Given elasticity of demand = \(\frac{p}{x^3}\)
\(\Rightarrow \frac { -p }{ x } .\frac { dx }{ dp } =\frac { p }{ { x }^{ 3 } } \)
\(\Rightarrow \frac { { -x }^{ 3 }dx }{ x } =p.\frac { dp }{ p } \)
\(\Rightarrow -{ x }^{ 2 }dx=dp\)
\(\Rightarrow -\int { { x }^{ 2 }dx } =\int { dp } \)
\(\Rightarrow -\frac { { x }^{ 3 } }{ 3 } +k=p\)
When p = 2, x = 3
\(\Rightarrow -\frac { { 3 }^{ 3 } }{ 3 } +k=2\)
k = 2 + 9 ⇒ k = 11
∴ (1) becomes
\(P=-\frac { { x }^{ 3 } }{ 3 } +11\)
= 11−\(\frac { { x }^{ 3 } }{ 3 } \)

Given f(x) = 1800x^-0.4
Since additional 400 units are required, the limits are from x = 500 to x = 900.
\(\therefore \int { f(x)dx } =\int _{ 500 }^{ 900 }{ 1800{ x }^{ -0.4 } } dx\)
\(=1800{ \left[ \frac { { x }^{ -0.4+1 } }{ -0.4+1 } \right] }_{ 500 }^{ 900 }\)
\(\\ =1800{ \left( \frac { { x }^{ 0.6 } }{ 0.6 } \right) }_{ 500 }^{ 900 }\)
\(\\ =3000{ \left( { x }^{ 0.6 } \right) }_{ 500 }^{ 900 }\)
= 3000( (900)^0·6 - (500)^0.6)
= 3000 (59.22 - 41.63)
[∵ (900)^0·6 = 59.22 & (500)^0·6 = 41.63]
= 3000(17.59)
= 52,770 
Hence, 52,770 hours are required to manufacture additional 400 units.

Given demand function P_d = 20 - 5x and
Supply function P_s = 4x + 8
Under market equilibrium p_s = P_d
⇒ 20-5x = 4x+8
⇒ 20-8 = 4x+5x
⇒ 12 = 9x
\(\Rightarrow x=\frac{\not 12}{\not 9}=\frac{4}{3}\)
When \({ x }_{ 0 }=\frac { 4 }{ 3 } ,{ p }_{ 0 }=20-5\left( \frac { 4 }{ 3 } \right) =20-\frac { 20 }{ 3 } \)
\(=\frac { 60-20 }{ 3 } =\frac { 40 }{ 3 } \)
\(\therefore { p }_{ 0 }{ x }_{ 0 }=\frac { 40 }{ 3 } \times \frac { 4 }{ 3 } =\frac { 160 }{ 9 } \)
Consumer Surplus (CS)
\(=\int _{ 0 }^{ x }{ f(x)dx } -{ p }_{ 0 }{ x }_{ 0 }\)
\(=\int _{ 0 }^{ \frac { 4 }{ 3 } }{ (20-5x)dx } \)
\(={ \left[ 20x-\frac { { 5x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ \frac { 4 }{ 3 } }-\frac { 160 }{ 9 } \)
\(=20\left( \frac { 4 }{ 3 } \right) -\frac { 5 }{ 2 } \left( \frac { 16 }{ 9 } \right) -\frac { 160 }{ 9 } \)
\(=\frac { 80 }{ 3 } -\frac { 40 }{ 9 } -\frac { 160 }{ 9 } \)
\(=\frac { 240-40-160 }{ 9 } =\frac { 40 }{ 9 } \)
\(\therefore CS=\frac { 40 }{ 9 }\)units
Producer's Surplus 
\((PS)={ p }_{ 0 }{ x }_{ 0 }-\int _{ 0 }^{ x }{ g(x)dx } \)
\(=\frac { 160 }{ 9 } -\int _{ 0 }^{ \frac { 4 }{ 3 } }{ (4x+8)dx } \)
\(=\frac { 160 }{ 9 } -{ \left[ \frac { { 4x }^{ 2 } }{ 2 } +8x \right] }_{ 0 }^{ \frac { 4 }{ 3 } }\)
\(=\frac { 160 }{ 9 } -\left( 2\left( \frac { 16 }{ 9 } \right) +8\left( \frac { 4 }{ 3 } \right) \right) \)
\(=\frac { 160 }{ 9 } -\left( \frac { 32 }{ 9 } +\frac { 32 }{ 3 } \right) \)
\(=\frac { 160 }{ 9 } -\frac { 32 }{ 9 } -\frac { 32 }{ 3 } \)
\(PS=\frac { 160-32-96 }{ 9 } =\frac { 32 }{ 9 } \)units

