(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.