3}\frac{\text{k}^3}{(54+4\pi)^\frac{1}{2}}$

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Question 34 Marks

A tin can manufacturer designs a cylindrical tin can for a company making sanitizer and disinfector. The tin can is made to hold 3 litres of sanitizer or disinfector.

Based on the above in formation, answer the following questions.

If r cm be the radius and h cm be the height of the cylindrical tin can, then the surface area expressed as a function of r as.

$2\pi\text{r}^2$
$2\pi\text{r}^2+6000$
$2\pi\text{r}^2+\frac{5000}{\text{r}}$
$2\pi\text{r}^2+\frac{6000}{\text{r}}$

The radius that will minimize the cost of the material to manufacture the tin can is.

$\sqrt[3]{\frac{600}{\pi}}\text{cm}$
$\sqrt{\frac{500}{\pi}}\text{cm}$
$\sqrt[3]{\frac{1500}{\pi}}\text{cm}$
$\sqrt{\frac{1500}{\pi}}\text{cm}$

The height that will minimize the cost of the material to manufacture the tin can is.

$\sqrt[3]{\frac{600}{\pi}}\text{cm}$
$2\sqrt[3]{\frac{1500}{\pi}}\text{cm}$
$\sqrt{\frac{1500}{\pi}}$
$2\sqrt{\frac{1500}{\pi}}$

If the cost of material used to manufacture the tin can is $₹\frac{100}{\text{m}^2}$ and $\sqrt[3]{\frac{1500}{\pi}}\approx7.8,$ then minimum cost is approximately.

₹ 11.538
₹ 12
₹ 13
₹ 14

To minimize the cost of the material used to manufacture the tin can, we need to minimize the.

Volume.
Curved surface area.
Total surface area.
Surface area of the base.

Answer

$2\pi\text{r}^2+\frac{6000}{\text{r}}$

Solution:
Given, r cm is the radius and 11cm is the height of required cylindrical can. Given that, volume = 3 l = $3000cm^3$
($\therefore 11 l = 1000cm^3$)
$\Rightarrow\pi\text{r}^2\text{h}=3000\Rightarrow\text{h}=\frac{3000}{\pi\text{r}^2}$
Now, the surface area, as a function of r is given by
$\text{S}\big(\text{r}\big)=2\pi\text{r}^2+2\pi\text{r}^2+2\pi\text{r}\bigg(\frac{3000}{\pi\text{r}^2}\bigg)$
$=2\pi\text{r}^2+\frac{6000}{\text{r}}$

Solution:
$\text{Now},\text{S}'\big(\text{r}\big)=2\pi\text{r}^2+\frac{6000}{\text{r}}$
$\Rightarrow\text{S}\big(\text{r}\big)=4\pi\text{r}-\frac{6000}{\text{r}^2}$
To find critical points, put S'(r) = 0
$\Rightarrow\frac{4\pi\text{r}^3-6000}{\text{r}^2}=0$
$\Rightarrow\text{r}^3=\frac{6000}{4\pi}\Rightarrow\text{r}=\bigg(\frac{1500}{\pi}\bigg)^\frac{1}{3}$
$\text{Also},\text{S}''\big(\text{r}\big)\mid_\text{r}=\sqrt[3]{\frac{1500}{\pi}}=4\pi+\frac{12000\times\pi}{1500}$
$=4\pi+8\pi=12\pi>0$
Thus, the critical point is the point of minima.

(b) $2\sqrt[3]{\frac{1500}{\pi}}\text{cm}$

Solution:
The cost of material for the tin can is minimized when $\text{r}=\sqrt[3]{\frac{1500}{\pi}}\text{cm}$ and the height is
$\frac{3000}{\pi\bigg(\sqrt[3]{\frac{1500}{\pi}}\bigg)^2}=2\sqrt[3]{\frac{1500}{\pi}}\text{cm}$

(a) ₹ 11.538

Solution:
We have minimum surface area $=\frac{2\pi\text{r}^3+6000}{\text{r}}$
$=\frac{2\pi.\frac{1500}{\pi}+6000}{\sqrt[3]{\frac{1500}{\pi}}}=\frac{9000}{7.8}=1153.84\text{cm}^2$
Cost of $1m^2$ material = ₹ 100
$\therefore$ Cost of $1cm^2$ material = $₹\ \frac{1}{100}$
$\therefore$ minimum cost $=₹\frac{1153.84}{100}=₹\ 11.538$

Solution:
To minimize the cost, we need to minimize the total surface area.

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Question 44 Marks

An open water tank of aluminium sheet of negligible thickness, with a square base and vertical sides, is to be constructed in a farm for irrigation. It should hold $32000$ l of water, that comes out from a tube well.

Based on above information, answer the following questions.

If the length, width, and height of the open tank be $x, x$ and $y$ $m$ respectively, then total surface area of tank is.

$(x^2 + 2xy)m^2$
$(2x^2 + 4xy)m^2$
$(2x^2 + 2xy)m^2$
$(2x^2 + 8xy)m^2$

The relation between $x$ and $y$ is.

$x^2y = 32$
$xy^2 = 32$
$x^2y^2 = 32$
$xy = 32$

The outer surface area of tank will be minimum when depth of tank is equal to.

Half of its width.
Its width.
$\big(\frac{1}{4}\big)^\text{th}$ of its Width
$\big(\frac{1}{3}\big)^\text{rd}$ of its Width

The cost of material will be least when width of tank is equal to.

Half of its depth
Twice of its depth
$\big(\frac{1}{4}\big)^\text{th}$ of its depth
Thrice of its depth

If cost of aluminium sheet is $ ₹ \frac{360}{\text{m}^2}$ then the minimum cost for the construction of tank will be.

$₹\ 15, 000$
$₹\ 16, 280$
$₹\ 17, 280$
$₹\ 18, 280$

Answer

(d) $(2x^2 + 8xy)m^2$^

Solution:
Since the tank is open from the top, therefore the total surface area is
$= ($Outer $+$ Inner$)$ surface area
$= 2(x \times x + 2(xy + yx)) = 2(x^2 + 2(2xy)) = (2x^2 + 8xy)m^2$^

(a) $x^2y = 32$

Solution:
Since, volume of tank should be $32000l.$
$\therefore x^2y m^3 = 32000l = 32m^3$
$[\therefore$ I litre $= 0.001m^3]$
So, $x^2y = 32$

Solution:
Let S be the outer surface area of tank. Then, $S = x^2 + 4xy$
$\Rightarrow\text{S}(\text{x})=\text{x}^2+4\text{x}\frac{32}{\text{x}^2}=\text{x}^2+\frac{128}{\text{x}}$
$\Rightarrow\frac{\text{dS}}{\text{dx}}=2\text{x}-\frac{128}{\text{x}^2}\text{ and }\frac{\text{d}^2\text{S}}{\text{dx}^2}=2+\frac{256}{\text{x}^3}$
For maximum or minimum values of S, consider $\frac{\text{dS}}{\text{dx}}=0$
$\Rightarrow2\text{x}=\frac{128}{\text{x}^2}$
$\Rightarrow\text{x}^3=64$
$\Rightarrow\text{x}=4\text{m}$
$\text{At}\text{ x }=4\frac{\text{d}^2\text{S}}{\text{dx}^2}=2+\frac{256}{4^3}=2+4=6>0$
$\therefore S$ is minimum when $x = 4$
Now as $x - y = 32,$ therefore $y = 2$ Thus $x = 2y$

(b) Twice of its depth

Solution:
Since, surface area is minimum when $x = 2y,$ therefore cost of material will be least when $x = 2y$. Thus, cost of material will be least when width is equal to twice of its depth Since, surface area is minimum when $x = 2y,$ therefore cost of material will be least when $x = 2y.$ Thus, cost of material will be least when width is equal to twice of its depth.

