Given Curve is y² = 27x³.
\({ x }^{ 3 }=\frac { { y }^{ 2 } }{ 27 } \)
\(x={ \left( \frac { { y }^{ 2 } }{ 27 } \right) }^{ \frac { 1 }{ 3 } }=\frac { { y }^{ \frac { 2 }{ 3 } } }{ 3 } \)
∴ Area \(=\frac { 1 }{ 3 } \int _{ 1 }^{ 2 }{ { y }^{ \frac { 2 }{ 3 } } } dy=\frac { 1 }{ 3 } { \left( \frac { { y }^{ \frac { 2 }{ 3 } +1 } }{ \frac { 2 }{ 3 } +1 } \right) }_{ 1 }^{ 2 }\)
\(={ \left( \frac { 1 }{ 3 } .\frac { { y }^{ \frac { 5 }{ 3 } } }{ \frac { 5 }{ 3 } } \right) }_{ 1 }^{ 2 }\)
\(=\frac { 1 }{ 3 } \times \frac { 3 }{ 5 } { \left( { y }^{ \frac { 5 }{ 3 } } \right) }_{ 1 }^{ 2 }=\frac { 1 }{ 5 } \left( { 2 }^{ \frac { 5 }{ 3 } }-{ 1 }^{ \frac { 5 }{ 3 } } \right) \)
\(=\frac { 1 }{ 5 } \left( { 2 }^{ \frac { 5 }{ 3 } }-1 \right) \)sq.units

Given y = 8x²-4x+6
Area \(=\int _{ 0 }^{ 2 }{ ({ 8x }^{ 2 }-4x+6)dx } \)
\(={ \left( \frac { 8x^{ 3 } }{ 3 } -\frac { { 4x }^{ 2 } }{ 2 } +6x \right) }_{ 0 }^{ 2 }\)
\(={ \left( \frac { 8x^{ 3 } }{ 3 } -{ 2x }^{ 2 }+6x \right) }_{ 0 }^{ 2 }\)
\(=\left[ \frac { 8(8) }{ 3 } -2(4)+6(2) \right] -0\)
\(=\frac { 64+12 }{ 3 } =\frac { 76 }{ 3 } \)
Area = \(\frac { 76 }{ 3 } \)sq.units

Given MR = 6 − 3x² − x³
⇒ ഽMR =ഽ(6 − 3x² − x³)dx
\(\Rightarrow \mathrm{R}=6 x-\frac{\not{3} x^{3}}{\not3}-\frac{x^{4}}{4}+k\)
\(\Rightarrow 6x-{ x }^{ 3 }-\frac { { x }^{ 4 } }{ 4 } +k\)
When x = 0, R = 0 ⇒ k = 0
\(\Rightarrow R=6x-{ x }^{ 3 }-\frac { { x }^{ 4 } }{ 4 } \)
Demand function \(P=\frac { R }{ x } =6-{ x }^{ 2 }-\frac { { x }^{ 3 } }{ 4 } \)

Given demand function P_d = 1600 - x² and
Supply function P_s = 2x² + 400
Under perfect competition p_d = p_s
⇒ 1600 - x² = 2x² + 400
⇒ 1600-400 = 2x² + x²
⇒ 1200 = 3x²
\(\Rightarrow \frac { 1200 }{ 3 } ={ x }^{ 2 } \Rightarrow { x }^{ 2 }=400\)
\(\Rightarrow x=\pm \sqrt { 400 } =+20\quad or-20\)
Since x cannot be negative, x₀ = 20
∴ p₀ = 1600 - (20)² = 1600 - 400
= 1200
∴ p₀x₀ = (1200) (20) = 24000
Producer's Surplus PS = P_ox_o -\(\int _{ 0 }^{ x }{ g(x)dx } \)
\(=24000-\int _{ 0 }^{ 20 }{ ({ 2x }^{ 2 }+400)dx } \)
\(=24000-{ \left[ \frac { { 2x }^{ 3 } }{ 3 } +400x \right] }_{ 0 }^{ 20 }\)
\(=24000-\left[ \frac { { 2(20) }^{ 3 } }{ 3 } +400(20) \right] \)
\(=24000-\left[ \frac { 16000 }{ 3 } +8000 \right] \)
\(=24000-\left[ \frac { 16000+24000 }{ 3 } \right] \)
\(=24000-\left[ \frac { 40000 }{ 3 } \right] \)
\(=\frac { 72000-40000 }{ 3 } \)
PS = \( \frac{32000}{3}\)units