Given C'(x) = 20 + \(\frac{x}{20}\)
R'(x) = 30
C'(x) = 20 + \(\frac{x}{20}\)
\(\Rightarrow \int { C'(x) } =\int { \left( 20+\frac { x }{ 20 } \right) dx } \)
\(=20x+\frac { { x }^{ 2 } }{ 40 } +{ k }_{ 1 }\)
Since fixed cost is Rs. 100
When x = 0, C = 100 ⇒ k₁ = 100
\(\therefore \ C(x)=20x+\frac { { x }^{ 2 } }{ 40 } +100\) ...(1)
Also, R'(x) = 30
\(\Rightarrow \int { R'(x) } =\int { 30 } dx\)
⇒ R(x) = 30x + k₂
When x = 0, R = 0 ⇒ k₂ = 0
∴ R(x) = 30x ...(2)
Profit function = R(x) - C(x)
\(=30x-\left( 20x+\frac { { x }^{ 2 } }{ 40 } +100 \right) \)
\(=30x-20x-\frac { { x }^{ 2 } }{ 40 } -100\)
\(P=10x-\frac { { x }^{ 2 } }{ 40 } -100\)

Given demand function P_d = 25 - 3x and
Supply function P_s= 5 + 2x
At market equilibrium, P_d = P_s
⇒ 25-3x = 5+2x
⇒ 25-5 = 2x+3x
⇒ 20 = 5x
⇒ x = \(\frac{20}{5}\)
⇒ x₀ = 4
When x₀ = 4, p₀ = 25-3(4)
= 25-12 = 13
p₀ = 13
∴p₀x₀ = 13(4) = 52
∴ Consumer's surplus
\(CS=\int _{ 0 }^{ x }{ f(x) } dx-{ p }_{ 0 }{ x }_{ 0 }\)
\(\\ =\int _{ 0 }^{ 4 }{ (25-3x)dx-52 } \)
\(={ \left[ 25x-\frac { { 3x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ 4 }-52\)
\(=25(4)-\frac { 3\left( { 4 }^{ 2 } \right) }{ 2 } -52\)
=100-24-52
=100-76
C.S = 24 units
Producer's Surplus
\((PS)={ p }_{ 0 }{ x }_{ 0 }-\int _{ 0 }^{ x }{ g(x)dx } \)
\(=52-\int _{ 0 }^{ 4 }{ (5+2x)dx } \)

= 52 - (5(4) + 4²}
= 52 - (20 + 16)
= 52 - 36
PS = 16 units

Given demand equation is \(x=\sqrt { 100-p } \) and
Supply equation is \(x=\frac { p }{ 2 } -10\)
At market equilibrium, \(\sqrt { 100-p } =\frac { p }{ 2 } -10\)  Squaring both sides,
\(100-p={ \left( \frac { p }{ 2 } -10 \right) }^{ 2 }\)
\(\frac { { p }^{ 2 } }{ 4 } -10p+p=0\)
\(\frac { { p }^{ 2 } }{ 4 } -9p=0\)
\({ p }^{ 2 }-36p=0\)
p(p-36) = 0
p = 0 or p = 36
Since p cannot be zero, p = 36
\(\therefore \ { x }_{ 0 }=\sqrt { 100-36 } =\sqrt { 64 } =8\)
∴ p₀x₀ = 36 x 8 = 288
Given demand equation is \(x=\sqrt { 100-p } \)
X² = 100-P(Squaring both sides)
p = 100-x²
Consumer's Surplus
\(CS=\int _{ 0 }^{ x }{ f(x) } dx-{ p }_{ 0 }{ x }_{ 0 }\)
\(CS=\int _{ 0 }^{ 8 }{ (100-{ x }^{ 2 }) } dx-288\)
\(={ \left[ 100x-\frac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 8 }-288\)
\(=100(8)-\frac { { 8 }^{ 3 } }{ 3 } -288\)
\(=800-\frac { 512 }{ 3 } -288\)
\(=512-\frac { 512 }{ 3 } \)
\(=\frac { 1536-512 }{ 3 } \)
\(CS=\frac { 1024 }{ 3 } \)units
Supply equation is given as
\(x=\frac { p }{ 2 } -10\)
\(x=\frac { p-20 }{ 2 } \)
2x = p - 20
p = 2x + 20
∴ Producer's surplus
\(PS={ p }_{ 0 }{ x }_{ 0 }-\int _{ 0 }^{ x }{ g(x)dx } \)
\(=288-\int _{ 0 }^{ 8 }{ (2x+20)dx } \)
\(=288-{ \left( \frac { { 2x }^{ 2 } }{ 2 } +20x \right) }_{ 0 }^{ 8 }\)
\(=288-{ \left( { x }^{ 2 }+20x \right) }_{ 0 }^{ 8 }\)
= 288 - (8² + 20(8))
= 288 - (64 + 160)
= 288 - 224
PS = 64 units

