(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) (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) (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 = \(y-1.5=\frac{24}{x}-1.5\) 
\(=\frac{24}{4}-1.5=6-1.5=4.5 \mathrm{ft}\)