Solution:
Since, minimum surface area $= x^2 + 4xy = 4^2 + 4 \times 4 \times 2 = 48m^2 $ and cost per $m^2 = ₹\ 360$
minimum cost is $= ₹\ (48 \times 360) = ₹\ 17280.$

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Question 54 Marks

Western music concert is organised every year in the stadium that can hold 36000 spectators. With ticket price of ₹ 10, the average attendance has been 24000. Some financial expert estimated that price of a ticket should be determined by the function.
$\text{p}(\text{x})=15-\frac{\text{x}}{3000}$ where x is the number of tickets sold.

Based on the above information, answer the following questions.

The revenue, R as a function of x can be represented as.

$15\text{x}-\frac{\text{x}^2}{3000}$
$15-\frac{\text{x}^2}{3000}$
$15\text{x}-\frac{1}{3000}$
$15\text{x}-\frac{\text{x}}{3000}$

The range of x is.

[24000, 36000]
[0, 24000]
[0, 36000]
None of these

The value of x for which revenue is maximum, is.

20000
21000
22500
25000

When the revenue is maximum, the price of the ticket is.

₹ 5
₹ 5.5
₹ 7
₹ 7.5

How many spectators should be present to maximize the revenue?

21500
21000
22000
22500

Answer

(a) $15\text{x}-\frac{\text{x}^2}{3000}$

Solution:

Let p be the price per ticket and x be the number of tickets sold. Then, revenue function $\text{R}\big(\text{x}\big)=\text{p}\times\text{x}=\bigg(15-\frac{\text{x}}{3000}\bigg)\text{x}=15\text{x}-\frac{\text{x}^2}{3000}$

[0, 36000]

Solution:

Since, more than 36000 tickets cannot be sold. So, range of x is [0, 36000]

Solution:

We have, $\text{R}\big(\text{x}\big)=15\text{x}-\frac{\text{x}^2}{3000}$

$\Rightarrow\text{R}\big(\text{x}\big)=15-\frac{\text{x}^2}{1500}$

For maxima/ minima, put R'(x) = 0

$\Rightarrow15-\frac{\text{x}}{1500}=0$

$\Rightarrow\text{x}=2250$

Also $\text{R}''\big(\text{x}\big)=-\frac{1}{1500}<0$

(d) ₹ 7.5

Solution:

Maximum revenue will be at x = 22500

$\therefore$ Price of a ticket $=15-\frac{22500}{3000}=15-7.5=₹\ 7.5$

(d) 22500

Solution:

Number of spectators will be equal to number of tickets sold

$\therefore$ Required number of spectators = 22500

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Question 64 Marks

Nitin wants to construct a rectangular plastic tank for his house that can hold $80\ ft^3$ of water. The top of the tank is open. The width of tank will be $5$ ft but the length and heights are variables. Building the tank cost? $₹\ 20$ per sq. foot for the base and $₹\ 10$ per sq. foot for the side.

Based on the above information, answer the following questions.

In order to make a least expensive water tank, Nitin need to minimize its.

Volume
Base
Curved surface area
Cost

Total cost of tank as a function of h can be represented as.

$c(h) = 100h - 320 - 1600/ h$
$c(h) = 100h - 320h - 720h^2$
$c(h) = 100 + 320h + 1600h^2$
$\text{c}\big(\text{h}\big)=100\text{h}+320+\frac{1600}{\text{h}}$

Range of $h$ is.

$(3, 5)$
$\big(0,\infty\big)$
$(0, 8)$
$(0, 3)$

Value of hat which $c(h)$ is minimum, is.

$4$
$5$
$6$
$6, 7$

The cost of least expensive tank is.

$₹\ 1020$
$₹\ 1100$
$₹\ 1120$
$₹\ 1220$

Answer

(d) Cost

Solution:
In order to make least expensive water tank, Nitin need to minimize its cost.

(d) $\text{c}\big(\text{h}\big)=100\text{h}+320+\frac{1600}{\text{h}}$

Solution:
Let l ft be the length and h ft be the height of the tank. Since breadth is equal to $5$ ft. (Given)
$\therefore$ Two sides will be 5h sq. feet and two sides will be $l/ h$ sq. feet. So, the total area of the sides is $(10h + 2l/ h)ft^2$
Cost of the sides is $ ₹\ 10$ per sq. foot. So, the cost to build the sides is $(10h + 2l/ h) \times 10\ ₹ (100h +20l/ h)$
Also, cost of base $= (5l) \times 20 = ₹\ 100l$
$\therefore$ Total cost of the tank in ₹ is given by
$c = 100h + 20l/ h + 100l$
Since, volume of tank $= 80\ ft^3$
$\therefore5l\text{h}=80\text{ft}^3\therefore\text{l}=\frac{80}{5\text{h}}=\frac{16}{\text{h}}$
$\therefore\text{c}\big(\text{h}\big)=100\text{h}+20\Big(\frac{16}{\text{h}}\Big)\text{h}+100(\frac{16}{\text{h}}\Big)$
$=100\text{h}+320+\frac{1600}{\text{h}}$

(b) $\big(0,\infty\big)$

Solution:
Since, all side lengths must be positive.
$\therefore\text{h}>0\ \text{and}\frac{16}{\text{h}}>0$
Since $\frac{16}{\text{h}}>0$ whenever $h > 0$
$\therefore$ Range of h is $(0,\infty)$

(a) $4$

Solution:
To minimize cost, $\frac{\text{dc}}{\text{dh}}=0$
$\Rightarrow100-\frac{1600}{\text{h}^2}=0$
$\Rightarrow100\text{h}^2=1600$
$\Rightarrow\text{h}^2=16$
$\Rightarrow\text{h}=\pm4$
$\Rightarrow\text{h}=4$ [$\therefore$ height can not be negative]

Solution:
Cost of least expensive tank is given by
$\text{c}(4)=400+320+\frac{1600}{4}$
$=720+400=₹\ 1120$

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Question 74 Marks

A poster is to be formed for a company advertisement. The top and bottom margins of poster should be 9cm and the side margins should be 6cm. Also, the area for printing the advertisement should be $864cm^2$.

Based on the above information, answer the following questions.

If a cm be the width and b cm be the height of Poster, then the area of poster, expressed in terms of a and b, is given by.

$648 + 18a + 12b$
$18a + 12b$
$584 + 18a + 12b$
None of these

The relation between a and b is given by.

$\text{a}=\frac{648+12\text{b}}{\text{b}-18}$
$\text{a}=\frac{12\text{b}}{\text{b}-18}$
$\text{a}=\frac{12\text{b}}{\text{b}+18}$
$\text{None of these}$

Area of poster in terms of bis given by.

$\text{a}=\frac{12\text{b}^2}{\text{b}-18}$
$\text{a}=\frac{648\text{b}+12\text{b}^2}{\text{b}-18}$
$\text{a}=\frac{648\text{b}+12\text{b}^2}{\text{b}+18}$
$\text{a}=\frac{12\text{b}^2}{\text{b}+18}$

The value of b, so that area of the poster is minimized, is.

The value of a, so that area of the poster is minimized, is.

Answer

(a) 648 + 18a + 12b

Solution:

Let A be the area of the poster, then
$A = 864 + 2(a·9) + 2(b·6) - 4(6·9)$
$= 864 + 18a + 12b - 216 = 648 + 18a + 12b$

(a) $\text{a}=\frac{648+12\text{b}}{\text{b}-18}$

Solution:

Clearly, $A = a·b$
$\therefore 648 + 18a + 12b = ab$
$\Rightarrow ab - 18a = 648 + 12b$
$\Rightarrow a(b - 18) = 648 + 12b$
$\Rightarrow\text{a}=\frac{648+12\text{b}}{\text{b}-18}$

(b) $\text{a}=\frac{648\text{b}+12\text{b}^2}{\text{b}-18}$

Solution:
Since, A = a·b, therefore
$\text{A}=\bigg(\frac{648+12\text{b}}{\text{b}-18}\bigg)\text{b}=\frac{648\text{b}+12\text{b}^2}{\text{b}-18}\bigg[\therefore\text{a}=\frac{648+12\text{b}}{\text{b}-18}\bigg]$

(a) 54

Solution:
Clearly
$\text{A}'\big(\text{b}\big)=\frac{\big(\text{b}-18\big)\big(648+2\text{ab}\big)-\big(648\text{b}+12\text{b}^2\big)}{\big(\text{b}-18\big)^2}$
$=\frac{12(\text{b}^2-36\text{b}-972)}{\big(\text{b}-18\big)^2}$
For minimum, consider A'(b) = 0
$\Rightarrow b^2 - 36b - 972 = 0$
$\Rightarrow b^2 - 54b + 18b - 972 = 0$
$\Rightarrow b(b - 54) + 18(b - 54) = 0$
$\Rightarrow b = -18 or b = 54$
$\therefore$ b is height, therefore can't be negative. So, $b = 54$.