Given demand function \({ p }_{ d }=\frac { 8 }{ x+1 } -2\) and 
Supply function \({ p }_{ s }=\frac { x+3 }{ 2 } \)
Under perfect competition, P_d = P_s
\(\Rightarrow \frac { 8 }{ x+1 } -2=\frac { x+3 }{ 2 } \)
\(\Rightarrow \frac { 8-2(x+1) }{ x+1 } =\frac { x+3 }{ 2 } \)
\(\\ \Rightarrow \frac { 8-2x-2 }{ x+1 } =\frac { x+3 }{ 2 } \)
\(\Rightarrow \frac { 6-2x }{ x+1 } =\frac { x+3 }{ 2 } \)
⇒ 12-4x = (x+1)(x+3)
⇒ 12-4x = x²+3x+x+3
⇒ 12-4x = x²+4x+3
⇒ x²+4x+3-12+4x = 0
⇒ x²+8x-9 = 0
⇒ (x+9)(x-1) = 0
⇒ x = -9 or x = 1

Since cannot be negative, x = 1
\(\therefore { p }_{ 0 }=\frac { x+3 }{ 2 } =\frac { 1+3 }{ 2 } =\frac { 4 }{ 2 } =2\)
∴ p₀x₀ = 2(1) = 2
Consumer's Surplus
\((CS)=\int _{ 0 }^{ x }{ f(x)dx-{ p }_{ 0 }{ x }_{ 0 } } \)
\(=\int _{ 0 }^{ 1 }{ \left( \frac { 8 }{ x+1 } -2 \right) } dx-2\)
\(={ \left[ 8 \log(x+1)-2x \right] }_{ 0 }^{ 1 }-2\)
= 8 log (2) - 8log (1) - 2 - 2
= 8 log2-8(0)-4
[∵ log 1 = 0]
= (8 log 2-4) units
Producer's Surplus PS = \({ p }_{ 0 }{ x }_{ 0 }-\int _{ 0 }^{ x }{ g(x)dx } \)
\(=2-\int _{ 0 }^{ 1 }{ \frac { x+3 }{ 2 } dx } \)
\(=2-\frac { 1 }{ 2 } { \left[ \frac { { x }^{ 2 } }{ 2 } +3x \right] }_{ 0 }^{ 1 }\)
\(=2-\frac { 1 }{ 2 } \left[ \frac { { 1 }^{ 2 } }{ 2 } +3(1) \right] \)
\(=2-\frac { 1 }{ 2 } \left( \frac { 1 }{ 2 } +3 \right) =12-\frac { 1 }{ 2 } \left( \frac { 7 }{ 2 } \right) \)
\(=2-\frac { 7 }{ 4 } =\frac { 8-7 }{ 2 } \)
PS = \(\frac{1}{4}\)units

Given P_d = 18− 2x − x² ; P_s = 2x − 3
We know that at equilibrium prices p_d = p_s
18− 2x − x² = 2x – 3
x² + 4x −21 = 0
(x − 3) (x + 7) = 0
x = –7 or 3
The value of x cannot be negative, x = 3
When x₀ = 3
ஃ p₀ = 18 − 2(3)  − (3)² = 3
CS = \(\int _{ 0 }^{ { x }_{ o } }{ f(x) } \) dx - x₀p₀
= \(\int _{ 0 }^{ 3 }{ (18-2x-{ x }^{ 2 }) } \)dx - 3 x 3
= \({ \left[ 18x-{ x }^{ 2 }-\frac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 3 }\)- 9
= 18(3) - (3)² - \(\left( \frac { { 3 }^{ 3 } }{ 3 } \right) \) - 9
CS = 27 units
PS = x₀P₀ - \(\int _{ 0 }^{ { x }_{ o } }{ g(x) } \)
= (3 \(\times\) 3) - \(\int _{ 0 }^{ 3 }{ (2x-3) } \)
= 9 - \(({ { { x }^{ 2 }-3x) } }_{ 0 }^{ 3 }\)
= 9 units
Hence at equilibrium price,
(i) the consumer’s surplus is 27 units
(ii) the producer’s surplus is 9 units.

