(i) (b) : Since, side of square is of length 20 cm therefore \(\begin{equation} x \in(0,10) \end{equation}\) .
(ii) (a) : 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 x (20 - 2x) x (20 - 2x
(iii) (d) : We have, V = x(20 - 2X)²
\(\begin{equation} \therefore \frac{d V}{d x}=x \cdot 2(20-2 x)(-2)+(20-2 x)^{2} \end{equation}\) 
= (20 - 2x)( -4x + 20 - 2x) = (20 - 2x)(20 - 6x)
Now, \(\begin{equation} \frac{d V}{d x}=0 \Rightarrow 20-2 x=0 \text { or } 20-6 x=0 \end{equation}\) 
\(\begin{equation} \Rightarrow x=10 \text { or } \frac{10}{3} \end{equation}\) 
(iv) (c) : We have, V = x(20 - 2X)²
and \(\begin{equation} \frac{d V}{d x}=(20-2 x)(20-6 x) \end{equation}\) 
\(\begin{equation} \Rightarrow \frac{d^{2} V}{d x^{2}}=(20-2 x)(-6)+(20-6 x)(-2) \end{equation}\) 
= (-2)[60 - 6x + 20 - 6x] = (-2)[80 - 12x] = 24x - 160
For \(\begin{equation} x=\frac{10}{3}, \frac{d^{2} V}{d x^{2}}<0 \end{equation}\) 
and for \(\begin{equation} x=10, \frac{d^{2} V}{d x^{2}}>0 \end{equation}\) 
So, volume will be maximum when \(\begin{equation} x=\frac{10}{3} \end{equation}\) .
(v) (d) : We have, V = x(20 - 2x)²,which will be maximum when \(\begin{equation} x=\frac{10}{3} \end{equation}\) .
\(\begin{equation} \therefore \ \text { Maximum volume }=\frac{10}{3}\left(20-2 \times \frac{10}{3}\right)^{2} \end{equation}\) 
\(\begin{equation} =\frac{10}{3} \times \frac{40}{3} \times \frac{40}{3}=\frac{16000}{27} \mathrm{~cm}^{3} \end{equation}\)

(i) (b) :To create a garden using 200 ft fencing, we
need to maximise its area.
(ii) (c) : Required relation is given by 2x + y = 200.
(iii) (c) : Area of garden as a function of x can be represented as
\(A(x)=x \cdot y=x(200-2 x)=200 x-2 x^{2}\)
(iv) (a) : \(\begin{equation} A(x)=200 x-2 x^{2} \Rightarrow A^{\prime}(x)=200-4 x \end{equation}\)
For the area to be maximum A'(x) = 0
\(\begin{equation} \Rightarrow 200-4 x=0 \Rightarrow x=50 \mathrm{ft} \end{equation}\)
(v) (c) : Maximum-area of the garden
= 200(50) - 2(50)2 = 10000 - 5000 = 5000 sq. ft

(i) (a) : Let x be the number of extra days after 1^st July.
\(\therefore\) Price = Rs.(300 - 3Xx) = Rs.(300 - 3x)
Quantity = 80 quintals + x(1 quintal per day)
= (80 + x) quintals
(ii) (b) : R(x) = Quantity x Price
= (80 + x) (300 - 3x) = 24000 - 240x + 300x -3x²
= 24000 + 60x - 3x²
(iii) (a) : We have, R(x) = 24000 + 60x - 3x²
\(\begin{equation} \Rightarrow R^{\prime}(x)=60-6 x \Rightarrow R^{\prime \prime}(x)=-6 \end{equation}\)
For R(x) to be maximum, R'(x) = 0 and R"(x) < 0
\(\begin{equation} \Rightarrow 60-6 x=0 \Rightarrow x=10 \end{equation}\)
(iv) (a) : Shyams 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.
(v) (c) : Maximum revenue is collected by Shyams father when x = 10
\(\therefore\) Maximum revenue = R(10)
= 24000 + 60(10) - 3(10)2 = 24000 + 600 - 300 = 24300

