the required trapezium is as given in figure. Draw perpendiculars DP and CQ on AB. Let AP = x cm. Note that \(\Delta \)APD \(\cong \) \(\Delta \) BQC.
Therefore, QB = x cm. Also, by pythagoras theorem DP = QC = \(\sqrt { 100-{ x }^{ 2 } } \)
Let A be the area of the trapezium.

Then, A \(\equiv \) A(x)
\(=\frac { 1 }{ 2 } (sum\ of\ parallel\ sides)\times (height)\)
\(=\frac { 1 }{ 2 } (2x+10+10)\sqrt { 100-{ x }^{ 2 } } \)
= (x + 10)\(\sqrt { 100-{ x }^{ 2 } } \)
\(or\ A'(x)=(x+10)\frac { (-2x) }{ 2\sqrt { 100-{ x }^{ 2 } } } +(\sqrt { 100-{ x }^{ 2 }) } \)
\(=\frac { -2{ x }^{ 2 }-10x+100 }{ \sqrt { 100-{ x }^{ 2 } } } \)
Now, A'(x) = 0 gives 2x² + 10x - 100 = 0,
i.e., x = 5 and x = -10
So, x = 5.
Now, A''(x) = \(\frac { \sqrt { 100-{ x }^{ 2 } } (-4x-10)-(-2x^{ 2 }-10x+100)\frac { (-2x) }{ 2\sqrt { 100-{ x }^{ 2 } } } }{ 100-{ x }^{ 2 } } \)
\(=\frac { { 2x }^{ 3 }-300x-1000 }{ { (100-{ x }^{ 2 }) }^{ \frac { 1 }{ 2 } } } \)
(on simplification)
or \(A''(5)=\frac { { 2(5) }^{ 3 }-300(5)-1,000 }{ (100-(5{ ) }^{ 2 })^{ \frac { 3 }{ 2 } } } \)
\(=\frac { -2,250 }{ 75\sqrt { 75 } } =\frac { -30 }{ \sqrt { 75 } } <0\)
Thus, area of trapezium is maximum at x = 5 and the maximum area is given by
A(5) = (5 + 10)\(\sqrt { 100-{ (5) }^{ 2 } } \)
= 15\(\sqrt { 75 } =75\sqrt { 3 } { cm }^{ 2 }\)

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 ABCD be a rectangle inscribed in a given circle with centre at 0 and radius a.
Let AB = 2x and BC = 2y

Then, OA² = OM² + AM² 
\(\Rightarrow a^2=y^2+x^2\)
\(\Rightarrow y=\sqrt{a^2-x^2}\)
Let A be the area of the rectangle.
\(\therefore A=4xy=4x\sqrt{x^2-x^2}\)
\(\Rightarrow \ \ \frac{dA}{dx}=4\{\frac{a^2-2x^2}{\sqrt{a^2-x^2}}\}\)
For maximum or minimum value of A,
\(\frac{dA}{dx}=0\)
\(=4\{\frac{a^2-2x^2}{\sqrt{a^2-x^2}}\}=0\Rightarrow\ \ x=\frac{a}{\sqrt{2}}\)
Now, \(\frac{d^2A}{dx^2}=4\frac{d}{dx}\{(a^2-2x^2)(a^2-x^2)^{-1/2}\}\)
\(\Rightarrow \frac{d^2A}{dx^2}=4[-4x(a^2-x^2)^{-1/2}+(a^2-2x^2)\times(-1/2)(a^2-x^2)^(-3/2)(-2x)]\)
\(=[\frac{-4x}{\sqrt{a^2-x^2}}+\frac{x(a^2-2x^2)}{(a^2-x^2)^{3/2}}]\)
\(\therefore\ (\frac{d^2A}{dx^2})_{x=\frac{a}{\sqrt2}}=-16<0\)
Thus A is maximum when \(x=\frac{a}{\sqrt2}\)
putting \(x=\frac{a}{\sqrt2}\) in (i) \(y=\frac{a}{\sqrt2}\)
Therefore \(x=y=\frac{a}{\sqrt2}\)
Hence area is maximum when x = y ⇒ 2x = 2y
i.e., the rectangle is a square.

let radius, height and slant height of cone be r, h and l respectively.
\(\therefore r^2+h^2=l^2\)