Given MC = \(\frac{dC}{dx}\alpha x\)
\(\Rightarrow \frac { dC }{ dx } ={ k }_{ 1 }x\)
\(\Rightarrow dC={ k }_{ 1 }xdx\)
\(\Rightarrow \int { dC={ k }_{ 1 }\int { x } dx } \)
\(\Rightarrow C={ k }_{ 1 }\frac { { x }^{ 2 } }{ 2 } +{ k }_{ 2 }...(1)\)
Given fixed cost is Rs. 5000
∴ When x = 0, C = 5000
⇒ 5000 = k₁(0) + k₂ = 5000
∴ (1)becomes C=k₁\(\frac{x^2}{2}+5000\) ...(2)
Also it is given that when x = 50, C = Rs. 5625
\(\therefore (2)5625={ k }_{ 1 }\frac { { x }^{ 2 } }{ 2 } +5000\)
\(\Rightarrow 5625-5000={ k }_{ 1 }\times \frac { { (50) }^{ 2 } }{ 2 } \)
\(\Rightarrow 625={ k }_{ 1 }\times \frac { (50)\times (50) }{ 2 } \)

\(C=\frac { 1 }{ 2 } \left( \frac { { x }^{ 2 } }{ 2 } \right) +5000\)
⇒C = \(\frac{x^2}{4}\) + 5000

Given C'(x) = 5 + 0.13x
R'(x) = 18
Fixed cost is Rs. 120
C'(x) = 5 + 0.13x
⇒ ∫C'(x) = ∫(5+0.13x)dx
⇒ C(x) = 5x \(+\frac { 0.13{ x }^{ 2 } }{ 2 } +{ k }_{ 1 }\)
Since fixed cost is Rs.120 ⇒ k₁ = 120
\(\therefore C(x)=5x+\frac { 0.13{ x }^{ 2 } }{ 2 } +120\) ...(1)
Also R'(x) = 18
⇒ ∫R'(x) = ∫18dx
⇒ R(x) = 18x+k₂
When x = 0, R = 0 ⇒ k₂= 0
∴ R(x) = 18x   ...(2)
Profit function ⇒ P(x) = R(x) - C(x)
\(=18x-5x-\frac { 0.13{ x }^{ 2 } }{ 2 } -120\)
[from (1) & (2)]
P(x) = 13x-0.065x²-120

Given marginal revenue function.
\(MR=\frac { DR }{ dx } =20{ e }^{ \frac { -x }{ 10 } }\left( 1-\frac { x }{ 10 } \right) \)
\(dR={ 20e }^{ \frac { -x }{ 10 } }\left( 1-\frac { x }{ 10 } \right) dx\)
\(R=\int { { 20e }^{ \frac { -x }{ 10 } } } \left( 1-\frac { x }{ 10 } \right) dx\)
We know that \(\int { { e }^{ ax }[af(x)+f'(x)]dx={ e }^{ ax }f(x) } +c\)
Here \(\\ a=\frac { -1 }{ 10 } ,f(x)=x,\quad f'(x)=1\)
\(R=20\int { { e }^{ \frac { -x }{ 10 } } } \left[ -\frac { 1 }{ 10 } x+1 \right] dx=20{ e }^{ \frac { -x }{ 10 } }x+k\)
\(\Rightarrow R=20{ e }^{ \frac { -x }{ 10 } }x+k\quad ...(1)\)
When x = 0, R = 0
0 = 0 + k ⇒ k = 0
(1) becomes 
\(R=20x{ e }^{ \frac { -x }{ 10 } }\)
Demand function P
\(=\frac { R }{ x } =\frac { 20x{ e }^{ \frac { -x }{ 10 } } }{ x } \)
\(\Rightarrow P=20{ e }^{ \frac { -x }{ 10 } }\)

\(\eta _{ d }=\frac { p+2{ p }^{ 2 } }{ 100-p-{ p }^{ 2 } } \)
\(\frac { -p }{ x } \frac { dx }{ dp } =\frac { p(2p+1) }{ 100-p-{ p }^{ 2 } } \)
\(\frac { -dx }{ x } =\frac { -(2p+1) }{ { p }^{ 2 }+p-100 } dp\)
\(\int { \frac { dx }{ x } } =\int { \frac { 2p+1 }{ { p }^{ 2 }+p-100 } } dp\)
log x = log(p² + p = 100) + log k
ஃ x = k(p² + p −100)
When x = 70, p = 5,
70 = k(25 + 5 − 100)
⇒ k = –1
Hence x = 100 − p − p²
R = px
Revenue = p(100 – p – p²)