(i) (a) : Let x be the charges per bike per day and n be the number of bikes rented per day.
R(x) = n x x = (2000 - lOx) x = -10x² + 2000x
(ii) (b) : We have, R(x) = 2000x - 10x²
\(\Rightarrow R^{\prime}(x)=2000-20 x\) 
For R(x) to be maximum or minimum, R'(x) = 0
\(\Rightarrow 2000-20 x=0 \Rightarrow x=100\) 
Also,\(R^{\prime \prime}(x)=-20<0\) 
Thus, R(x) is maximum at x = 100
(iii) (c) : If company charge ~ 200 or more, they will not rent any bike. Therefore, revenue collected by him will be zero.
(iv) (c) : If x = 105, number of bikes rented per day is given by
n = 2000 - 10 x 105 = 950
(v) (d) : At x = 100, R(x) is maximum
\(\therefore\) Maximum revenue = R(100)
= -10(100)² + 2000(100) = Rs. 1,00,000

(i) (c) : 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)
[\(\therefore\)Rs. 500/month is the maintenance charges for each occupied unit]
(ii) (c) : 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 + 250x - 500 
= (50 - x) (9500 + 250 x) = 250(50 - x) (38 + x)
(iii) (b) : Clearly, if P = 10500, then 
\(10500=10000+250 x \Rightarrow x=2 \Rightarrow N=48\) 
(iv) (a) : Also, if P = 11000, then
\(11000=10000+250 x \Rightarrow x=4\) and so profit
(v) (b) : 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
\(\Rightarrow 12-2 x=0 \Rightarrow x=6\) 
Thus, price per apartment is, P = 10000 + 1500 = 11500
Hence, the rent that maximizes the profit is Rs. 11500.

(i) (a) : Let p be the price per ticket and x be the number of tickets sold.
Then, revenue function \(R(x)=p \times x=\left(15-\frac{x}{3000}\right) x\) 
\(=15 x-\frac{x^{2}}{3000}\) 
(ii) (c) : Since, more than 36000 tickets cannot be sold.
So, range of x is [0, 36000].
(iii) (c) : We have, \(R(x)=15 x-\frac{x^{2}}{3000}\) 
\(\Rightarrow R^{\prime}(x)=15-\frac{x}{1500}\) 
For maxima/minima, put R'(x) = 0
\(\Rightarrow \quad 15-\frac{x}{1500}=0 \Rightarrow x=22500\) 
Also,\(R^{\prime \prime}(x)=-\frac{1}{1500}<0\) 
(iv) (d) : Maximum revenue will be at x = 22500
\(\therefore \text { Price of a ticket }=15-\frac{22500}{3000}=15-7.5=Rs.7.5\) 
(v) (d) : Number of spectators will be equal to number of tickets sold.
\(\therefore\) Required number of spectators = 22500