(b) 36

Solution:
$\text{Since},\text{a}=\frac{648+12\text{b}}{\text{b}-18}$
$\therefore\text{a}=\frac{648+12\times54}{54-18}=\frac{648+648}{36}=36$

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Question 84 Marks

In a street, two lamp posts are 600 feet apart. The light intensity at a distanced from the first (stronger) lamp post is $\frac{1000}{\text{d}^2}$ the light intensity at distance d from the second (weaker) lamp post is $\frac{125}{\text{d}^2}$ (in both cases the light intensity is inversely proportional to the square of the distance to the light source). The combined light intensity is the sum of the two light intensities coming from both lamp posts.

Based on the above information, answer the following questions.

If you are in between the lamp posts, at distance x feet from the stronger tight, then the formula for the combined light intensity coming from both lamp posts as function of x, is

$\frac{1000}{\text{x}^2}+\frac{125}{\text{x}^2}$
$\frac{1000}{\big(600-\text{x}^2\big)}+\frac{125}{\text{x}^2}$
$\frac{1000}{\text{x}^2}+\frac{125}{\big(600-\text{x}^2\big)}$
$\text{None of these}$

The maximum value of x can not be.

100
200
600
None of these

The minimum value of x can not be.

0
100
200
None of these

If l(x) denote the combined tight intensity, then l(x) will be minimum when x =

The darkest spot between the two lights is.

At a distance of 200 feet from the weaker lamp post.
At distance of 200 feet from the stronger lamp post.
At a distance of 400 feet from the weaker lamp post.
None of these

Answer

Solution:

Since, the distance is x feet from the stronger light, therefore the distance from the weaker light will be 600 - x

So, the combined light intensity from both lamp posts is given by $\frac{1000}{\text{x}^2}+\frac{125}{(600-\text{x})^2}$

Solution:

Since, the person is in between the lamp posts, therefore x will lie in the interval (0, 600). So, maximum value of x can't be 600.

(a) 0

Solution:

Since, 0 < x < 600, therefore minimum value of x cant be 0.

(b) 400

Solution:

$\text{We have l (x)}=\frac{1000}{\text{x}^2}+\frac{125}{(600-\text{x})^2}$

$\Rightarrow \text{l}'(\text{x})=\frac{-2000}{\text{x}^3}+\frac{250}{(600-\text{x})^2}\text{and}$

$\Rightarrow \text{l}'(\text{x})=\frac{6000}{\text{x}^4}+\frac{750}{(600-\text{x})^4}$

For maxima/ minima, l'(x) = 0

$\Rightarrow \frac{2000}{\text{x}^3}=\frac{250}{(600-\text{x})^3}$

$\Rightarrow8(600-\text{x})^3=\text{x}^3$

Taking cube root on both sides, we get 2(600 - x) = x

⇒ 1200 = 3x

⇒ x = 400

Thus, l(x) is minimum when you are at 400 feet from the strong intensity lamp post.

(a) At a distance of 200 feet from the weaker lamp post.

Solution:

Since, l(x) is minimum when x = 400 feet, therefore the darkest spot between the two light is at a distance of 400 feet from stronger lamp post, i.e., at a distance of 600 - 400 = 200 feet from the weaker lamp post.

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Question 94 Marks

The Government declare that farmers can get $₹\ 300$ per quintal for their onions on $1^{st}$ July and after that, the price will be dropped by $₹\ 3$ per quintal per extra day. Shyams father has $80$ quintal of onions in the field on $1^{st}$ July, and he estimates that crop is increasing at the rate of 1 quintal per day.

Based on the above information, answer the following questions.

If $x$ is the number of days after $1^{st}$ July, then price and quantity of onion respectively can be expressed as.

$₹(300 - 3x), (80 + x)$ quintals
$₹(300 - 3x), (80 - x)$ quintals
$₹(300 + x), 80$ quintals
None of these

Revenue $R$ as a function of $x$ can be represented as.

$R(x) = 3x^2 - 60x - 24000$
$R(x) = 3x^2 + 60x - 24000$
$R(x) = 3x^2 + 40x - 16000$
$R(x) = 3x^2 - 60x - 14000$

Find the number of days after $1^{st}$ July, when Shyam's father attain maximum revenue.

$10$
$20$
$12$
$22$

On which day should Shyam's father harvest the onions to maximise his revenue?

$11^{th}$ July
$20^{th}$ July
$12^{th}$ July
$22^{nd}$ July

Maximum revenue is equal to.

$₹\ 20,000$
$₹\ 24,000$
$₹\ 24,300$
$₹\ 24,700$

Answer

(a) $₹(300 - 3x), (80 + x)$ quintals

Solution:
Let $x$ be the number of extra days after $1^{st}$ July.
$\therefore$ Price $= ₹(300 - 3 \times x) = ₹(300 - 3x)$
Quantity $= 80$ quintals $+ x (1$ quintal per day$) = (80 + x)$ quintals

(b) $R(x) = 3x^2 + 60x - 24000$

Solution:
$R(x) =$ Quantity $\times$ Price
$= (80 + x)(300 - 3x) = 24000 - 240x + 300x - 3x^2$
$= 24000 + 60x - 3x^2$

(a) $10$

Solution:
We have, $R(x) = 24000 + 60x - 3x^2$
$\Rightarrow R'(x) = 60 - 6x \Rightarrow R''(x) = -6$
For $R(x)$ to be maximum, $R'(x) = 0$ and $R"(x) < 0$
$\Rightarrow 60 - 6x = 0 \Rightarrow x = 10$

(a) $11^{th}$ July

Solution:
Shyam's father will attain maximum revenue after $10$ days. So, he should harvest the onions after $10$ days of $1^{st}$ July i.e., on $11^{th}$ July.

Solution:
Maximum revenue is collected by Shyam's father when $x = 10$
$\therefore$ Maximum revenue $= R(10)$
$= 24000 + 60(10) - 3(10)^2 = 24000 + 600 - 300 = 24300$

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Question 104 Marks

A real estate company is going to build a new residential complex. The land they have purchased can hold at most $4500$ apartments. Also, if they make x apartments, then the monthly maintenance cost for the whole complex would be as follows: Fixed cost $= ₹ 50,00,000$. Variable cost $= (160x - 0.04x^2)$

Based on the above information, answer the following questions.

The maintenance cost as a function of $x$ will be.

$160x - 0.04x^2$
$5000000$
$5000000 + 160x - 0.04x^2$
None of these

If $C(x)$ denote the maintenance cost function, then maximum value of $C(x)$ occur at $x =$

$0$
$2000$
$4500$
$5000$

The maximum value of $C(x)$ would be.

$₹\ 5225000$
$₹\ 5160000$
$₹\ 5000000$
$₹\ 4000000$

The number of apartments, that the complex should have in order to minimize the maintenance cost, is.

$4500$
$5000$
$1750$
$3500$

If the minimum maintenance cost is attain, then the maintenance cost for each apartment would be.