\(\frac { { E }_{ y } }{ { E }_{ x } } =\frac { x }{ x-2 } \)
\(\frac { { xd }_{ y } }{ { yd }_{ x } } =\frac { x }{ x-2 } \)
\(\frac { dy }{ y } =\frac { x }{ x-2 } .\frac { dx }{ x } \)
\(\int { \frac { dy }{ y } } =\int { \frac { dx }{ x-2 } } \)
log y = log(x − 2) + log k
y = k(x–2)
when x = 6, y = 16 ⇒ 16 = k(6–2)
k = 4
y = 4 (x–2)

Given elasticity of demand \(=\frac { (4-x) }{ x } \)
We know that \({ \eta }^{ 2 }=\frac { -p }{ x } \frac { dx }{ dp } \)

\(\Rightarrow \frac { dx }{ 4-x } =\frac { -dp }{ p } \)
\(\Rightarrow \int { \frac { dx }{ 4-x } =-\int { \frac { dp }{ p } } } \)
\(\Rightarrow log\frac { |4-x| }{ -1 } =-logp+logk\)
\(\Rightarrow -log|4-x|=log\left( \frac { k }{ p } \right) \)
\(\Rightarrow log{ (4-x) }^{ -1 }=log\left( \frac { k }{ p } \right) \)
\(\Rightarrow \frac { 1 }{ 4-x } =\frac { k }{ p } \Rightarrow p=k(4-x)\) ...(1)
When x = 2 and p = 4
4 = k(4-2) ⇒ 4 = k(2)
⇒\(\frac{4}{2}\)= k
⇒ k = 2
∴ (1) becomes
p = 2(4-x)
⇒ p = 8-2x
Also, R = px
⇒ R = (8 - 2x)x
⇒ R = 8x-2x²
Hence, the demand function is 8 - 2x and the revenue function is 8x - 2x².

\(\frac { EY }{ Ex } =\frac { -7x }{ (1-2x)(2+3x) } \)
\(\frac { 7 }{ (2x-1)(3x+2) } =\frac { A }{ 2x-1 } +\frac { B }{ 3x+2 } \)
7 = A(3x+2)+B(2x-1)
\(Put\ x=\frac { -2 }{ 3 } 7=B\left( \frac { -4 }{ 3 } -1 \right) 7=B\left( \frac { -7 }{ 3 } \right) \)
\(Put\ x=\frac { 1 }{ 2 } 7=A\left( \frac { 3 }{ 2 } +2 \right) \Rightarrow 7=A\left( \frac { 7 }{ 2 } \right) \)
\(\therefore \frac { 7 }{ (2x-1)(3x+2) } =\frac { 2 }{ 2x-1 } -\frac { 3 }{ 3x+2 } \)
Also, it is given that x = 2, when y \(=\frac{3}{8}\)
\(\Rightarrow \frac { x }{ y } \frac { dy }{ dx } =\frac { -7x }{ (1-2x)(2+3x) } \)

\(=\frac { -7dx }{ (1-2x)(2+3x) } \)
\(\Rightarrow \frac { dy }{ y } =\frac { 7dx }{ (2x+1)(3x+2) } \)
\(\int { \frac { dy }{ y } =\int { \frac { 7dx }{ (2x-)(3x+2) } } } \)
\(\int { \frac { dy }{ y } =\int { \left( \frac { 2 }{ 2x-1 } -\frac { 3 }{ 3x+2 } \right) dx } } \)
\(log\quad y=2\int { \frac { 1 }{ 2x-1 } dx-3\int { \frac { dx }{ 3x+2 } } } \)
\(=2\frac { log|2x-1| }{ 2 } -3\frac { log|3x+2| }{ 3 } +logc\)
\(=log\quad y-log\quad c=log\left| \frac { 2x-1 }{ 3x+2 } \right| \)
\(\Rightarrow log\left| \left( \frac { y }{ c } \right) \right| =log\left| \frac { 2x-1 }{ 3x+2 } \right| \)
\(\Rightarrow \frac { y }{ c } =\frac { 2x-1 }{ 3x+2 } y=c\left( \frac { 2x-1 }{ 3x+2 } \right) \) ....(1)
When \(x=2,y=\frac { 3 }{ 8 } \)
\(\Rightarrow \frac { 3 }{ 8 } =c\left( \frac { 4-1 }{ 8 } \right) \)
\(\frac { 3 }{ 8 } =c\left( \frac { 3 }{ 8 } \right) =c=1\)
\(y=\left( \frac { 2x-1 }{ 3x+2 } \right) \)
\(\Rightarrow y= \frac { 2x-1 }{ 3x+2 }\)