(i) (d) : Given, r cm is the radius and h cm is the height of required cylindrical can
Given that, volume = 3 l= 3000 cm³ \(\left(\because 1 l=1000 \mathrm{~cm}^{3}\right)\) 
\(\Rightarrow \pi r^{2} h=3000 \Rightarrow h=\frac{3000}{\pi r^{2}}\) 
Now, the surface area, as a function of r is given by
\(S(r)=2 \pi r^{2}+2 \pi r h=2 \pi r^{2}+2 \pi r\left(\frac{3000}{\pi r^{2}}\right)\) 
\(=2 \pi r^{2}+\frac{6000}{r}\) 
(ii) (c) : Now, \(S(r)=2 \pi r^{2}+\frac{6000}{r}\) 
\(\Rightarrow S^{\prime}(r)=4 \pi r-\frac{6000}{r^{2}}\) 
To find criti£al points, put S'(r) = 0
\(\Rightarrow \frac{4 \pi r^{3}-6000}{r^{2}}=0\) 
\(\Rightarrow r^{3}=\frac{6000}{4 \pi} \Rightarrow r=\left(\frac{1500}{\pi}\right)^{1 / 3}\) 
Also, \(\left.S^{\prime \prime}(r)\right|_{r=} \sqrt[3]{\frac{1500}{\pi}}=4 \pi+\frac{12000 \times \pi}{1500}\) 
\(=4 \pi+8 \pi=12 \pi>0\) 
Thus, the critical point is the point of minima.
(iii) (b) : The cost of material for the tin can is minimized when \(r=\sqrt[3]{\frac{1500}{\pi}} \mathrm{cm}\) and the height is  \(\frac{3000}{\pi\left(\sqrt[3]{\frac{1500}{\pi}}\right)^{2}}=2 \sqrt[3]{\frac{1500}{\pi}} \mathrm{cm} .\) .
(iv) (a) : We have,minimum surface area = \(\frac{2 \pi r^{3}+6000}{r}\) .
\(=\frac{2 \pi \cdot \frac{1500}{\pi}+6000}{\sqrt[3]{\frac{1500}{\pi}}}=\frac{9000}{7.8}=1153.84 \mathrm{~cm}^{2}\) 
Cost of 1 m² material = Rs.100
\(\therefore \ \text { Cost of } 1 \mathrm{~cm}^{2} \text { material }=Rs. \frac{1}{100}\) 
\(\therefore \ \text { Minimum cost }=Rs. \frac{1153.84}{100}=Rs. 11.538\) 
(v) (c) : To minimize the cost we need to minimize the total surface area.

(i) (a) : 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
 
(ii) (a) : Clearly, A = a-b
\(\therefore \quad 648+18 a+12 b=a b\) 
\(\Rightarrow a b-18 a=648+12 b \Rightarrow a(b-18)=648+12 b\) 
\(\Rightarrow \quad a=\frac{648+12 b}{b-18}\) 
(iii) (b) : Since, A = a-b, therefore
\(A=\left(\frac{648+12 b}{b-18}\right) \cdot b=\frac{648 b+12 b^{2}}{b-18} \quad\left[\because a=\frac{648+12 b}{b-18}\right]\) 
(iv) (a) : Clearly,
\(A^{\prime}(b)=\frac{(b-18)(648+24 b)-\left(648 b+12 b^{2}\right)}{(b-18)^{2}}\) 
\(=\frac{12\left[b^{2}-36 b-972\right]}{(b-18)^{2}}\) 
For minimum, consider A'(b) = 0 
\(\Rightarrow b^{2}-36 b-972=0\) 
\(\Rightarrow b^{2}-54 b+18 b-972=0\) 
\(\Rightarrow b(b-54)+18(b-54)=0\) 
\(\Rightarrow b=-18 \text { or } b=54\) 
\(\therefore\) b is height, therefore can't be negative.
So, b = 54.
(v) (b) : Since, \(a=\frac{648+12 b}{b-18}\) 
\(\therefore \ a=\frac{648+12 \times 54}{54-18}=\frac{648+648}{36}=36\) 

(i) (d) : In order to make least expensive water tank, Nitin need to minimize its cost.
(ii) (d) : Let 1ft 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
1h sq. feet. So, the total area of the sides is (10 h + 2 1h)ft²
Cost of the sides is Rs.10 per sq. foot. So, the cost to build the sides is (10h + 21h) x 10 = Rs.(100h + 20lh)
Also, cost of base = (5l) x 20 = Rs. 100 l
\(\therefore\) Total cost of the tank in Rs. is given by
c = 100 h + 20 I h + 100 l
Since, volume of tank = 80ft³
\(\therefore \quad 5 l h=80 \mathrm{ft}^{3} \quad \therefore l=\frac{80}{5 h}=\frac{16}{h}\) 
\(\therefore \quad c(h)=100 h+20\left(\frac{16}{h}\right) h+100\left(\frac{16}{h}\right)\) 
\(=100 h+320+\frac{1600}{h}\) 
(iii) (b) : Since, all side lengths must be positive
\(\therefore \quad h>0\) and \(\frac{16}{h}>0\) 
Since, \(\frac{16}{h}>0, \text { whenever } h>0\) 
\(\therefore \text { Range of } h \text { is }(0, \infty)\) 
(iv) (a) : To minimize cost, \(\frac{d c}{d h}=0\) 
\(\Rightarrow \quad 100-\frac{1600}{h^{2}}=0\) 
\(\Rightarrow 100 h^{2}=1600 \Rightarrow h^{2}=16 \Rightarrow h=\pm 4\) 
\(\Rightarrow h=4\)  [\(\therefore\) height can not be negative]
(v) (c) : Cost of least expensive tank is given by
\(c(4)=400+320+\frac{1600}{4}\) 
= 720 + 400 = Rs. 1120