$₹\ 1091.11$
$₹\ 1200$
$₹\ 1000$
$₹\ 2000$

Answer

Solution:
Let $C(x)$ be the maintenance cost function, then $C(x) = 5000000 + 160x - 0.04x^2$

(b) $2000$

Solution:
We have, $C(x) = 5000000 + 160x - 0.04x^2$
Now, $C'(x) = 160 - 0.08x$
For maxima/ minima, put $C'(x) = 0$
$\Rightarrow 160 = 0.08x$
$\Rightarrow x = 2000$

(b) $₹\ 5160000$

Solution:
Clearly, from the given condition we can see ...that we only want critical points that are in the interval $[0, 4500]$
Now, we have $C(O) = 5000000$
$C(2000) = 5160000$
And $C(4500) = 4910000$
$\therefore$ Maximum value of $C(x)$ would be $₹\ 5160000$

(a) $4500$

Solution:
The complex must have $4500$ apartments to minimise the maintenance cost.

(a) $₹\ 1091.11$

Solution:
The minimum maintenance cost for each apartment woud be $₹\ 1091.11$

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Question 114 Marks

Megha wants to prepare a handmade gift box for her friend's birthday at home. For making lower part of box, she takes a square piece of cardboard of side $20cm$.

Based on the above information, answer the following questions.

If x cm be the length of each side of the square cardboard which is to be cut off from corners of the square piece of side 20cm, then possible value of x will be given by the interval.

[0, 20]
(0, 10)
(0, 3)
None of these

Volume of the open box formed by folding up the cutting corner can be expressed as.

$\text{V}=\text{x}(20-2\text{x})(20-2\text{x)}$
$\text{V}=\frac{\text{x}}{2}(20+\text{x})(20-\text{x})$
$\text{V}=\frac{\text{x}}{3}(20-\text{x})(20+\text{x})$
$\text{V}=\text{x}(20-2\text{x})(20-\text{x)}$

The values of x for which $\frac{\text{dV}}{\text{dX}}=0$, are.

3, 4
$0,\frac{10}{3}$
0, 10
$10,\frac{10}{3}$

Megha is interested in maximizing the volume of the box. So, what should be the side of the square to be cut off so that the volume of the box is maximum?

12cm
8cm
$\frac{10}{3}\text{cm}$
2cm

The maximum value of the volume is.

$\frac{17000}{27}\text{cm}^3$
$\frac{11000}{27}\text{cm}^3$
$\frac{8000}{27}\text{cm}^3$
$\frac{16000}{27}\text{cm}^3$

Answer

(b) (0, 10)

Solution:

Since, side of square is of length 20cm therefore $\text{x}\in$ (0, 10)

(a) $\text{V}=\text{x}(20-2\text{x})(20-2\text{x)}$

Solution:
Clearly, height of open box = x cm Length of open box = 20 - 2x and width of open box = 20 - 2x
$\therefore$ Volume (V) of the open box
= x × (20 - 2x) × (20 - 2x)

(d) $10,\frac{10}{3}$

Solution:
We have, $V = x(20 - 2x)^2$
$\therefore\frac{\text{dV}}{\text{dX}}=\text{x}2(20-2\text{x})+(20-2\text{x})^2$
= (20 - 2x)(-4x + 20 - 2x) = (20 - 2x)(20 - 6x)
Now $\frac{\text{dV}}{\text{dX}}=0$
$\Rightarrow 20 - 2x = 0 or 20 - 6x = 0$
$\therefore\text{x}=10\ \text{or}\ \frac{10}{3}$

Solution:
We have, $V = x(20 - 2x)^2$
and $\frac{\text{dV}}{\text{dX}}$ = (20 - 2x)(20 - 6x)
$\Rightarrow\frac{\text{d}^2\text{V}}{\text{dX}^2}$ = (20 - 2x)(-6) + (20 - 6x)(-2)
= (-2)[60 - 6x + 20 - 6x) = (-2)[80 - 12x) = 24x - 160
For $\text{x}=\frac{10}{3},\frac{\text{d}^2\text{V}}{\text{dX}^2}<0$
and For $\text{x}=\frac{10}{3},\frac{\text{d}^2\text{V}}{\text{dX}^2}>0$
So, volume will be maximum when $\text{x} = \frac{10}{3}$

(d) $\frac{16000}{27}\text{cm}^3$

Solution:
We have, V = x(20 - 2x)2, which will be maximum when $\text{x} = \frac{10}{3}$
$\therefore\text{Maximum volume}=\frac{10}{3}\Big(20-2\times\frac{10}{3}\Big)^2$
$=\frac{10}{3}\times\frac{40}{3}\times\frac{40}{3}=\frac{16000}{27}\text{cm}^3$

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Question 124 Marks

A student Arun is running on a playground along the curve given by $y = x^2 + 7$. Another student Manila standing at point $(3, 7)$ on playground wants to hit Arun by paper ball when Arun is nearest to Manila.

Based on above information, answer the following questions.

Arun's position at any value of x will be.

$(x^2, y - 7)$
$(x^2, y + 7)$
$(x, x^2 + 7)$
$(x^2, x - 7)$

Distance (say D) between Arun and Manila will be.

$(\text{x}-1)(2\text{x}^2+2\text{x}+3)$
$(\text{x}-3)^2+\text{x}^4$
$\sqrt{(\text{x}-3)+\text{x}^4}$
$\sqrt{(\text{x}-1)(2\text{x}^2+2\text{x}+3)}$

For which real value(s) of x, first derivative of $D^2$ w.r.t, x will Vanish?

Find the position of Arun when Manila will hit the paper hall.

(5, 32)
(1, 8)
(3, 7)
(3, 16)

The minimum value of D is.

$3$
$\sqrt{3}$
$5$
$\sqrt{5}$

Answer

$(c) (x, x^2 + 7)$

Solution:
For all values of $x, y = x^2 + 7$
$\therefore$ Arun's position at any point of x will be $(x, x^2 + 7)$

Solution:
Distance between Arun and Manila, i.e., D

$=\sqrt{(\text{x}-3)^2+(\text{x}^2+7-7)^2}$
$=\sqrt{(\text{x}-3)^2+\text{x}^4}$

(a) 1

Solution:
We have, $\text{D}=\sqrt{(\text{x}-3)^2+\text{x}^4}$
$\therefore\text{D}^2=(\text{x}-3)^3+\text{x}^4$
Now, $\frac{\text{d}}{\text{dx}}(\text{D}^2)=2(\text{x}-3)+4\text{x}^3=0$
$\Rightarrow4\text{x}^3+2\text{x}-6=0$
$\Rightarrow2\text{x}^3+\text{x}-3=0$
$\Rightarrow(\text{x}-1)(2\text{x}+2\text{x}+3)=0$
$\therefore\text{x}=1$
$[\therefore2\text{x}^2+2\text{x}+3=0\text{ will give imaginary values]}$

(b) (1, 8)

Solution:
We have, D $=\sqrt{(\text{x}-3)^2+\text{x}^4}$
$\text{D}'(\text{x})=\frac{2(\text{x}-3)+4\text{x}^3}{2\sqrt{(\text{x}-3)^2+\text{x}^4}}=0$
$\Rightarrow2\text{x}^3+\text{x}-3=0$
$\therefore\text{x}=1$

Clearly, D'' (x) at x = 1 is > 0
Value of x for which D will be minimum is 1 For x = 1, y = 8
Thus, the required position is (1, 8)

(d) $\sqrt{5}$

Solution:
Minimum value of $\text{D}=\sqrt{(1-3)^3+(1)^4}$
$=\sqrt{4+1}=\sqrt{5}$

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Question 134 Marks

Mr. Sahil is the owner of a high rise residential society having 50 apartments. When he set rent at ₹ 10000/ month, all apartments are rented. If he increases rent by ₹ 250/ month, one fewer apartment is rented. The maintenance cost for each occupied unit is ₹ 500/ month.

Based on the above information, answer the following questions.

If P is the rent price per apartment and N is the number of rented apartment, then profit is given by.

NP
(N - 500)P
N(P - 500)
None of these

If x represent the number of apartments which are not rented, then the profit expressed as a function of x is.