The present value of payment for t year = \(\int _{ 0 }^{ t }{ { 72000e }^{ -0.09t } } dt\)
Present value of 12 years = \(\int _{ 0 }^{ t }{ { 72000e }^{ -0.09t } } dt\)
= 72000\({ \left[ \frac { { e }^{ -0.09t } }{ { -0.09 } } \right] }_{ 0 }^{ \\ 12 }\)
= \(\frac { 72000 }{ -0.09 } \left[ { e }^{ -0.09(12) }-{ e }^{ 0 } \right] \)
= \(-8,00,000[{ e }^{ -1.08 }-{ e }^{ 0 }]\)
= −8,00,000 [0.3396 −1]
= 5,28,320
Cost of the machine = 5,00,000 − 30,000
= 4,70,000
Hence it not advisable to rent the machine
It is better to buy the machine.

Given, Marginal revenue R'(x) = 10 + e^−0.05x
Total revenue from sale of 100 units is
\(R=\int _{ 0 }^{ 100 }{ (10+{ e }^{ -0.05x }) } dx\)
= \({ \left[ 10x+\frac { { e }^{ -0.05x } }{ -0.05 } \right] }_{ 0 }^{ 100 }\)
= \(\left( 1000-\frac { { e }^{ 5 } }{ 0.05 } \right) -\left( 0-\frac { 100 }{ 5 } \right) \)
= 1000 + 20 − (20 × 0.0067)
= 1019.87
Total revenue = 1019.87 × 1000
= Rs. 10,19,870

Given MC = 8 + 6x
\(C(x)=\int { (8+6x)dx } \) + k₁
= 8x + 3x² + k_{1 ...}(1)
But given when x = 0, C = 0 ⇒ k₁ = 0
ஃ C(x) = 8x + 3x²  ....(2)
Given that MR = 24
R(x) = \(\int { MR } \) dx + k₂
= \(\int { 24 } \) dx + k₂
= \(\int { 24 } \) + k₂
Revenue = 0, when x = 0 ⇒ k₂ = 0
R(x) = 24x ...(3)
Total Profit functions P(x) = R(x) – C(x)
P(x) = 24x − 8x − 3x²
= 16x − 3x²

Given C(x) = \(\int { C' } (x)dx+\)k₁
= \(\int { \left( 50+\frac { x }{ 50 } \right) } \)dx +k₁
C(x) = 50x + \(\frac{x^2}{100}\)+k₁
When quantity produced is zero, then the fixed cost is 200.
i.e. When x = 0, c = 200
⇒ k₁ = 200
Cost function is C(x) = 50x + \(\frac{x^2}{100}\) + 200        (1)
The Revenue R'(x) = 60
R(x) = \(\int { R' } (x)dx\) + k₂
\(\int { 60 } dx\)+ k₂
= 60x + k₂
When no product is sold, revenue = 0
i.e. When x = 0, R = 0
Revenue R(x) = 60x          (2)
Profit P = Total Revenue – Total cost
= 60x - 50x - \(\frac { { x }^{ 2 } }{ 100 } \) - 200
= 10x - \(\frac { { x }^{ 2 } }{ 100 } \) - 200
\(\frac { dp }{ dx } =10-\frac { x }{ 50 } \)
To get profit maximum, \(\frac { dp }{ dx } \) = 0 ⇒ x = 500
\(\frac { { d }^{ 2 }P }{ { dx }^{ 2 } } =\frac { -1 }{ 50 } <0\)
ஃ Profit is maximum when x = 500 and
Maximum Profit is P = 10(500) - \(\frac { { (500) }^{ 2 } }{ 100 } \) - 200
= 5000 – 2500 – 200
= 2300
Profit = Rs.  2,300.

