Let the price of low-grade steel be Rs. p per tonne. Then the price of high-grade steel is Rs. 2p per tonne.
The total receipt per day is given by \(R=px+py=px+2p\left( \frac { 40-5x }{ 10-x } \right) \). Hence the problem is to maximise R . Now, simplifying and differentiating R with respect to x , we get
\(R=p\left( \frac { 80-{ x }^{ 2 } }{ 10-x } \right) \)
\(\frac { dR }{ dx } =p\left( \frac { { x }^{ 2 }-20x+80 }{ (10-x)^{ 2 } } \right) \)
\(\frac { dR }{ dx } =-\frac { 40P }{ \left( 10-x \right) ^{ 3 } } \)
Now, \(\frac { dR }{ dx } =0\Rightarrow { x }^{ 2 }-20x+80=0\) and hence \(x=10\pm 2\sqrt { 5 } \) 
At \(x=10-2\sqrt { 5 } ,\frac { { d }^{ 2 }R }{ { dx }^{ 2 } } <0\) and hence R will be maximum. If x \(x=10-2\sqrt { 5 } \) then \(y=5-\sqrt { 5 } \)  
Therefore the steel plant must produce low-grade and high-grade steels respectively in tonnes per day are \(10-2\sqrt { 5 } \)  and \(5-5\sqrt { 5 } \)

Let x = length of the cut on each side of the little squares.
V = the volume of the folded box.
The length of the base after two cuts along each edge of size x is 12 − 2x. The depth of the box  after folding is x, so the volume is V = x \(\times\) (12 - 2x)² Note that,
when x = 0 or 6, the volume is zero and hence there cannot be a box. Therefore the problem is to maximize, V = x \(\times\) (12 - 2x)², x ∈ (0, 6) 
 \(\frac { dV }{ dx } ={ (12-2x) }^{ 2 }-4x(12-2x)\)
= (12 − 2x)(12 − 6x).

\(\frac { dV }{ dx } \) = 0 gives the stationary points x = 2, 6. Since 6 ∉ (0, 6) the only stationary point is at x = 2∈(0, 6). Further, \(\frac { dV }{ dx } \)- changes its sign from postive to negative when passing through x = 2 .
Therefore at x = 2 the volume V is local maximum. The local maximum volume value is V = 128 units. Hence the maximum cut can only be 2 units.

The given function is a polynomial of degree 4. Now,

f′(x) = (x −1)³ + 3(x −1)² . (x − 5) 
= 4(x-1)².(x-4)
f"(x) = 4(x-1)²+2(x-1).(x-4)
= 12(x −1)  (x − 3)
Now,
f''(x) = 0 ⇒ x = 1, x = 3
The intervals of concavity are tabulated in the table 7.7.

Interval	(-∞, 1)	(1, 3)	(3, ∞)
Sign of f'(x)	+	-	+
Concavity	concave up	concave down	concave up

The curve is concave upwards on (∞, 1) and (3, ∞) .
The curve is concave downwards on (1, 3) .
As f′′(x) changes its sign when it passes through x = 1 and x = 3, (1, f(1)) = (1, 0) and (3, f(3)) = (3, −16) are points of inflection for the graph y = f(x). This may be observed from the adjoining figure of the curve f′′(x) .

Given AC = b and CD = a
Let hand r be the height and radius of the cylinder
ஃ AB = AC - BC = b - h
In the diagram