(50 - x)(38 + x)
(50 + x)(38 - x)
250(50 - x)(38 + x)
250(50 + x)(38 - x)

If P = 10500, then N =

If P = 11,000, then the profit is.

₹ 4,83,000
₹ 5,00,000
₹ 5,05,000
₹ 6,50,000

The rent that maximizes the total amount of profit is.

₹ 11000
₹ 11500
₹ 15800
₹ 16500

Answer

Solution:

If P is the rent price per apartment and N is the number of rented apartment, the profit is given by

NP - 500 N = N(P - 500)

[$\therefore$ ₹ 500/month is the maintenance charges for each occupied unit]

Solution:

Now, if x be the number of non-rented apartments, then N = 50 - x and P = 10000 + 250 x

Thus, profit = N(P - 500) = (50 - x)(10000 + 250 x - 500)

= (50 - x)(9500 + 250 x) = 250(50 - x)(38 + x)

(b) 48

Solution:

Clearly, if P = 10500, then 10500 = 10000 + 250x

⇒ x = 2 ⇒ N = 48

(a) ₹ 4,83,000

Solution:

Also, if P = 11000, then

11000 = 10000 + 250 x ⇒ x = 4 and so profit

P(4) = 250(50 - 4)(38 + 4) = ₹ 4,83,000

(b) ₹ 11500

Solution:

We have, P(x) = 250(50 - x) (38 + x)

Now, P(x) = 250[50 - x - (38 + x)] = 250[12 - 2x)

For maxima/minima, put p'(x) = 0

⇒ 12 - 2x = 0

⇒ x = 6

Thus, price per apartment is, P = 10000 + 1500 = 11500 Hence, the rent that maximizes the profit is ₹ 11500.

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Question 144 Marks

Shobhit's father wants to construct a rectangular garden using a brick wall on one side of the garden and wire fencing for the other three sides as shown in figure. He has 200 ft of wire fencing.

Based on the above information, answer the following questions.

To construct a garden using 200 ft of fencing, we need to maximise its.

Volume
Area
Perimeter
Length of the side

If x denote the length of side of garden perpendicular to brick wall and y denote the length of side parallel to brick wall, then find the relation representing total amount of fencing wire.

$x + 2y = 150$
$x + 2y = 50$
$y + 2x = 200$
$y + 2x = 100$

Area of the garden as a function of x, say A(x), can be represented as.

$200 + 2x^2$
$x - 2x^2$
$200x - 2x^2$
$200 - x^2$

Maximum value of A(x) occurs at x equals.

50 ft
30 ft
26 ft
31 ft

Maxi mum area of garden will be.

2500 sq. ft
4000 sq. ft
5000 sq. ft
6000 sq. ft

Answer

(b) Area

Solution:
To create a garden using 200 ft fencing, we need to maximise its area.

Solution:
Required relation is given by 2x + y = 200.

Solution:
Area of garden as a function of x can be rep resented as
$A(x) = x·y = x(200 - 2x) = 200x - 2x^2$

(a) 50 ft

Solution:
$A(x) = 200x - 2x^2 \Rightarrow A'(x) = 200 - 4x$
For the area to be maximum $A'(x) = 0$
$\Rightarrow 200 - 4x =0$
$\Rightarrow x = 50 ft$

Solution:
Maximum area of the garden = $200(50) - 2(50)^2 = 10000 - 5000 = 5000$ sq. ft

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Question 154 Marks

Kyra has a rectangular painting canvas having a total area of $24\ ft^2$ which includes a border of $0.5$ ft. on the left right and a border of $0.75$ ft. on the bottom, top inside it.

Based on the above information, answer the following questions.

If Kyra wants to paint in the maximum area, then she needs to maximize.

Area of outer rectangle.
Area of inner rectangle.
Area of top border.
None of these.

If x is the length of the outer rectangle, then area of inner rectangle in terms of x is.

$(\text{x}+3)\Big(\frac{24}{\text{x}}-2\Big)$
$(\text{x}-1)\Big(\frac{24}{\text{x}}+1.5\Big)$
$(\text{x}-1)\Big(\frac{24}{\text{x}}-1.5\Big)$
$(\text{x}-1)\Big(\frac{24}{\text{x}}\Big)$

Find the range of x.

$(1, \infty)$
$(1, 16)$
$(-\infty, 16)$
$(-1, 16)$

If area of inner rectangle is maximum, then x is equal to.

2 ft.
3 ft.
4 ft.
5 ft.

If area of inner rectangle is maximum, then length and breadth of this rectangle are respectively.

3 ft, 4.5 ft.
4.5 ft, 5 ft.
1 ft, 2 ft.
2 ft, 4 ft.

Answer

(b) Area of inner rectangle

Solution:
In order to paint in the maximum area, Kyra needs to maximize the area of inner rectangle.

Solution:
Let x be the length and y be the breadth of outer rectangle.
$\therefore$ Length of inner rectangle = $x - 1$
and breadth of inner rectangle = y - 1.5
$\therefore$ A(x) = (x - 1)(y - 1.5) [$\therefore$ xy = 24 (given)
$=\big(\text{x}-1\big)\bigg(\frac{24}{\text{x}}-1.5\bigg)$

(b) $(1, 16)$

Solution:
Dimensions of rectangle (outer/inner) should be positive.
$\therefore\text{x}-1>0\text{ and }\frac{24}{\text{x}}-1.5>0$
$\therefore\text{x}>1\text{ and }{\text{x}}<16$
$\therefore$ Range of x is (1, 16)

Solution:
We have $\text{A}'\text{(x)}=(\text{x}-1)\bigg(\frac{24}{\text{x}}-1.5\bigg)$
$\Rightarrow\text{A}'(\text{x)}=(\text{x}-1)\Big(\frac{-24}{\text{x}^2}\Big)+\bigg(\frac{24}{\text{x}}-1.5\bigg)$
$=\frac{24}{\text{x}^2}-1.5\ \text{and}\ \text{A}''(\text{x})=\frac{-48}{\text{x}^3}$
For A(x) to be maximum or minimum, A'(x) = 0
$\Rightarrow-1.5+\frac{24}{\text{x}^2}=0$
$\Rightarrow\text{x}^2=16$
$\Rightarrow\text{x}=\pm4$
$\therefore\text{x}=4$ [Since, length can't be negative]
Also, $\text{A}''(4)=\frac{-48}{4^3}<0$
Thus, at x = 4, area is maximum.

(a) 3 ft, 4.5 ft.

Solution:
If area of inner rectangle is maximum, then Length of inner rectangle = x - 1 = 4 - 1 = 3 ft.
And breadth of inner rectangle $=\text{y}-1.5=\frac{24}{\text{x}}-1.5$
$=\frac{24}{4}-1.5=6-1.5=4.5\ \text{ft}$

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Question 164 Marks

In a society there is a garden in the shape of rectangle inscribed in a circle of radius 10m as shown in given figure.

Based on the above information, answer the following questions.

If 2x and 2y denotes the length and breadth in metres, of the rectangular part, then the relation between the variables is.

$x^2 - y^2 = 10$
$x^2 + y^2 = 10$
$x^2 + y^2 = 100$
$x^2 - y^2 = 100$

The area (A) of green grass, in terms of x, is given by.

$2\text{x}\sqrt{100-\text{x}^2}$
$4\text{x}\sqrt{100-\text{x}^2}$
$2\text{x}\sqrt{100+\text{x}^2}$
$4\text{x}\sqrt{100+\text{x}^2}$

The maximum value of A is.

$100\text{m}^2$
$200\text{m}^2$
$400\text{m}^2$
$1600\text{m}^2$

The value of length of rectangle, when A is maximum, is.

$10\sqrt{2}\text{m}$
$20\sqrt{2}\text{m}$
$20\text{m}$
$5\sqrt{2}\text{m}$

The area of gravelling path is.

