Let AB be the lamp post and a man CD be at a distance x
from the lamp post and let CE = y be his shadow

Given that 
\( \frac{d x}{d t}=6 \mathrm{~km} / \mathrm{h}, A B=6 \mathrm{~m}=\frac{6}{1000} \mathrm{~km} \text { and }\)
\( C D=2 \mathrm{~m}=\frac{2}{1000} \mathrm{~km}[\because 1 \mathrm{~km}=1000 \mathrm{~m}] \) 
Here,\(\Delta B A E \sim \Delta D C E\) 
\( \Rightarrow \frac{6}{\frac{1000}{2}}{\frac{1000}{y}}=\frac{x+y}{y} \Rightarrow 3=\frac{x+y}{y}\)
\( \Rightarrow 3 y =x+y \Rightarrow 3 y-y=x \Rightarrow 2 y=x \) 
On differentiating both sides w.r.t. t, we get
\(2 \frac{d y}{d t}=\frac{d x}{d t} \)
\( \Rightarrow 2 \frac{d y}{d t}=6 \quad\left[\because \frac{d x}{d t}=6\right] \) 
\(\Rightarrow \frac{d y}{d t}=\frac{6}{2}=3 \mathrm{~km} / \mathrm{h}\) 

Let 'r' and 'h' be the radius and height respectively of the cylinder.
\(\therefore \) S, the surface area = \(\pi r^{ 2 }+2\pi rh\)
\(\Rightarrow \)  \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)....(1)
And V,the volume = \(\pi r^{ 2 }h\)
i.e. \(V=\pi r^{ 2 }\left( \frac { S-\pi r^{ 2 } }{ 2\pi r } \right) \)  [Using (1)]
\(\Rightarrow \) \(V=\frac { r }{ 2 } (S-\pi r^{ 2 })\)
\(\Rightarrow \) \(V=\frac { r }{ 2 } (Sr-\pi r^{ 2 })\)
\(\therefore \)  \(\frac { dV }{ dr } =\frac { 1 }{ 2 } (S-3\pi r^{ 2 })\)...(2)
and   \(\frac { d^{ 2 }V }{ dr } =-\frac { 3\pi }{ 2 } (2r)=-3\pi r\)   (3)
For greatest volume \(\frac { dV }{ dr } =0\Rightarrow -\frac { 1 }{ 2 } (S-3\pi r^{ 2 })=0\)
\(\Rightarrow \) \(S=3\pi r^{ 2 }\Rightarrow r=\sqrt { \frac { S }{ 3\pi } } \)
Putting in (3),
\(\frac { d^{ 2 }V }{ dr^{ 2 } } =-3\pi \sqrt { \frac { S }{ 3\pi } } =-\sqrt { 3\pi S } \) which is -ve.
Thus V is the greatest when \(r=\sqrt { \frac { s }{ 3\pi } } \)
And from (1),
\(h=\frac { S-\pi \left( \frac { S }{ 3\pi } \right) }{ 2\pi \sqrt { \frac { S }{ 3\pi } } } =\frac { \frac { 2S }{ 3 } }{ \frac { 2 }{ \sqrt { 3 } } \sqrt { \pi S } } =\sqrt { \frac { s }{ 3\pi } } \)
Hence, height = Radius of the base.                   

Let 'x' and 'y' be the side of the rectangle.
By the equation, perimeter = p(say)  [Given]
\(\Rightarrow 2x+2y=p\Rightarrow y=\frac { 1 }{ 2 } (p-2x)\)..(1)
Now area, A = xy = x .\(\frac { 1 }{ 2 } (p-2x)\)   [Using (1)]
\(\frac { 1 }{ 2 } (p-2x^{ 2 })\)
For A to be largest, \(\frac { dA }{ dx } =0\) and \(\frac { d^{ 2 }A }{ dx } \) is -ve
Here  \(\frac { dA }{ dx } \quad =\frac { 1 }{ 2 } (p-4x)\)
Now  \(\frac { dA }{ dx } =0\) gives \(\frac { 1 }{ 2 } (p-4x)=0\)
\(\Rightarrow p-4x=0\Rightarrow p=4x\)
And  \(\frac { d^{ 2 }A }{ dx^{ 2 } } \) is -ve
Thus A is the largest when p = 4x.
From(1), \(y=\frac { 1 }{ 2 } (4x-2x)\Rightarrow y=\frac { 1 }{ 2 } (2x)\Rightarrow y=x\)
Hence area is largest when x=y i.e when rectangle becomes a square.

Let 'x' and 'y' be two positive numbers.
By the question  x + y = 24....(1)
Let  \(s=x^{ 2 }+y^{ 2 }\)
\(\Rightarrow \)\(s=x^{ 2 }+(24-x)^{ 2 }\) [Using (1)]
\(=x^{ 2 }+576+x^{ 2 }-48x\)
\(=2x^{ 2 }-48x+576\)
\(\therefore \) \(\frac { dS }{ dx } =4x-48\)
For S to be maximum, \(\frac { dS }{ dx } =0,\)which gives:
4x - 48 = 0\(\Rightarrow 4x=48\Rightarrow x=12\)
Now  \(\frac { d^{ 2 }S }{ dx^{ 2 } } =4,\) which is +ve for x = 12
Hence S is minimum when the two positive numbers are 12 and (54-12) i.e. 12 and 12.  