ΔABE - ΔACD
\(\therefore \frac { AB }{ BE } =\frac { AC }{ CD } \) 
\(\Rightarrow \frac { b-h }{ r } =\frac { b }{ a } \)
 \(\Rightarrow r=\frac { a\left( b-h \right) }{ b } \)
Volume of the cylinder = \({ \pi r }^{ 2 }h\)
= \(\pi \left( \frac { a }{ b } \left( b-b \right) ^{ 2 }h \right) \) 
= \(\pi \frac { { a }^{ 2 } }{ { b }^{ 2 } } \left( b-h \right) ^{ 2 }h\)
= \(\pi \frac { { a }^{ 2 } }{ { b }^{ 2 } } \left( { b }^{ 2 }+{ h }^{ 2 }-2bh \right) h\)
 \(V=\frac { { \pi a }^{ 2 } }{ { b }^{ 2 } } \left( { b }^{ 2 }{ h }^{ 2 }+{ h }^{ 3 } \right) \)
Differentiating with respect to 'h',
\(\frac { dV }{ dh } =\frac { { a }^{ 2 } }{ { b }^{ 2 } } \pi \left( { b }^{ 2 }+{ 3h }^{ 2 }-4bh \right) \)
\(\frac { dV }{ dh } =0\Rightarrow { b }^{ 2 }-4bh+{ 3j }^{ 2 }=0\)
\(\Rightarrow (b-3h)+(b-h)=0\Rightarrow b-3h=0\quad \)or b = h
\(\Rightarrow h=\frac { b }{ 3 } asb\neq h\) 
\(\frac { dV }{ { dh }^{ 2 } } =\frac { { a }^{ 2 } }{ { b }^{ 2 } } \pi \left( 6\left( \frac { b }{ 3 } \right) -4b \right) \)
\(\left( \frac { { d }^{ 2 }V }{ { dh }^{ 2 } } \right) _{ at\ h=\frac { b }{ 3 } }=\frac { { a }^{ 2 } }{ { b }^{ 2 } } \pi \left( 6\left( \frac { b }{ 3 } \right) -4b \right) \)
= \(\frac { { a }^{ 2 } }{ { b }^{ 2 } } \pi (-2b)<0\) 
ஃ V is maximum at \(h=\frac { b }{ 3 } \)
ஃ Maximum volume of the cylinder
\(V=\frac { { \pi a }^{ 2 } }{ { b }^{ 2 } } \left( b-h \right) ^{ 2 }h\) 
= \(\frac { { \pi a }^{ 2 } }{ { b }^{ 2 } } \left( b-\frac { b }{ 3 } \right) ^{ 2 }\left( \frac { b }{ 3 } \right) \left[ \therefore h=\frac { b }{ 3 } \right] \)
= \(\frac { { \pi a }^{ 2 } }{ { b }^{ 2 } } \left( \frac { 2b }{ 3 } \right) ^{ 2 }\left( \frac { b }{ 3 } \right) \)
\(=\frac{\pi a^{2}}{\not b^{2}}\left(\frac{4\not b^{2}}{9}\right)\left(\frac{b}{3}\right)=\frac{4}{9}\left(\frac{1}{3} \pi a^{2} b\right)\)
= \(\frac { 4 }{ 9 } \)(Volume of the cone)
Hence, volume of the largest cylinder that can be hidden underneath is \(\frac { 4 }{ 9 } \) times volume of the cone.

Let x and y be the length, breadth of the printed rectangular page

Given xy = 24
\(\Rightarrow y=\frac { 24 }{ x } \)    ..(1)
Length of the page with margin
= x+1+1 = x+2
breadth of the page with margin
= y + 1.5 + 1.5 = y + 3
Area of the page = (x + 2) (y + 3)
Let f(x) = (x + 2) (y + 3)
= \((x+2)\left( \frac { 24 }{ x } +3 \right) \) 
= \(24+3x+\frac { 48 }{ x } +30\) 
= \(3x+\frac { 48 }{ x } +30\) 
\(f'\left( x \right) =3-\frac { 48 }{ { x }^{ 2 } } \)
\(\Rightarrow { x }^{ 2 }=16\Rightarrow x=\pm 4\)
ஃ The critical number are 4,-4
\(f''\left( x \right) =-48\left( \frac { -2 }{ { x }^{ 3 } } \right) =\frac { 96 }{ { x }^{ 3 } } \) 
\(f''\left( 4 \right) =\frac { 96 }{ 64 } >0\) 
ஃ f(x) is minimum when x = 4
When \(x=4,y=\frac { 24 }{ 4 } =6\)  [From (1)]
ஃ Length of the page = x + 2 = 4 + 2 = 9 cm
Breadth of the page = y + 3 = 6 + 3 = 6 cm