V(volume) \(=\frac{\pi}{3}r^2h\)
\(\Rightarrow V^2=\frac{\pi^2r^4h^2}{9}\)
\(\Rightarrow h^2=\frac{9V^2}{\pi^2r^4}\)
\(A=\pi r l,z=A^2=\pi^2r^2l^2\)
\(=\pi^2r^2(r^2+h^2)\)
\(let \ \ A^2=z\)
\(\therefore\ z=\pi^2r^2[r^2+\frac{9V^2}{\pi^2r^4}]\)
\( z=\pi^2[r^4+\frac{9V^2}{\pi^2r^2}]\)
\( \frac{dz}{dr}=\pi^2[4r^3-\frac{18V^2}{\pi^2r^3}]\)
\(\therefore \ \ \ \frac{dz}{dr}=0\)
\(\Rightarrow\ r=6\sqrt{\frac{9V^2}{2\pi^2}}\)
at \( r=6\sqrt{\frac{9V^2}{2\pi^2}}\)
\(\frac{d^2z}{dr^2}=\pi^2(12r^2+\frac{54V^2}{\pi^2r^4 })>0\)
\(\therefore\) curved surface area is minimum iff
\(2\pi^2r^6=9V^2\)
i.e., \(2\pi^2r^6=\pi^2r^4h^2\)

Given, volume of the box = 1024 cm³.Let length of the side ofsquare base be x cm and height ofthe box be y cm.
= x cm,
height of box = h cm.
Volume of box = 1024 cm³
\(\Rightarrow x.x.h=1024\)
\(\therefore h=\frac{1024}{x^2}\)
C (cost of box) = 5(2x²) + 2.5(4xh)
= 10x² + 10hx
\(=10x^2+\frac{10,240}{x}\)
\(\Rightarrow \frac{dC}{dx}=20x-\frac{10,240}{x^2}\)
Solving \(\frac{dC}{dx}=0,\)  we get x³ = 512   ஃ x = 8
\(=20+\frac{1(10240)}{x^3}>0\)
Thus, cost of box is least at x = 8 and least cost of
box is:
\(C(8)=10(8)^2+\frac{10240}{8}\)
= Rs. 1,920

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.

f(x) = sin3x - cos3x, 0 < x < π 
\(\Rightarrow f'(x)=3\cos3x+3\sin3x\)
\(=3(\cos3x+\sin3x)\)
Put \(f'(x)=0\)
\(\Rightarrow \cos3x+\sin3x=0\)
\(\Rightarrow \sin3x=-\cos3x\)
\(\Rightarrow -\tan3x=1\)
\(\Rightarrow \tan3x=-1\)
As 0\(\pi\), 0 < 3x > 3\(\pi\)
ஃ tan 3x is negative for the following values:
\(3x=\frac{3\pi}{4}\)
\(\Rightarrow x=\frac{\pi}{4}\)
\(3x=\pi+\frac{3\pi}{4}=\frac{7\pi}{4}\)
\(\Rightarrow x=\frac{7\pi}{12}\)
\(3x=\frac{7\pi}{4}+\pi=\frac{11\pi}{4}\)
\(\Rightarrow x=\frac{11\pi}{12}\)
Hence we have intervals:
Hence, \(f(x)-\sin3x-\cos3x\) is strictly increasing in the intervals \((0,\frac{\pi}{4})\cup(\frac{7\pi}{12},\frac{11\pi}{12}) \) and strictly decreasing in intervals \((\frac{\pi}{4},\frac{7\pi}{12})\cup(\frac{11\pi}{12},\pi)\)

Slope of tangent = \(\frac{dy}{dx}=2x-2\)
(i) Tangent parallel to \(2x-y+9=0\)
Slope of line = m₁(say)
ஃ m₁ = 2
ஃ  They are parallel
\(\therefore \frac{dy}{dx}=m_1\)
\(\therefore\ 2x-2=2\)
\(x=2,\ y=7\)
Equation of tangent through (2, 7) and parallel to the given line is \(y-7=2(x-2)\Rightarrow y=2x+3\)
(ii) Tangent perpendicular to \(5y-15x=13\)
Slope of line = m₂(say)
ஃ m₂ = 3
ஃ  They are perpendicular
\(\therefore \frac{dy}{dx}=-\frac{1}{m}\)
\(\therefore (2x-2)\cdot3=-1\)
\(\therefore x=\frac{5}{6},y=\frac{217}{36}\)
Equation of tangent through (\(\frac{5}{6},\frac{217}{36}\)) and perpendicular to the line is \(y=-\frac{217}{36}=-\frac{1}{3}(x-\frac{5}{6})\)
\(\Rightarrow y=\frac{-x}{3}+\frac{227}{36}\)
\(12x+36y=227\)