The given curve is : x = 1 \(cos\theta ,y-=\theta -sin\theta \)
\(\therefore \frac { dx }{ d\theta } =sin\theta ,\frac { dy }{ d\theta } =1-cos\theta \)
\(\therefore \frac { dy }{ dx } =\frac { dy/d\theta }{ dx/d\theta } =\frac { 1-cos\theta }{ sin\theta } =\frac { 2sin^{ 2 }\frac { \theta }{ 2 } }{ 2sin\frac { \theta }{ 2 } cos\frac { \theta }{ 2 } } \)
\(=\frac { sin\frac { \theta }{ 2 } }{ cos\frac { \theta }{ 2 } } =tan\frac { \theta }{ 2 } \)
At \(\theta \) \(=\frac { \pi }{ 4 } \)    \(x=1-cos\frac { \pi }{ 4 } =1-\frac { 1 }{ \sqrt { 2 } } \)
\(y=\frac { \pi }{ 4 } -sin\frac { \pi }{ 4 } =\frac { 1 }{ \sqrt { 2 } } \)
and  \(\frac { dy }{ dx } =tan\frac { \pi }{ 8 } \)

We have:\(f(x)=4x^{ 3 }-18x^{ 2 }+27x-7\)
\(\therefore f'(x)=12x^{ 2 }-36x+27\)
 \( =3(4x^{ 2 }-12x+9)=3(2x-3)^{ 2 }\)
Now \(f'(x)=0\Rightarrow 3(2x-3)^{ 2 }=0(2x-3)^{ 2 }=0\)
\(\Rightarrow x=\frac { 3 }{ 2 } \)  (Critical Point)
Since \(f'(x)>0\) for all \(x<\frac { 3 }{ 2 } \)Hence 'f' has neither maximum nor minimum. and for all \(x<\frac { 3 }{ 2 } \)
\(\therefore x=\frac { 3 }{ 2 } \) is a point of inflexion.

We have: \(f(x)=cos^{ 2 }x+sinx,x\in [0,\pi ]\)
\(\therefore f'(x)=2cosx(-sinx)+cosx\)
For extreme values \(f'(x)=0\)
\(\Rightarrow cosx(-2sinx+1)=0\)
\(\Rightarrow Cosx=0or-2sinx+1=0\)
\(\Rightarrow x=\frac { \pi }{ 2 } orsinx=\frac { 1 }{ 2 } \Rightarrow x=\frac { \pi }{ 6 } \)
\(f(0)=cos^{ 2 }\frac { \pi }{ 6 } +sin\frac { \pi }{ 6 } \)
\(f\left( \frac { \pi }{ 6 } \right) =cos^{ 2 }\frac { \pi }{ 6 } +sin\frac { \pi }{ 6 }\)
\(=\left( \frac { \sqrt { 3 } }{ 2 } \right) ^{ 2 }+\frac { 1 }{ 2 } \)
\(=\frac { 3 }{ 4 } +\frac { 1 }{ 2 } =\frac { 5 }{ 4 } \)
\(f\left( \frac { \pi }{ 2 } \right) =cos^{ 2 }\frac { \pi }{ 2 } +sin\frac { \pi }{ 2 } =0+1=1\)
\( f(\pi )=cos^{ 2 }\pi +sin\pi \)
\(=(-1)^{ 2 }+0=1\)
Hence absolute max and min values are \(\frac { 5 }{ 4 } \)and 1 respectively.

We have : \(f(x)\)  = tan x - 4x.
\(f'(x)=sec^{ 2 }x-4\)
When  \((\frac { -\pi }{ 3 } )\)
Thus for \( (\frac { -\pi }{ 3 } )\)  
Hence 'f' is strictly decreasing on \(\left( -\frac { \pi }{ 3 } ,\frac { \pi }{ 3 } \right) \)

We have:
\(f(x)=x^{ 3 }-3x^{ 2 }+3x-100\quad \)
\(\therefore f(x)=3x^{ 2 }-6x+3=3(x-1)^{ 2 }\)
Which is for all \(x\in R\)
Hence,f(x) is increasing function on R.

Marginal cost is the rate of change of total cost with respect to output.
Now \(C(x)=0.007x^{ 3 }-0.003x^{ 2 }+15x+4000\)
\(\therefore \) Marginal Cost (MC)=\(\frac { dC }{ dx } \) 
=\((0.007)(3x^{ 2 })-0.003(2x)+15\)
When  x = 17, MC = (0.007) (3\(17)^{ 2 })\)- 0.003 (2(17)) + 15
= 6.069 - 0.102 + 15 = 20.967
Hence the reqd marginal cost is Rs. 21 (nearly).

Let \(x_{ 1 }\quad and\quad x_{ 2 }\in R\)
Now  \(x_{ 1 }<{ x }_{ 2 }\)
\(\Rightarrow 7x_{ 1 }<7x_{ 2 }\)
\( \Rightarrow 7x_{ 1 }-3<7x_{ 2 }-3\)
\(\Rightarrow f(x_{ 1 })\)
Hence,\(f\) is strictly increasing on \(R\)

Let x be the length of a side, V be the volume and S be the surface area of the cube. Then, V = x³ and S = 6x², where x is a function of time t.
Here \(\frac { dV }{ dt } =9cm3/s \)
\(9=\frac{d \mathrm{~V}}{d t}=\frac{d}{d t}\left(x^3\right)=\frac{d}{d x}\left(x^3\right) \cdot \frac{d x}{d t}\)(By Chain Rule)
\( =3 x^2 \cdot \frac{d x}{d t} \\ \frac{d x}{d t} =\frac{3}{x^2} \ \ ...(1) \)
\(\frac{d S}{d t}=\frac{d}{d t}\left(6 x^2\right)=\frac{d}{d x}\left(6 x^2\right) \cdot \frac{d x}{d t}\)(By Chain Rule)
\(=12 x \cdot\left(\frac{3}{x^2}\right)=\frac{36}{x}\)(Using (1))
Hence when,  \(x=10 \mathrm{~cm}, \frac{d S}{d t}=3.6 \mathrm{~cm}^2 / \mathrm{s}\)