(i) (c) :Let Sbe the sum of volume of parallelopiped and sphere, then
\(S=x(2 x)\left(\frac{x}{3}\right)+\frac{4}{3} \pi r^{3}=\frac{2 x^{3}}{3}+\frac{4}{3} \pi r^{3}\)   ...(i)
(ii) (a) : Since, sum of surface area of box and sphere is given to be constant k².
\(\therefore \quad 2\left(x \times 2 x+2 x \times \frac{x}{3}+\frac{x}{3} \times x\right)+4 \pi r^{2}=k^{2}\) 
\(\Rightarrow 6 x^{2}+4 \pi r^{2}=k^{2}\)
\(\Rightarrow x^{2}=\frac{k^{2}-4 \pi r^{2}}{6} \Rightarrow x=\sqrt{\frac{k^{2}-4 \pi r^{2}}{6}}\)  ...(2)
(iii) (b) : From (1) and (2), we get
\(S=\frac{2}{3}\left(\frac{k^{2}-4 \pi r^{2}}{6}\right)^{3 / 2}+\frac{4}{3} \pi r^{3}\) 
\(=\frac{2}{3 \times 6 \sqrt{6}}\left(k^{2}-4 \pi r^{2}\right)^{3 / 2}+\frac{4}{3} \pi r^{3}\) 
\(\Rightarrow \quad \frac{d S}{d r}=\frac{1}{9 \sqrt{6}} \frac{3}{2}\left(k^{2}-4 \pi r^{2}\right)^{1 / 2}(-8 \pi r)+4 \pi r^{2}\) 
\(=4 \pi r\left[r-\frac{1}{3 \sqrt{6}} \sqrt{k^{2}-4 \pi r^{2}}\right]\) 
For maximum /minimum \(\frac{d S}{d r}=0\) 
\(\Rightarrow \frac{-4 \pi r}{3 \sqrt{6}} \sqrt{k^{2}-4 \pi r^{2}}=-4 \pi r^{2}\) 
\(\Rightarrow k^{2}-4 \pi r^{2}=54 r^{2}\) 
\(\Rightarrow r^{2}=\frac{k^{2}}{54+4 \pi} \Rightarrow r=\sqrt{\frac{k^{2}}{54+4 \pi}}\)   ...(3)
(iv) (d) : Since, \(x^{2}=\frac{k^{2}-4 \pi r^{2}}{6}=\frac{1}{6}\left[k^{2}-4 \pi\left(\frac{k^{2}}{54+4 \pi}\right)\right]\) [From (2) and (3)] 
\(=\frac{9 k^{2}}{54+4 \pi}=9\left(\frac{k^{2}}{54+4 \pi}\right)=9 r^{2}=(3 r)^{2}\) 
\(\Rightarrow x=3 r\) 
(v) (c) : Minimum value of S is given by 
\(\frac{2}{3}(3 r)^{3}+\frac{4}{3} \pi r^{3}\) 
\(=18 r^{3}+\frac{4}{3} \pi r^{3}=\left(18+\frac{4}{3} \pi\right) r^{3}\) 
\(=\left(18+\frac{4}{3} \pi\right)\left(\frac{k^{2}}{54+4 \pi}\right)^{3 / 2}\) [Using (3)]
\(=\frac{1}{3} \frac{k^{3}}{(54+4 \pi)^{1 / 2}}\)