Given demand function P_d = 85 - 5x and 
Supply function p₅ = 3x - 35
At equilibrium prices,P_d = P_s
⇒ 85 - 5x = 3x - 35
⇒ 85 + 35 = 3x + 5x
⇒ 120 = 8x
⇒ x = \(\frac{120}{8}\) = 15
When x₀ = 15, p₀ = 85-5(15)
= 85 - 75 = 10
p₀ = 10
∴ p₀x₀ = 15\(\times\)10 = 150
Consumer's Surplus
\(Cs=\int _{ 0 }^{ x }{ f(x) } dx-{ p }_{ 0 }{ x }_{ 0 }\)
\(=\int _{ 0 }^{ 15 }{ (85-5x)dx-150 } \)
\(={ \left[ 85x-\frac { 5{ x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ 15 }-150\)
\(=85(15)-5\frac { { (15) }^{ 2 } }{ 2 } -150\)
\(=1275-\frac { 1125 }{ 2 } -150\)
= 1275 - 562.5 - 150
= 1275 - 712.5
Cs = 562.5

Given marginal cost function = \(\frac { a }{ \sqrt { ax+b } } \)
\(\Rightarrow MC=\frac { a }{ \sqrt { ax+b } } \Rightarrow \frac { dC }{ dx } =\frac { a }{ \sqrt { ax+b } } \)
\(\Rightarrow dC=\frac { a }{ \sqrt { ax+b } } dx\)
\(\Rightarrow \int { dC } =a\int { \frac { dx }{ \sqrt { ax+b } } } \)
\(\Rightarrow C=a\int { { (ax+b) }^{ \frac { -1 }{ 2 } }dx } \)
\(\Rightarrow C=\not a \frac{(a x+b)^{\frac{-1}{2}+1}}{\left(-\frac{1}{2}+1\right) \not a}+k\)
\(\Rightarrow C=\frac { { (ax+b) }^{ \frac { 1 }{ 2 } } }{ 1 } +k\)
\(\Rightarrow C=2\sqrt { ax+b } +k\) ...(1)
Since the cost of output is zero.
C = 0, when x = 0
\(\therefore (1)\rightarrow 0=2\sqrt { 0+b } +k\)
\(\Rightarrow 0=2\sqrt { b } +k\)
\(\Rightarrow k=-2\sqrt { b } \)
∴(1)becomes
\(C=2\sqrt { ax+b } -2\sqrt { b } \)

Given \(MC=300{ x }^{ \frac { 2 }{ 5 } }\)
\(\Rightarrow \frac { dC }{ dx } =300{ x }^{ \frac { 2 }{ 5 } }\)
\(\int { dC } =300\int { { x }^{ \frac { 2 }{ 5 } }dx } \)
\(\Rightarrow C=300\frac { { x }^{ \frac { 2 }{ 5 } +1 } }{ \frac { 2 }{ 5 } +1 } +k\)
\(\Rightarrow C=300\frac { { x }^{ \frac { 7 }{ 5 } } }{ \frac { 7 }{ 5 } } +k\)
\(\Rightarrow C=300\times \frac { 5 }{ 7 } { x }^{ \frac { 7 }{ 5 } }+k\) ...(1)
Given fixed cost is zero ⇒ k = 0
∴(1) becomes,
\(C=\frac { 1500 }{ 7 } { x }^{ \frac { 7 }{ 5 } }+0\Rightarrow C=\frac { 1500 }{ 7 } { x }^{ \frac { 7 }{ 5 } }\)
Average cost function \((AC)=\frac { C }{ x } \)
\(\Rightarrow AC=\frac { 1500 }{ 7 } \frac { { x }^{ \frac { 7 }{ 5 } } }{ x } \)
\(\Rightarrow AC=\frac { 1500 }{ 7 } { x }^{ \frac { 7 }{ 5 } -1 }\)
\(\Rightarrow AC=\frac { 1500 }{ 7 } { x }^{ \frac { 2 }{ 5 } }\)