$100(\pi+2)\text{m}^2$
$100(\pi-2)\text{m}^2$
$200(\pi+2)\text{m}^2$
$200(\pi-2)\text{m}^2$

Answer

Solution:
Length, AB = 2x Breadth, BC = 2y
Also, radius, OA = 10
$\therefore\text{AC}=20$
$\text{in }\triangle\text{ABC},\text{AB}+\text{BC}^2=\text{AC}^2$
$\Rightarrow(2\text{x})^2+(2\text{y}^2)=(20)^2$
$\Rightarrow\text{x}^2+\text{y}^2=100$

(b) $4\text{x}\sqrt{100-\text{x}^2}$

Solution:
Area of green grass = Area of rectangular part
$\therefore\text{A}=2\text{x}.2\text{y}$ $[\therefore\text{Area of rectangle} = \text{length }\times \text{breadth}]$
$=4\text{xy}=4\text{x}\sqrt{100-\text{x}^2}$
$[\therefore\text{x}^2+\text{y}^2=100]$

(b) $200\text{m}^2$

Solution:
We have, $\text{A}=4\text{x}\sqrt{100-\text{x}^2}$
$\frac{\text{dA}}{\text{dx}}=\frac{4\text{x}(-2\text{x})}{2\sqrt{100-\text{x}^2}}+\sqrt{100-\text{x}^2.4}$
$\frac{-4\text{x}^2+4(100-\text{x}^2)}{\sqrt{100-\text{x}^2}}$
For maximum value, $\frac{\text{dA}}{\text{dx}}=0$
$\Rightarrow-4\text{x}^2+4000-4\text{x}^2=0$
$\Rightarrow-8\text{x}+400=0$
$\Rightarrow\text{x}^2=50$
$\Rightarrow\text{x}=5\sqrt{2}$
$\text{At}\ \text{x}=5\sqrt{2}$
$\text{A}=4\text{x}\sqrt{100-\text{x}^2}$
$=4\times5\sqrt{2}.\sqrt{100-50}=4\times5\sqrt{2}\times5\sqrt{2}=200\text{m}^2$

$10\sqrt{2}\text{m}$

Solution:
Length of rectangle for which A is maximum
$=2\times5\sqrt{2}=10\sqrt{2}$

(b) $100(\pi-2)\text{m}^2$

Solution:
Area of gravelling path $=\pi(10)^2-200$
$=100(\pi-2)\text{m}^2$

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Question 174 Marks

An owner of an electric bike rental company have determined that if they charge customers $₹\ x$ per day to rent a bike, where $50 < x < 200,$ then number of bikes (n), they rent per day can be shown by linear function $n(x) = 2000 - 10x.$ If they charge $₹\ 50$ per day or less, they will rent all their bikes. If they charge $₹\ 200$ or more per day, they will not rent any bike.

Based on the above information, answer the following questions.

Total revenue $R$ as a function of $x$ can be represented as.

$2000x - 10x^2$
$2000x + 10x^2$
$2000 - 10x$
$2000 - 5x^2$

If $R(x)$ denote the revenue, then maximum value of $R(x)$ occur when $x$ equals.

$10$
$100$
$1000$
$50$

At $x = 260$, the revenue collected by the company is.

$₹\ 10$
$₹\ 500$
$₹\ 0$
$₹\ 1000$

The number of bikes rented per day, if $x = 105$ is.

$850$
$900$
$950$
$1000$

Maximum revenue collected by company is.

$₹\ 40,000$
$₹\ 50,000$
$₹\ 75,000$
$₹\ 1,00,000$

Answer

(a) $2000x - 10x^2$

Solution:
Let $x$ be the charges per bike per day and n be the number of bikes rented per day.
$R(x) = n \times x = (2000 - 10x) x = -10x^2 + 2000x$

(b) $100$

Solution:
We have, $R(x) = 2000x - 10x^2$
$\Rightarrow R'(x) = 2000 - 20x$
For $R(x)$ to be maximum or minimum, $R'(x) = 0$
$\Rightarrow 2000 - 20x = 0 \Rightarrow x = 100$
Also, $R"(x) = -20 < 0$
Thus, $R(x)$ is maximum at $x = 100$

Solution:
If company charge $₹\ 200$ or more, they will not rent any bike. Therefore, revenue collected by him will be zero.

Solution:
If $x = 105,$ number of bikes rented per day is given by
$n = 2000 - 10 \times 105 = 950$

(d) $₹\ 1,00,000$

Solution:
At $x = 100, R(x)$ is maximum.
$\therefore$ Maximum revenue $= R(100)$
$= -10(100)^2 + 2000(100) = ₹ 1,00,000$

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Question 184 Marks

Two multi-storey buildings $($represented by $AP$ and $BQ)$ are on opposite side of a $20m$ wide road at point $A$ and $B$ respectively. There is a point $R$ on road as shown in figure.

Based on the above information, answer the following questions.

Area of trapezium $ABQP$ is.

$380$ sq. m
$280$ sq. m
$320$ sq. m
$430$ sq. m

The length $PQ$ is.

$20.5m$
$19.80m$
$20.88m$
$21m$

Let there be a quantity $S$ such that $S = RP^2 + RQ^2,$ then $S$ is given by.

$2x^2 - 40x - 1140$
$2x^2 + 40x + 1140$
$2x^2 - 40x + 1140$
$2x^2 + 40x - 1140$

Find the value of $x$ for which value of $S$ is minimum.

$10$
$0$
$4$
$-10$

For minimum value of S, find the value of $PR$ and $RQ.$

$18.50\ m, 19.36\ m$
$18.86\ m, 24.17\ m$
$17.56\ m, 23.29\ m$
None of these

Answer

(a) $380\ sq. m$

Solution:
Area of trapezium $=\frac{1}{2}\times$ (sum of parallel 2 sides) × distance between parallel sides
$=\frac{1}{2}\times(16+22)\times20=380\text{m}^2$

Solution:
$\text{PQ}^2=6^2+(20)^2=36+400=436$
$\therefore\text{PQ}=\sqrt{436}=20.88\text{m}$

Solution:
$\text{S}=\text{RP}^2+\text{RQ}^2$
Since $\text{RP}^2=(16)^2+\text{x}^2=256+\text{x}^2$
and $\text{RQ}^2=(22)^2+(20-\text{x})^2=484+400+\text{x}^2-40\text{x}$
$\therefore\text{S}=2\text{x}-40\text{x}+1140$

(a) $10$

Solution:
We have $\text{S}(\text{x})=2\text{x}^2-40\text{x}+1140$
$\therefore\text{S}'(\text{x})=4\text{x}-40$
For minimum value of $x, S(x) = 0$
$\Rightarrow4\text{x}-40=0$
$\Rightarrow\text{x}=10$
Clearly, at $x = 10, S (x) = 4 > 0$

(b) $18.86m, 24.17\ m$

Solution:
At $x = 10, PR^2 = 16^2 + x^2$
$= (16)^2 + (10)^2 = 256 + 100 = 356$
$\therefore\text{PR}=\sqrt{356}=18.86\text{m}$
Also $\text{RQ}^2=(22)^2+(20-10)^2=484+100=584$
$\therefore\text{RQ}=\sqrt{584}=24.17\text{m}$

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Question 194 Marks

An architecture design a auditorium for a school for its cultural activities. The floor of the auditorium is rectangular in shape and has a fixed perimeter P.

Based on the above information, answer the following questions.

If x and y represents the length and breadth of the rectangular region, then relation between the variable is.

$x + y = P$
$x^2 + y^2 = P^2$
$2(x + y) = P$
$x + 2y = P$

The area (A) of the rectangular region, as a function of x, can be expressed as.

$\text{A}=\text{px}+\frac{\text{x}}{2}$
$\text{A}=\frac{\text{px}+\text{x}^2}{2}$
$\text{A}=\frac{\text{px}-\text{2x}^2}{2}$
$\text{A}=\frac{\text{x}^2}{2}+\text{px}^2$

School's manager is interested in maximising the area of floor 'A' for this to be happen, the value of x should be.

$\text{P}$
$\frac{\text{P}}{2}$
$\frac{\text{P}}{3}$
$\frac{\text{P}}{4}$

The value of y, for which the area of floor is maximum, is.