f (x) is defined and differentiable for all x∈(-∞, ∞)
\(f'\left( x \right) =\frac { 1 }{ 2 } \left( { e }^{ x }+{ e }^{ -x } \right) \)
\(f''(x)=\frac { 1 }{ 2 } \left( { e }^{ x }-{ e }^{ -x } \right) \)
f"(x) = 0
\(\Rightarrow \frac { 1 }{ 2 } \left( { e }^{ x }-{ e }^{ -x } \right) =0\Rightarrow { e }^{ x }-{ e }^{ -x }=0\) 
\(\Rightarrow { e }^{ x }={ e }^{ -x }\Rightarrow { e }^{ x }=\frac { 1 }{ { e }^{ x } } \) 
\(\Rightarrow { e }^{ 2x }=1\Rightarrow { e }^{ 2x }={ e }^{ 0 }\) 
\(\Rightarrow 2x=0\Rightarrow x=0\) 
The possible intervals are (-∞,0) and (0,∞)

Intervel	(-∞, 0)	(0, ∞)
Sign of f"(x)	Say x = -1 \(\cfrac { 1 }{ 2 } \left( { e }^{ -1 }-{ e }^{ 1 } \right) =-ve\)	Say x = 1 \(\cfrac { 1 }{ 2 } \left( { e }^{ -1 }-{ e }^{ 1 } \right) =+ve\)
Concavity	Concave down	Concave up

ஃ f(x) is concave up in (0, ∞) and concave down in (∞, 0).
Since f"(x) changes its position from negative to positive, when it passes through x = 0 the points of inflection is (0,1(0))
\(f(0)=\frac { 1 }{ 2 } \left( { e }^{ o }-{ e }^{ o } \right) =\frac { 1 }{ 2 } \left( 1-1 \right) =0\) 
ஃ (0, 0) is the point of inflection.

f (x) is defined and differentiable for all x ∈ (0, 2π).
f'(x) = sin x (-sin x) + cos x (cos x)
= cos² X - sin² x
= cos2x
f'(x) = 0
\(\Rightarrow cosx=0cos=\frac { \pi }{ 2 } ,cos\frac { 3\pi }{ 2 } ,cos\frac { 5\pi }{ 2 } ,cos\frac { 7\pi }{ 2 } \)
\(\Rightarrow 2x=\frac { \pi }{ 2 } \frac { \pi }{ 2 } ,\frac { 3\pi }{ 2 } ,\frac { 5\pi }{ 2 } ,\frac { 7\pi }{ 2 } \) 
\(\Rightarrow x=\frac { \pi }{ 4 } ,\frac { 3\pi }{ 4 } ,\frac { 5\pi }{ 4 } ,\frac { 7\pi }{ 4 } \)
The stationary points are at

\(x=\frac { \pi }{ 4 } ,\frac { 3\pi }{ 4 } ,\frac { 5\pi }{ 4 } ,\frac { 7\pi }{ 4 } \) 
\(\left( 0,\frac { \pi }{ 4 } \right) ,\left( \frac { \pi }{ 4 } ,\frac { 3\pi }{ 4 } \right) \left( \frac { 3\pi }{ 4 } ,\frac { 5\pi }{ 4 } \right) \left( \frac { 5\pi }{ 4 } ,\frac { 7\pi }{ 4 } \right) \left( \frac { 7\pi }{ 4 } ,2\pi \right) \)