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 \({ p }_{ d }=\frac { 8 }{ x+1 } -2\) 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 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 supply function is P = x + x + 3 and X_o = 4
∴ P_o = 4² + 4 + 3
= 16+ 4 + 3 = 23
∴ p₀x₀ = 23(4) = 92
Producer's surplus
\(PS={ p }_{ 0 }{ x }_{ 0 }-\int _{ 0 }^{ x0 }{ g(x)dx } \)
\(=92-\int _{ 0 }^{ 4 }{ \left( { x }^{ 2 }+x+3 \right) dx } \)
\(=92-{ \left( \frac { { x }^{ 3 } }{ 3 } +\frac { { x }^{ 2 } }{ 2 } +3x \right) }_{ 0 }^{ 4 }\)
\(=92-\left( \frac { { 4 }^{ 3 } }{ 3 } +\frac { { 4 }^{ 2 } }{ 2 } +3(4) \right) \)
\(=92-\left( \frac { 64 }{ 3 } +8+12 \right) \)
\(=92-\frac { 64 }{ 3 } -20=72-\frac { 64 }{ 3 } \)
\(=\frac { 216-64 }{ 3 } \)
\(PS=\frac { 152 }{ 3 } \) units

Given C'(x) = 85 + \(\frac{375}{x^2}\).
We know C(x) ഽC'(x) + k
The cost of producing 10 incremental units after 15 units have been produced
\(C'(x)=85+\frac { 375 }{ { x }^{ 2 } } \)
\(C(x)=\int { C'(x) } dx\)
\(\int _{ 15 }^{ 25 }{ C'(x) } dx\)
\(\int _{ 15 }^{ 25 }{ \left( 85+\frac { 375 }{ { x }^{ 2 } } \right) } dx\)
\({ \left[ 85+\frac { 375 }{ { x } } \right] }_{ 15 }^{ 25 }\)
\(\left( 85(25)-\frac { 375 }{ 25 } \right) -\left( 85(15)-\frac { 375 }{ 15 } \right) \)
= (2125 - 15) - (1275 - 25)
= 2110 - 1250 = Rs. 860

Given MR = 9 - 6x² + 2x
⇒ഽMR=ഽ(9 - 6x² + 2x)sx
\(\Rightarrow R=9x-\frac { { 6x }^{ 3 } }{ 3 } +\frac { { 2x }^{ 2 } }{ 2 } +k\)
⇒ R = 9x - 2x³ + x² + k
When x = 0, R = 0 ⇒ k = 0
∴ R = 9x - 2x³ + x²

The Y-axis is the ordinate at x = 0.
The area bounded by the ordinates at x = 0, x = 2 and the given curve is
\(A=\int _{ a }^{ b }{ y } dx\)
\(=\int _{ 0 }^{ 2 }{ ({ 4x }^{ 2 }-8x+6)dx } \)
\(={ \left[ \frac { { 4x }^{ 3 } }{ 3 } -\frac { 8{ x }^{ 2 } }{ 2 } +6x \right] }_{ 0 }^{ 2 }\)
\(=4\left( \frac { 8 }{ 3 } \right) -4(4)+6(2)\)
\(\frac { 32 }{ 3 } -6+12=\frac { 32 }{ 3 } -4\)
\(=\frac { 32-12 }{ 3 } =\frac { 20 }{ 3 } \)
∴ Area \(=\frac { 20 }{ 3 } \)sq.units

Area under the curve is
\(A=\int _{ c }^{ d }{ xdy } =\int _{ 2 }^{ 4 }{ \sqrt { 4y } dy } \)
\(=2\int _{ 2 }^{ 4 }{ { y }^{ \frac { 1 }{ 2 } }dy } =2{ \left( \frac { { y }^{ \frac { 3 }{ 2 } } }{ \frac { 3 }{ 2 } } \right) }_{ 2 }^{ 4 }\)
\(=\frac { 4 }{ 3 } \left( { y }^{ \frac { 3 }{ 2 } } \right) \)
\(=\frac { 4 }{ 3 } \left( { 4 }^{ \frac { 3 }{ 2 } }-{ 2 }^{ \frac { 3 }{ 2 } } \right) \)
\(=\frac { 4 }{ 3 } \left( 4\sqrt { 4 } -2\sqrt { 2 } \right) \)
\(=\frac { 4 }{ 3 } (8-2\sqrt { 2 } )\)sq.units