(i) (c) : Let C(x) be the maintenance cost function, then C(x) = 5000000 + 160x - 0.04x²
(ii) (b) : We have, C(x) = 5000000 + 160x - 0.04x²
Now, C(x) = 160 - 0.08x
For maxima/minima, put C'(x) = 0
\(\Rightarrow\) 160 = 0.08x
\(\Rightarrow\) x = 2000
(iii) (b) : 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(0) = 5000000
C(2000) = 5160000
and C(4500) = 4910000
\(\therefore\) Maximum value of C(x)would be Rs.5160000
(iv) (a) : The complex must have 4500 apartments to minimise the maintenance cost.
(v) (a) : The minimum maintenance cost for each apartment woud be Rs.1091.11

(i) (b) : In order to paint in the maximum area, Kyra needs to maximize the area of inner rectangle.
(ii) (c) : 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
\(\therefore A(x)=(x-1)(y-1.5)\) \([\because x y=24 \text { (given) }]\) 
\(=(x-1)\left(\frac{24}{x}-1.5\right)\) 
(iii) (b) : Dimensions of rectangle (outer/inner) should be positive.
\(\therefore \ x-1>0\) and \(\frac{24}{x}-1.5>0\) 
\(\Rightarrow x>1\) and \(x<16\) 
(iv) (c) : We have, \(A(x)=(x-1)\left(\frac{24}{x}-1.5\right)\) 
and \(A^{\prime \prime}(x)=\frac{-48}{x^{3}}\) 
For A(x) to be maximum or minimum, A'(x) = 0
\(\Rightarrow -1.5+\frac{24}{x^{2}}=0 \Rightarrow x^{2}=16 \Rightarrow x=\pm 4\) 
\(\therefore \ x=4\) [Since, length can't be negative]
Also,\(A^{\prime \prime}(4)=\frac{-48}{4^{3}}<0\) 
Thus, at x = 4, area is maximum.
(v) (a) : If area of inner rectangle is maximum, then Length of inner rectangle = x-1 = 4 - 1 = 3 ft
And breadth of inner rectangle = \(y-1.5=\frac{24}{x}-1.5\) 
\(=\frac{24}{4}-1.5=6-1.5=4.5 \mathrm{ft}\)

(i) (d) : 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+2000 x-x^{2}, 0 
(ii) (a) : Clearly, at x = 500
R(500) = 15000000 + 2000(500) - (500)²
= 15000000 + 1000000 - 250000 = Rs.15750000
(iii) (c) : Since, 15000000 + 2000x - x² = 15640000 (Given)
\(\Rightarrow x^{2}-2000 x+640000=0\) 
\(\Rightarrow \quad x^{2}-1600 x-400 x+640000=0\) 
\(\Rightarrow x(x-1600)-400(x-1600)=0 \Rightarrow x=400,1600\) 
(iv) (a) : \(\frac{d R}{d x}=2000-2 x\) and \(\frac{d^{2} R}{d x^{2}}=-2<0\) 
For maximum revenue, \(\frac{d R}{d x}=0 \Rightarrow x=1000\) 
\(\therefore\) Required amount = Rs. 1000
(v) (b) : Maximum revenue = R(1000)
= (3000 + 1000) (5000 - 1000)
= 4000 x 4000 = ~ 16000000