MR = a (x + b)^-2 − c
\(R=\int { { (x+b) }^{ 2 } } \)dx - c \(\int { dx } \)
R = \(\frac { a(x+{ b })^{ -2 } }{ -1 } \) - cx + k
R = \(\frac { a }{ (x+b) } \) - cx + k
When x = 0, R = 0
\(0=\frac { a }{ b } -c(0)+k\)
\(k=\frac { a }{ b } \)
\(R=-\frac { a }{ x+b } \) - cx+\(\frac{a}{b}\)
\(=\frac { -ab+a(x+b) }{ b(x+b) } -cx\quad \)
\(R=\frac { ax }{ b(x+b) } -cx\)
Demand function \(P=\frac { R }{ x } \)
\(P=\frac { a }{ b(x+b) } -c\)
\(P+c=\frac { a }{ b(x+b) } \)
\(b(x+b)=\frac { a }{ P+c } \)
\(x=\frac { a }{ b(P+c) } -b\)

Assume that F(t) is the total sales after t months, sales rate is \(\frac { d }{ dt } F(t)=f(t)\)
∴ F(t) = \(\int _{ 0 }^{ t }{ f(t) } dt\)
Total cumulative sales after 4 months.
F(4) = \(\int _{ 0 }^{ 4 }{ f(t) } dt\)
= \(\int _{ 0 }^{ 4 }{ 3000 } { e }^{ -0.3t }dt\)
= 3000 \({ \left[ \frac { { e }^{ -0.3t } }{ -0.3 } \right] }_{ 0 }^{ 4 }\)
= -10,000 \([{ e }^{ -1.2 }-{ e }^{ 0 }]\)
= −10,000 [0.3012 −1]
= 6988 units
(ii) Sales during the 5th month
= \(\int _{ 0 }^{ 4 }{ 3000 } { e }^{ -0.3t }dt\)
= 3000 \({ \left[ \frac { { e }^{ -0.3t } }{ -0.3 } \right] }_{ 0 }^{ 4 }\)
= -10,000 \([{ e }^{ -1.2 }-{ e }^{ 0 }]\)
= −10,000 [0.2231− 0.3012]
= 781 units
Total sales due to the advertisement campaign
= \(\int _{ 0 }^{ 4 }{ 3000 } { e }^{ -0.3t }dt=\frac { 3000 }{ -0.3 } { \left[ { e }^{ -0.3t } \right] }_{ 0 }^{ \infty }\)
= −10000 [0 −1]
= 10,000 untis.

Since y = x² is symmetric
about Y-axis, the required
Area = \(2\int _{ 0 }^{ 4 }{ x\quad dy } \)
When \(y={ x }^{ 2 }\Rightarrow x=\sqrt { y } \)
∴ Area \(=2\int _{ 0 }^{ 4 }{ \sqrt { y } dy } \)
\(=2\int _{ 0 }^{ 4 }{ { y }^{ \frac { 1 }{ 2 } }dy } \)
\(=2\times \frac { 2 }{ 2 } { \left[ { y }^{ \frac { 3 }{ 2 } } \right] }_{ 0 }^{ 4 }\)
\(=\frac { 4 }{ 3 } \left[ { 4 }^{ \frac { 3 }{ 2 } }-{ 0 }^{ \frac { 3 }{ 2 } } \right] \)
\(=\frac { 4 }{ 3 } \left[ { 4 }^{ \frac { 3 }{ 2 } } \right] \)
\(=\frac { 4 }{ 3 } \times { ({ 2 }^{ 2 }) }^{ \frac { 3 }{ 2 } }=\frac { 4 }{ 3 } \times { 2 }^{ 3 }\)
\(A=\frac { 4 }{ 3 } \times 8=\frac { 32 }{ 3 } \) sq.units

y² = 4 ax is the right open
⇒ y = \(\sqrt { 4ax } \) parabola.
The limits are from x = 0 to x = a
∴ Area \(=2\int _{ 0 }^{ a }{ ydx } \)
\(=2\int _{ 0 }^{ a }{ \sqrt { 4ax } } dx\)
\(=2\int _{ 0 }^{ a }{ 2\sqrt { a } \sqrt { x } dx } \)
\(=4\sqrt { a } \int _{ 0 }^{ a }{ \sqrt { x } dx } =4\sqrt { a } \int _{ 0 }^{ a }{ { x }^{ \frac { 1 }{ 2 } }dx } \)
\(=4\sqrt { 3 } { \left[ \frac { { x }^{ \frac { 3 }{ 2 } } }{ \frac { 3 }{ 2 } } \right] }_{ 0 }^{ a }=4\sqrt { a } \times \frac { 2 }{ 3 } { \left( { x }^{ \frac { 3 }{ 2 } } \right) }_{ 0 }^{ a }\)
\(=8\frac { \sqrt { a } }{ 3 } \left( { a }^{ \frac { 2 }{ 3 } }-0 \right) \)
\(=\frac { 8\sqrt { a } }{ 3 } (a\sqrt { a } )\)
\(=\frac { 8 }{ 3 } { a }^{ 2 }\) sq.units