$\frac{\text{P}}{2}$
$\frac{\text{P}}{3}$
$\frac{\text{P}}{4}$
$\frac{\text{P}}{16}$

Maximum area of floor is.

$\frac{\text{P}^2}{16}$
$\frac{\text{P}^2}{64}$
$\frac{\text{P}^2}{4}$
$\frac{\text{P}^2}{28}$

Answer

Solution:
Perimeter of floor = 2(Length + breadth)
⇒ P = 2(x + y)

Solution:
Area, A = length × breadth
⇒ A = xy
Since, P = 2(x + y)
$\Rightarrow\frac{\text{P}-2\text{x}}{2}=\text{y}$
$\therefore\text{A}=\text{x}\Big(\frac{\text{P}-2\text{x}}{2}\Big)$
$\Rightarrow\text{A}=\frac{\text{Px}-2\text{x}^2}{2}$

(d) $\frac{\text{P}}{4}$

Solution:
We have, $\text{A}=\frac{1}{2}(\text{Px}-2\text{x}^2)$
$\frac{\text{dA}}{\text{dx}}=\frac{1}{2}(\text{P}-4\text{x})=0$
$\Rightarrow\text{P}-4\text{x}=0\Rightarrow\text{x}=\frac{\text{P}}{4}$
Clearly, at $\text{x}=\frac{\text{P}}{4},\frac{\text{d}^2\text{A}}{\text{dx}^2}=-2<0$
$\therefore$ Area of maximum at $\text{x}=\frac{\text{P}}{4}$

Solution:
We have, $\text{y}=\frac{\text{P}-2\text{x}}{2}=\frac{\text{P}}{2}-\frac{\text{P}}{4}=\frac{\text{P}}{4}$

(a) $\frac{\text{P}^2}{16}$

Solution:
$\text{A}=\text{xy}=\frac{\text{P}}{4}\cdot\frac{\text{P}}{4}=\frac{\text{P}^2}{16}$

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Question 204 Marks

Rohan, a student of class XII, visited his uncle's flat with his father. He observe that the window of the house is in the form of a rectangle surmounted by a semicircular opening having perimeter 10m as shown in the figure.

Based on the above information, answer the following questions.

If x and y represents the length and breadth of the rectangular region, then relation between x and y can be represented as.

$\text{x}+\text{y}+\frac{\pi}{2}=10$
$\text{x}+\text{2y}+\frac{\pi\text{x}}{2}=10$
$\text{2x}+\text{2y}=10$
$\text{x}+\text{2y}+\frac{\pi}{2}=10$

The area (A) of the window can be given by.

$\text{A}=\text{x}-\frac{\text{x}^3}{8}-\frac{\text{x}^2}{2}$
$\text{A}=\text{5x}-\frac{\text{x}^2}{8}-\frac{\pi\text{x}^2}{8}$
$\text{A}=\text{x}+\frac{\pi\text{x}^3}{8}-\frac{\text{3x}^2}{8}$
$\text{A}=\text{5x}+\frac{\text{x}^3}{2}+\frac{\pi\text{x}^2}{8}$

Rohan is interested in maximizing the area of the whole window, for this to happen, the value of x should be.

$\frac{10}{2-\pi}$
$\frac{20}{4-\pi}$
$\frac{20}{4+\pi}$
$\frac{10}{2+\pi}$

Maximum area of the window is.

$\frac{30}{4+\pi}$
$\frac{30}{4-\pi}$
$\frac{50}{4-\pi}$
$\frac{50}{4+\pi}$

For maximum value of A, the breadth of rectangular part of the window is.

$\frac{10}{4+\pi}$
$\frac{10}{4-\pi}$
$\frac{20}{4+\pi}$
$\frac{20}{4-\pi}$

Answer

(b) $\text{x}+\text{2y}+\frac{\pi\text{x}}{2}=10$

Solution:

Given, perimeter of window = 10m

$\therefore$ x + y + y + perimeter of semicircle = 10

$\Rightarrow\text{x}+2\text{y}+\pi\frac{\text{2}}{2}=10$

(b) $\text{A}=\text{5x}-\frac{\text{x}^2}{8}-\frac{\pi\text{x}^2}{8}$

Solution:

$\text{A}=\text{x}\cdot\text{y}+\frac{1}{2}\pi\Big(\frac{\text{x}}{2}\Big)^2$

$=\text{x}\Big(5-\frac{\text{x}}{2}-\frac{\pi\text{x}}{4}\Big)+\frac{1}{2}\frac{\pi\text{x}^2}{4}$

$[\therefore\text{From }(\text{i)},\text{y}=5-\frac{\text{x}}{2}-\frac{\pi\text{x}}{4}]$

$=\text{5x}-\frac{\text{x}}{2}^2-\frac{\pi\text{x}^2}{4}+\frac{\pi\text{x}^2}{8}=5\text{x}-\frac{\text{x}}{2}^2-\frac{\pi\text{x}^2}{8}$

Solution:

We have, $\text{A}=5\text{x}-\frac{\text{x}^2}{2}-\frac{\pi\text{r}^2}{8}$

$\Rightarrow\frac{\text{dA}}{\text{dx}}=5-\text{x}-\frac{\pi\text{x}}{4}$

Now, $\Rightarrow\frac{\text{dA}}{\text{dx}}=0$

$\Rightarrow5=\text{x}+\frac{\pi\text{x}}{4}$

$\Rightarrow\text{x}(4+\pi)=20$

$\Rightarrow\text{x}=\frac{20}{4+\pi}$

$\Bigg[\text{Clearly},\frac{\text{d}^2\text{A}}{\text{dx}^2}<0\text{ at }\text{x}=\frac{20}{4+\pi}\Bigg]$

(d) $\frac{50}{4+\pi}$

Solution:

At $\text{x}=\frac{20}{\text{x}}=\frac{20}{4+\pi}$

$\text{A}=5\Big(\frac{20}{4+\pi}\Big)-\Big(\frac{20}{4+\pi}\Big)^2\frac{1}{2}-\frac{\pi}{8}\Big(\frac{20}{4+\pi}\Big)^2$

$=\frac{100}{4+\pi}-\frac{200}{(4+\pi)^2}-\frac{50\pi}{(4+\pi)^2}$

$\frac{(4+\pi)(100)-200-50\pi}{(4+\pi)^2}=\frac{400+100\pi-200-50\pi}{(4+\pi)^2}$

$\frac{200+50\pi}{(4+\pi)}=\frac{50(4+\pi)}{(4+\pi)}=\frac{50}{4+\pi}$

(a) $\frac{10}{4+\pi}$

Solution:

We have, $\text{y}=5-\frac{\text{x}}{2}-\frac{\pi\text{x}}{4}=5-\text{x}\Big(\frac{1}{2}+\frac{\pi}{4}\Big)$

$=5-\text{x}\Big(\frac{2+\pi}{4}\Big)=5-\Big(\frac{20}{4+\pi}\Big)\Big(\frac{2+\pi}{4}\Big)$

$=5-5\frac{(2+\pi)}{4+\pi}=\frac{20+5\pi-10-5\pi}{4+\pi}=\frac{10}{4+\pi}$

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Question 214 Marks

Show that $\text{f(x)}=\tan^{-1}(\sin\text{x}+\cos\text{x})$ is an increasing function in $\Big(0,\frac{\pi}{4}\Big).$

Answer

We have, $\text{f(x)}=\tan^{-1}(\sin\text{x}+\cos\text{x})$
$\therefore\ \text{f}'(\text{x})=\frac{1}{1+(\sin\text{x}+\cos\text{x})^2}\cdot(\cos\text{x}-\sin\text{x})$
$=\frac{1}{1+\sin^2\text{x}+\cos^2\text{x}+2\sin\text{x}.\cos\text{x}}(\cos\text{x}-\sin\text{x})$
$=\frac{1}{(2+\sin2\text{x})}(\cos\text{x}-\sin\text{x})$
$\big[\because\sin2\text{x}=2\sin\text{x}\cos\text{x and }\sin^2\text{x}+\cos^2\text{x}=1\big]$
For $\text{f}'(\text{x})\geq0.$
$\frac{1}{(2+\sin2\text{x})}\cdot(\cos\text{x}-\sin\text{x})\geq0$
$\Rightarrow\ \cos\text{x}-\sin\text{x}\geq0$ $\Big[\because(2+\sin2\text{x})\geq0\text{ in }\Big(0,\frac{\pi}{4}\Big)\Big]$
$\Rightarrow\ \cos\text{x}\geq\sin\text{x}$
Which is true, if $\text{x}\in\Big(0,\frac{\pi}{4}\Big)$
Hence, f(x) is an increasing function in $\Big(0,\frac{\pi}{4}\Big).$

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Question 224 Marks

Show that for $\text{a}\geq1,\text{ f(x)}=\sqrt{3}\sin\text{x}-\cos\text{x}-2\text{ax}+\text{b}$ is decreasing in R.