(i) (c) : 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 \(\frac{1000}{x^{2}}+\frac{125}{(600-x)^{2}}\) .
(ii) (c) : 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.
(iii) (a) : Since, \(0 ,therefore minimum value of x can't be 0.
(iv) (b) : We have, \(I(x)=\frac{1000}{x^{2}}+\frac{125}{(600-x)^{2}}\) 
\(\Rightarrow I^{\prime}(x)=\frac{-2000}{x^{3}}+\frac{250}{(600-x)^{3}}\) and 
\(\Rightarrow I^{\prime \prime}(x)=\frac{6000}{x^{4}}+\frac{750}{(600-x)^{4}}\) 
For maxima/minima, I'(x) = 0
\(\Rightarrow \frac{2000}{x^{3}}=\frac{250}{(600-x)^{3}} \Rightarrow 8(600-x)^{3}=x^{3}\) 
Taking cube root on both sides, we get
\(2(600-x)=x \Rightarrow 1200=3 x \Rightarrow x=400\) 
Thus, I(x) is minimum when you are at 400 feet from the strong intensity lamp post.
(v) (a) : Since, I(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. 

(i) (d) : Since the tank is open from the top therefore the total surface area is 
= (Outer + Inner) surface area
= 2(x X x + 2 (xy +yx) = 2(x^? + 2(2xy» = (2x² + 8xy) m²
 
(ii) (a) : Since, volume of tank should be 32000 I.
\(\therefore \ x^{2} y \mathrm{~m}^{3}=32000 l=32 \mathrm{~m}^{3}\) \(\left[\because 1 \text { litre }=0.001 \mathrm{~m}^{3}\right]\) 
So, x²y = 32
(iii) (a) : Let S be the outer surface area of tank.
Then, S = x²+ 4xy
\(\Rightarrow S(x)=x^{2}+4 x \cdot \frac{32}{x^{2}}=x^{2}+\frac{128}{x}\) \(\left[\because x^{2} y=32\right]\) 
\(\Rightarrow \quad \frac{d S}{d x}=2 x-\frac{128}{x^{2}} \text { and } \frac{d^{2} S}{d x^{2}}=2+\frac{256}{x^{3}}\) 
For maximum or minimum values of S, consider
\(\frac{d S}{d x}=0\) 
\(\Rightarrow \quad 2 x=\frac{128}{x^{2}} \Rightarrow x^{3}=64 \Rightarrow x=4, \mathrm{~m}\) 
At \(x=4, \frac{d^{2} S}{d x^{2}}=2+\frac{256}{4^{3}}=2+4=6>0\) 
\(\therefore\) S is minimum when x = 4
Now as x²y = 32, therefore y = 2
Thus, x = 2y
(iv) (b) : 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.
(v) (c) : Since, minimum surface area
\(=x^{2}+4 x y=4^{2}+4 \times 4 \times 2=48 \mathrm{~m}^{2}\) and
cost per m² = Rs. 360

(i) (c) : For all values of x, y = x² + 7
\(\therefore\) Arun's position at any point of x will be (x, x²+ 7)
(ii) (c) : Distance between
Arun and Manita, i.e., D
\(=\sqrt{(x-3)^{2}+\left(x^{2}+7-7\right)^{2}}\) 
\(=\sqrt{(x-3)^{2}+x^{4}}\) 
(iii) (a) : We have \(D=\sqrt{(x-3)^{2}+x^{4}}\) 
\(\therefore \quad D^{2}=(x-3)^{2}+x^{4}\) 
Now, \(\frac{d}{d x}\left(D^{2}\right)=2(x-3)+4 x^{3}=0\) 
\(\Rightarrow 4 x^{3}+2 x-6=0 \Rightarrow 2 x^{3}+x-3=0\) 
\(\Rightarrow \quad(x-1)\left(2 x^{2}+2 x+3\right)=0\) 
\(\therefore\) x = 1
(\(\therefore\) 2x² + 2x + 3 = 0 will give imaginary values)
(iv) (d) : We have,\(D=\sqrt{(x-3)^{2}+x^{4}}\) 
\(D^{\prime}(x)=\frac{2(x-3)+4 x^{3}}{2 \sqrt{(x-3)^{2}+x^{4}}}=0\) 
 