The equation y² = 16x represents a parabola (Open rightward)
Required Area = 2\(\int _{ a }^{ b }{ y } dx\)
\(=2\int _{ 0 }^{ 4 }{ \sqrt { 16x } } \ dx\)
\(=8\int _{ 0 }^{ 4 }{ { x }^{ \frac { 1 }{ 2 } } } dx=8{ \left[ { \frac { { x }^{ { \frac { 3 }{ 2 } } } }{ \frac { 3 }{ 2 } } } \right] }_{ 0 }^{ 4 }=\frac { 16 }{ 3 } \left( { \left( 4 \right) }^{ \frac { 3 }{ 2 } } \right) =\frac { 128 }{ 3 } \) sq.units

Equation of the required circle is \({ x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }\) (1)
put \(y=0\), \({ x }^{ 2 }={ a }^{ 2 }\)
⇒  \(x=\pm a\)
Since equation (1) is symmetrical about both the axes
The required area = 4 [Area in the first quadrant between the limit 0 and a.]
\(=4\int _{ 0 }^{ a }{ y } \ dx\)
\(=4\int _{ 0 }^{ a }\sqrt { { a }^{ 2 }-{ x }^{ 2 } } dx=4{ \left[ \frac { x }{ 2 } \sqrt { { a }^{ 2 }-{ x }^{ 2 } } +\frac { { a }^{ 2 } }{ 2 } \sin ^{ -1 }{ \frac { x }{ a } } \right] }_{ 0 }^{ a }\)
\(=4{ \left[ 0+\frac { { a }^{ 2 } }{ 2 } \sin ^{ -1 }{( \frac { a }{ a } )} \right] }=4{ \left[ \frac { { a }^{ 2 } }{ 2 } \sin ^{ -1 }{( 1)} \right] }=4.\frac { { a }^{ 2 } }{ 2 }\frac { {π} }{ 2 }\)
= πa² sq. units

\(y=\left| x+3 \right| =\begin{cases} x+3\quad if\quad x\ge -3\quad \\ -(x+3)\quad if\quad x<-3 \end{cases}\)
Required area = \(\int _{ b }^{ a }{ y } dx=\int _{ -6 }^{ 0 }{ y } dx\)
= \(\int _{ -6 }^{ -3 }{y}\ dx+\int _{ -3 }^{ 0 }{ y} dx\)
= \(\int _{ -6 }^{ -3 }{ -(x+3) } dx+\int _{ -3 }^{ 0 }{ (x+3) } dx\)
= \(-{ \left[ \frac { { (x+3) }^{ 2 } }{ 2 } \right] }_{ -6 }^{ -3 }{ +\left[ \frac { { (x+3) }^{ 2 } }{ 2 } \right] }_{ -3 }^{ 0 }\) 
\(=-\left[ 0-\frac { 9 }{ 2 } \right] +\left[ \frac { 9 }{ 2 } -0 \right] \)
= 9 sq. units

 

\({ y }^{ 2 }=8x\) (1)
Comparing this with the standard form \({ y }^{ 2 }=4x\),
4a = 8
a = 2
Equation of latus rectum is x = 2
Since equation (1) is symmetrical about x- axis
Required Area = 2[Area in the first quadrant between the limits x = 0 and x = 2]
\(=2\int _{ 0 }^{ 2 }{ y } dx\)
\(2\int _{ 0 }^{ 2 }{ \sqrt { 8xdx } } =2(2\sqrt { 2 } )\int _{ 0 }^{ 2 }{ { x }^{ 1/2 } } dx\)
\(=4\sqrt { 2 } { \left[ \frac { { 2x }^{ \frac { 3 }{ 2 } } }{ 3 } \right] }_{ 0 }^{ 2 }=4\sqrt { 2 } \times 2 \times \frac { { 2 }^{ \frac { 3 }{ 2 } } }{ 3 } \)
\(=\frac { 32 }{ 3 } \) sq. units.

Area = \(\int _{ 1 }^{ 4 }{ ydx } \)
= \(\int _{ 1 }^{ 4 }{ (4x+3)dx } \)
= \([{ { 2x }^{ 2 }+3x] }_{ 1 }^{ 4 }\) = 32 +12 − 2 − 3
= 39 sq.units