Increasing:\(\left( 0,\frac { \pi }{ 4 } \right) \cup \left( \frac { 5\pi }{ 4 } ,2\pi \right) \) ;Decreasing:\(\left( \frac { \pi }{ 4 } ,\frac { 5\pi }{ 4 } \right) \)

\(x=40 \cot \theta \text { and } \frac{d x}{d t}=2 \mathrm{~m} / \mathrm{s} \)
\(\frac{d x}{d t}=-40 \operatorname{cosec}^{2} \theta \cdot \frac{d \theta}{d t} \)
\(\therefore \frac{d \theta}{d t}=\frac{-2}{40\left(1+\cot ^{2} \theta\right)} \)
\(\left.\Rightarrow \frac{d \theta}{d t}\right]_{x=30}=\frac{-2}{40\left(1+\frac{9}{16}\right)}=\frac{-2 \times 16}{40 \times 25} \)
\(=\frac{-4}{125} \approx-\left(\frac{4}{125}\right)^{c} / \mathrm{s} .\)
Hence, the angle of elevation of the tower is decreasing at the rate of \(\left(\frac{4}{125}\right)^{e} / \mathrm{s}\)

The area A ofa circle with radius r is given by \(A=\pi r^{2}\)
\(\frac{d A}{d t}=2 \pi r \frac{d r}{d t} \Rightarrow 100=2 \pi r \times 5 \Rightarrow r=\frac{10}{\pi} \mathrm{cm} . \)
\({\left[\frac{d r}{d t}=5 \mathrm{~cm} / \mathrm{s}, \frac{d A}{d t}=100 \mathrm{~cm}^{2} / \mathrm{s}\right]}\) 
Hence, radius of the circle is \(r={10\over\pi}{cm}\)

Let radius of cone be x and its height be h.
\(\therefore\) OD = (h - r)

Volume of cone (V)
\(=\frac { 1 }{ 3 } \pi { x }^{ 2 }h\) ...(i)
In \(\Delta OCD,\quad { x }^{ 2 }+({ h-r) }^{ 2 }={ r }^{ 2 }or\quad { x }^{ 2 }={ r }^{ 2 }-{ (h-r) }^{ 2 }\)
\(\therefore V=\frac { 1 }{ 3 } \pi h\{ { r }^{ 2 }-(h-r{ ) }^{ 2 }\} \)
\(=\frac { 1 }{ 3 } \pi (-{ h }^{ 3 }+{ 2h }^{ 2 }r)\)
\(\Rightarrow \frac { dV }{ dh } =\frac { \pi }{ 3 } (-3{ h }^{ 2 }+4hr)\)
\(\therefore \quad \frac { dV }{ dh } =0\Rightarrow h=\frac { 4r }{ 3 } \)
\(\frac { { d }^{ 2 }V }{ { dh }^{ 2 } } =\frac { \pi }{ 3 } (-6h+4r)\)
\(=\frac { \pi }{ 3 } \left( -6\left( \frac { 4r }{ 3 } \right) +4r \right) \)
\(=-\frac { 4\pi r }{ 3 } <0\)
\(\therefore \ at\quad h=\frac { 4r }{ 3 } \), Volume is maximum
Maximum volume
\(=\frac { 1 }{ 3 } \pi .\left\{ -{ \left( \frac { 4r }{ 3 } \right) }^{ 3 }+2{ \left( \frac { 4r }{ 3 } \right) }^{ 2 }r \right\} \)
\(=\frac { 8 }{ 27 } .\left( \frac { 4 }{ 3 } \pi { r }^{ 3 } \right) \)
\(=\frac { 8 }{ 27 } \) (volume of sphere)

Let length and breadth of rectangle be x and y.
\(\therefore\ P=2y+x+\pi\frac{x}{2}\)
\(\therefore\ A=xy+\frac{1}{2}\pi\frac{x^2}{4}\)
\(=\frac{x}{2}[P-x-\frac{\pi x}{2}]+\frac{\pi x^2}{8}\)
\(=P\frac{x}{2}-\frac{x^2}{2}-\pi\frac{x^2}{4}+\pi\frac{x^2}{8}\)
​​​​​​
\(\frac{dA}{dx}=\frac{P}{2}-x-\frac{\pi x}{2}+\frac{\pi x}{4}\)
\(=\frac{P}{2}-x-\frac{\pi x}{4}\)
\(\frac{dA}{dx}=0\)
\(\Rightarrow\ x=\frac{2P}{4+\pi}\ and\ y=\frac{P}{4+\pi}\)
\(\frac{d^2A}{dx^2}=-1-\pi / 4<0\)
⇒Area is maximum when length
\(=\frac{2P}{4+\pi}\)
breath \(=\frac{P}{4+\pi}\)

Let l, b, h be the length, breadth and depth of the tank, respectively.
\(\therefore\ l\times b\times 3=75\)
\(\Rightarrow l\times b=25\)
Let C be the cost, then
C = 100(1x b) + 100[h(b + I)] 
\(=100(l\times \frac{25}{l})+300(\frac{25}{l}+l)\)
\(=2500+300(\frac{25}{l}+l)\)
Differentiating w.r.t. l,
\(\therefore\ \frac{dC}{dl}=0+300(\frac{-25}{l^2}+1)\)
Putting \(\frac{dC}{dl}=0\)
\(\Rightarrow 300(-\frac{25}{l^2}+1)=0\)
\(\Rightarrow l^2=25 \ or\ l=5\)
Getting \(\frac{d62C}{dl^2}=300(\frac{50}{l^3})\)
 \(\Rightarrow (\frac{d62C}{dl^2})_{at\ l=5}=\frac{15000}{125}>0\)
i.e., C is minimum when l = 5
\(\Rightarrow b=5\)
ஃ C = 100(25) + 300(10)
= 2,500 + 3,000
= 5,500
Hence the minimum cost is Rs. 5,500.

Given \(f(x)=\sin x-\cos x, \ \ \ \ 0
\(f'(x)=\cos x+\sin x,\ \ \ \ \ 0
\(\therefore\ f'(x)=0\)
\(\Rightarrow \cos x+\sin x=0\ \ or\ \tan x=-1\)
\(\therefore\ x=\frac{3\pi}{4},\frac{7\pi}{4}\)
Getting   \(f'(x)=\cos x+\sin x\)
i.e., \(f''(\frac{3\pi}{4})\) negative, so at x = \(\frac{3\pi}{4},f(x)\) is local maxima
Hence, local maximum value = \(\frac{1}{\sqrt2}+\frac{1}{\sqrt2}=\sqrt2\) and \(f''(\frac{7\pi}{2})=\cos\frac{7\pi}{4}-\sin\frac{7\pi}{4}=\frac{1}{\sqrt2}+\frac{1}{\sqrt2}\)
i.e., \(f''(\frac{7\pi}{2})\) is positive so \(x=\frac{7\pi}{4}\) is local minima
Hence, local minimum value \(=-\frac{1}{\sqrt2}-\frac{1}{\sqrt2}=-\sqrt2\)