\(\Rightarrow 2 x^{3}+x-3=0\) 
\(\Rightarrow x=1\) 
Clearly, D"(x) at x = 1 is > 0
\(\therefore\) Value of x for which Dwill be minimum is 1
For x = 1, y = 8.
Thus, the required position is (1, 8).
(v) (d) : Minimum value of \(D=\sqrt{(1-3)^{2}+(1)^{4}}\) 
\(=\sqrt{4+1}=\sqrt{5}\)

(i) (c) : Length, AB = 2x
Breadth, BC = 2y
Also, radius, OA = 10
\(\therefore\) AC = 20
In \(\Delta A B C, A B+B C^{2}=A C^{2}\) 
\(\Rightarrow (2 x)^{2}+(2 y)^{2}=(20)^{2}\) 
\(\Rightarrow x^{2}+y^{2}=100\) 
(ii) (b) : Area of green grass = Area of rectangular part
\(\therefore \ A=2 x \cdot 2 y\) [\(\therefore\) Area of rectangle = length x breadth]
\(=4 x y=4 x \sqrt{100-x^{2}}\) \(\left[\because x^{2}+y^{2}=100\right]\) 
(iii) (b) : We have, \(A=4 x \sqrt{100-x^{2}}\) 
\(\frac{d A}{d x}=\frac{4 x(-2 x)}{2 \sqrt{100-x^{2}}}+\sqrt{100-x^{2}} \cdot 4\) 
\(=\frac{-4 x^{2}+4\left(100-x^{2}\right)}{\sqrt{100-x^{2}}}\) 
For maximum value, \(\frac{d A}{d x}=0\) 
\(\Rightarrow-4 x^{2}+400-4 x^{2}=0\) 
\(\Rightarrow-8 x^{2}+400=0\) 
\(\Rightarrow x^{2}=50 \Rightarrow x=5 \sqrt{2}\) 
At  \(x=5 \sqrt{2}\) 
\(A=4 x \sqrt{100-x^{2}}\) 
\(=4 \times 5 \sqrt{2} \cdot \sqrt{100-50}=4 \times 5 \sqrt{2} \times 5 \sqrt{2}=200 \mathrm{~m}^{2}\) 
(iv) (a) : Length of rectangle for which A is maximum
\(=2 \times 5 \sqrt{2}=10 \sqrt{2}\) 
(v) (b) : Area of gravelling path = \(\pi(10)^{2}-200\) 
\(=100(\pi-2) \mathrm{m}^{2}\)

(i) (a) : Area of trapezium = \(\frac{1}{2}\) x (sum of parallel sides) x distance between parallel sides
\(=\frac{1}{2} \times(16+k 2) \times 20=380 \mathrm{~m}^{2}\) 
(ii) (c) : PQ² = 6² + (20)² = 36 + 400 = 436
\(\therefore \ P Q=\sqrt{436}=20.88 \mathrm{~m}\) 
(iii) (c) : We have, S = RP²+ RQ² ...(i)
Since, RP² = (16)² + x² = 256 + x² and RQ² = (22)² + (20 - x)² = 484 + 400 + x² - 40x
\(\therefore\) S = 2x²- 40x + 1140

(iv) (a) : We have, S(x) = 2x²- 40x + 1140
\(\therefore\) S'(x) = 4x - 40
For minimum value of x, S''(x) = 0
\(\Rightarrow\) 4x - 40 = 0 \(\Rightarrow\) x = 10
Clearly, at x = 10, S(x) = 4 > 0
(v) (b) : At x = 10, PR \(\Rightarrow\) = 16² + x²
= (16)² + (10)² = 256 + 100 = 356
\(\therefore \ P R=\sqrt{356}=18.86 \mathrm{~m}\)
Also, RQ² = (22)² + (20 - 10)² = 484 + 100 = 584
\(\therefore \ R Q=\sqrt{584}=24.17 \mathrm{~m}\)