Answer

We have,$\text{ f(x)}=\sqrt{3}\sin\text{x}-\cos\text{x}-2\text{ax}+\text{b}$
$\Rightarrow\ \text{f(x)}=\sqrt{3}\cos\text{x}-(-\sin\text{x})-2\text{a}$
$=\sqrt{3}\cos\text{x}+\sin\text{x}-2\text{a}$
$=2\Big[\frac{\sqrt{3}}{2}\cdot\cos\text{x}+\frac{1}{2}\cdot\sin\text{x}\Big]-2\text{a}$
$=2\Big[\cos\frac{\pi}{6}\cdot\cos\text{x}+\sin\frac{\pi}{6}\cdot\sin\text{x}\Big]-2\text{a}$
$=2\cos\Big(\frac{\pi}{6}-\text{x}\Big)-2\text{a}$ $\big[\therefore\cos(\text{A}-\text{B})=\cos\text{A}\cdot\text{B}+\sin\text{A}\cdot\sin\text{B}\big]$
$=2\Big[\cos\Big(\frac{\pi}{6}-\text{x}\Big)-\text{a}\Big]$
Since, $\cos\text{x}\in[-1,1]\text{ and a}\geq1$
$\therefore\ 2\Big[\cos\Big(\frac{\pi}{6}-\text{a}\Big)-\text{a}\Big]\leq0$
We know tha if $\text{f(x)}\leq0,$ then f(x) is a decreasing function.
Thus, f(x) is a decreasing function in R.

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Question 234 Marks

$x$ and $y$ are the sides of two squares such that $y = x - x^2.$ Find the rate of change of the area of second square with respect to the area of first square.

Answer

Given $x$ and $y$ are the sides of two squares such that $y = x - x^2.$
$\therefore$ Area of the first square, $A_1 = x^2$
and area of the second square$, A_2 = y^2 = (x - x^2)^2$
$\therefore\ \frac{\text{dA}_1}{\text{dt}}=\frac{\text{d}}{\text{dt}}(\text{x}^2)=2\text{x}\cdot\frac{\text{dx}}{\text{dt}}$
and $\frac{\text{dA}_2}{\text{dt}}=\frac{\text{dt}}{\text{dx}}(\text{x}-\text{x}^2)^2$
$=2(\text{x}-\text{x}^2)(1-2\text{x})\frac{\text{dx}}{\text{dt}}$
$\therefore\ \frac{\text{dA}_2}{\text{dx}_1}=\frac{\frac{\text{dA}_2}{\text{dt}}}{\frac{\text{dA}_1}{\text{dt}}}=\frac{(2\text{x}-2\text{x}^2)(1-2\text{x})\frac{\text{dx}}{\text{dt}}}{2\text{x}\frac{\text{dx}}{\text{dt}}}$
$=\frac{(1-2\text{x})2\text{x}(1-\text{x})}{2\text{x}}=(1-2\text{x})(1-\text{x})=2\text{x}^2-3\text{x}+1$

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Question 244 Marks

A spherical ball of salt is dissolving in water in such a manner that the rate of decrease of the volume at any instant is propotional to the surface. Prove that the radius is decreasing at a constant rate.

Answer

We have, rate of decrease of the volume of spherical ball of salt at any instant is surface. Let the radius of the spherical ball of the salt be r.
$\therefore$ Volume of the ball $(\text{V})=\frac{4}{3}\pi\text{r}^3$
and surface area $(\text{S})=4\pi\text{r}^2$
$\because\ \frac{\text{dV}}{\text{dT}}\propto\text{S}$
$\Rightarrow\ \frac{\text{d}}{\text{dt}}\Big(\frac{4}{3}\pi\text{r}^3\Big)\propto4\pi\text{r}^2$
$\Rightarrow\ \frac{4}{3}\pi3\text{r}^2\frac{\text{dr}}{\text{dt}}\propto4\pi\text{r}^2$
$\Rightarrow\ \frac{\text{dr}}{\text{dt}}\propto\frac{4\pi\text{r}^2}{4\pi\text{r}^2}$
$\Rightarrow\ \frac{\text{dr}}{\text{dt}}=\text{k.1}$ [where, k is the proportionality constant]
$\Rightarrow\ \frac{\text{dr}}{\text{dt}}=\text{k}$
Hence, the radius of ball is decreasing at a constant rate.

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Question 254 Marks

A telephone company in a town has $500$ subscribers on its list and collects fixed charges of $Rs. 300/-$ per subscriber per year. The company proposes to increase the annual subscription and it is believed that for every increase of Re $1/-$ one subscriber will discontinue the service. Find what increase will bring maximum profit?

Answer

Consider that company increases the annual subscription by $Rs. x$
So$, x$ subscribes will discontinue the service.
$\therefore$ Total revenue of company after the increment is given by
$R(x) = (500 - x)(300 + x)$
$= -x^2 + 200x + 150000$
$\Rightarrow R's) = -2x + 200$
$R'(x) = 0$
$\Rightarrow 2x = 200$
$\Rightarrow x = 100$
Also $R''(x) = -2 < 0$
So$, R(x)$ is maximum when $x = 100.$
Hence$,$ the company should increase the subscription fee $Rs. 100,$ so that it has maximum profit.

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Question 264 Marks

If the area of a circle increases at a uniform rate, then prove that perimeter varies inversely as the radius.

Answer

Let the radius of circle at any time t is r.
Then area of the circle at any time t is $\text{A}=\pi\text{r}^2$
$\therefore\ \frac{\text{d}}{\text{dt}}\text{A}=\frac{\text{d}}{\text{dt}}(\pi\text{r}^2)$
$\Rightarrow\ \frac{\text{dA}}{\text{dt}}=2\pi\text{r}\cdot\frac{\text{dr}}{\text{dt}}\ \ \dots(\text{i})$
Since, the area of a circle increases at a uniform rate, we have
$\frac{\text{dA}}{\text{dt}}=\text{k},$ Where, k is a constant ...(ii)
From (i) and (ii), we get
$2\pi\text{r}\cdot\frac{\text{dr}}{\text{dt}}=\text{k}$
$\Rightarrow\ \frac{\text{dr}}{\text{dt}}=\frac{\text{k}}{2\pi\text{r}}=\frac{\text{k}}{2\pi}\cdot\Big(\frac{1}{\text{r}}\Big)$
$\Rightarrow\ 2\pi\frac{\text{dr}}{\text{dt}}=\frac{\text{k}}{\text{r}}$
$\Rightarrow\ \frac{\text{d}(2\pi\text{r})}{\text{dt}}=\frac{\text{k}}{\text{r}}$
$\Rightarrow\ \frac{\text{dP}}{\text{dt}}=\frac{\text{k}}{\text{r}},$ where $\text{P}=2\pi\text{r}$
$\Rightarrow\ \frac{\text{dP}}{\text{dt}}\propto\frac{1}{\text{r}}$
Thus perimeter varies inversely as time radius.

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