Interval	\(\left( 0,\frac { \pi }{ 4 } \right) ,\)	\(\left( \frac { \pi }{ 4 } ,\frac { 3\pi }{ 4 } \right) \)	\(\left( \frac { 3\pi }{ 4 } ,\frac { 5\pi }{ 4 } \right) \)	\(\left( \frac { 5\pi }{ 4 } ,\frac { 7\pi }{ 4 } \right) \)	\(\left( \frac { 7\pi }{ 4 } ,2\pi \right) \)
Sign of f'(x)	Say \(x=\frac { \pi }{ 6 } \) cos \(cos2\times \cfrac { \pi }{ 6 } \) = \(=cos\cfrac { \pi }{ 3 } =\cfrac { 1 }{ 2 } \) +ve	Say \(x=\cfrac { \pi }{ 2 } \) \(cos2\times \cfrac { \pi }{ 2 } \)  = \(cos\pi =-1-ve\) -ve	Say y = π cos 2π = 1+ve	Say \(x=\cfrac { 3\pi }{ 2 } \) \(cos2\times \cfrac { 3\pi }{ 2 } \) = cos 3π = -1	Say x = 3200 cos 2 x 3200 = cos 640 = cos (360 + 280) = cos 2800 = cos (270 + 10) = sin 100 =+ve
monotonicity	Strictly increasing	Strictly decreasing	Strictly increasing	Strictly decreasing	Strictly increasing

\(\therefore\) f (x) is strictly increasing in \(\left( 0,\frac { \pi }{ 4 } \right) \)\(\left( \frac { 3\pi }{ 4 } ,5\frac { \pi }{ 4 } \right) \left( \frac { 7\pi }{ 4 } ,2\pi \right) \) and strictly decreasing in \(\left( \frac { \pi }{ 4 } ,\frac { 3\pi }{ 4 } \right) \left( \frac { 5\pi }{ 4 } ,\frac { 7\pi }{ 4 } \right) \) 
Since f'(x) changes its positionfrom positive to negative at \(x=\frac { \pi }{ 4 } ,\frac { 5\pi }{ 4 } \) there is a local
maximum at \(x=\frac { \pi }{ 4 } ,\frac { 5\pi }{ 4 } \) 
\(f\left( \frac { \pi }{ 4 } \right) =sin\frac { \pi }{ 4 } cos\frac { \pi }{ 4 } +5\) 
= \(\frac { 1 }{ \sqrt { 2 } } .\frac { 1 }{ \sqrt { 2 } } +5=\frac { 1 }{ 2 } +5=\frac { 11 }{ 2 } \) 
\(f\left( \frac { 5\pi }{ 4 } \right) =sin{ \frac { 5\pi }{ 4 } }cos\frac { \pi }{ 4 } +5\) 
= \(\left( \frac { -1 }{ \sqrt { 2 } } \right) \left( \frac { -1 }{ \sqrt { 2 } } \right) +5\) 
= \(\frac { 1 }{ 2 } +5=\frac { 11 }{ 2 } \) 
Also f'(x) changes its position from negative to positive at \(x=\frac { 3\pi }{ 4 } ,\frac { 7\pi }{ 4 } \) there is local mmimum
at \(x=\frac { 3\pi }{ 4 } ,\frac { 7\pi }{ 4 } \)
\(\therefore f\left( \frac { 3\pi }{ 4 } \right) =cos\frac { 3\pi }{ 4 } sin\frac { 3\pi }{ 4 } +5\)
= \(\left( \frac { -1 }{ \sqrt { 2 } } \right) \left( \frac { +1 }{ \sqrt { 2 } } \right) +5=5-\frac { 1 }{ 2 } =\frac { 9 }{ 2 } \) 
\(f\left( \frac { 7\pi }{ 4 } \right) =cos\frac { 7\pi }{ 4 } sin\frac { 7\pi }{ 4 } +5\) 
= \(cos\left( 2\pi -\frac { \pi }{ 4 } \right) sin\left( 2\pi -\frac { \pi }{ 4 } \right) +5\)
= \(\left( \frac { 1 }{ \sqrt { 2 } } \right) \left( \frac { -1 }{ \sqrt { 2 } } \right) +5\)
= \(\frac { -1 }{ 2 } +5=\frac { 9 }{ 2 } \)
= \(\frac { -1 }{ 2 } +5=\frac { 9 }{ 2 } \)