(i) (b) : Given, perimeter of window = 10 m 
\(\therefore\) x + y + y + perimeter of semicircle = 10
\(\Rightarrow x+2 y+\pi \frac{x}{2}=10\) 
(ii) (b) : \(A=x \cdot y+\frac{1}{2} \pi\left(\frac{x}{2}\right)^{2}\) 
\(=x\left(5-\frac{x}{2}-\frac{\pi x}{4}\right)+\frac{1}{2} \frac{\pi x^{2}}{4}\left[\because \text { From }(\mathrm{i}), y=5-\frac{x}{2}-\frac{\pi x}{4}\right]\) 
\(=5 x-\frac{x^{2}}{2}-\frac{\pi x^{2}}{4}+\frac{\pi x^{2}}{8}=5 x-\frac{x^{2}}{2}-\frac{\pi x^{2}}{8}\) 
(iii) (c) : We have, \(A=5 x-\frac{x^{2}}{2}-\frac{\pi x^{2}}{8}\) 
\(\Rightarrow \quad \frac{d A}{d x}=5-x-\frac{\pi x}{4}\) 
Now, \(\frac{d A}{d x}=0 \Rightarrow 5=x+\frac{\pi x}{4}\) 
\(\Rightarrow x(4+\pi)=20 \Rightarrow x=\frac{20}{4+\pi}\) 
\(\left[\text { Clearly, } \frac{d^{2} A}{d x^{2}}<0 \text { at } x=\frac{20}{4+\pi}\right]\) 
(iv) (d) : At \(x=\frac{20}{4+\pi}\) 
 \(A=5\left(\frac{20}{4+\pi}\right)-\left(\frac{20}{4+\pi}\right)^{2} \frac{1}{2}-\frac{\pi}{8}\left(\frac{20}{4+\pi}\right)^{2}\) 
\(=\frac{100}{4+\pi}-\frac{200}{(4+\pi)^{2}}-\frac{50 \pi}{(4+\pi)^{2}}\) 
\(=\frac{(4+\pi)(100)-200-50 \pi}{(4+\pi)^{2}}=\frac{400+100 \pi-200-50 \pi}{(4+\pi)^{2}}\) 
\(=\frac{200+50 \pi}{(4+\pi)^{2}}=\frac{50(4+\pi)}{(4+\pi)^{2}}=\frac{50}{4+\pi}\) 
(v) (a) : We have, \(y=5-\frac{x}{2}-\frac{\pi x}{4}=5-x\left(\frac{1}{2}+\frac{\pi}{4}\right)\) 
\(=5-x\left(\frac{2+\pi}{4}\right)=5-\left(\frac{20}{4+\pi}\right)\left(\frac{2+\pi}{4}\right)\) 
\(=5-5 \frac{(2+\pi)}{4+\pi}=\frac{20+5 \pi-10-5 \pi}{4+\pi}=\frac{10}{4+\pi}\)

(i) (c) : Perimeter of floor = 2(length + breadth)
\(\Rightarrow P=2(x+y)\) 
(ii) (c) : Area, A = length x breadth
\(\Rightarrow A=x y\) 
Since, P = 2(x + y)
\(\Rightarrow \frac{P-2 x}{2}=y\) 
\(\therefore \quad A=x\left(\frac{P-2 x}{2}\right) \Rightarrow A=\frac{P x-2 x^{2}}{2}\) 
(iii) (d) : We have, \(A=\frac{1}{2}\left(P x-2 x^{2}\right)\) 
\(\frac{d A}{d x}=\frac{1}{2}(P-4 x)=0\) 
\(\Rightarrow P-4 x=0 \Rightarrow x=\frac{P}{4}\) 
Clearly, at \(x=\frac{P}{4}, \frac{d^{2} A}{d x^{2}}=-2<0\) 
\(\therefore\) Area is maximum at \(x=\frac{P}{4}\) 
(iv) (c) : We have ,\(y=\frac{P-2 x}{2}=\frac{P}{2}-\frac{P}{4}=\frac{P}{4}\) 
(v) (a) : We have, \(A=x y=\frac{P}{4} \cdot \frac{P}{4}=\frac{P^{2}}{16}\)