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 demand function is p = 25 - x - X²
and p₀ = 19
⇒ 19 = 25-x-x²
⇒ x² + x - 6 = 0
⇒ (x + 3) (x - 2) = 0
⇒ x = -3 or x = 2
Since x cannot be negative x_o = 2
p_o x_o = 19(2) = 38
\(CS=\int _{ 0 }^{ 2 }{ f(x)dx-{ p }_{ 0 }{ x }_{ 0 } } \)
\(=\int _{ 0 }^{ 2 }{ \left( 25-x-{ x }^{ 2 } \right) dx-38 } \)
\(={ \left( 25x-\frac { { x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 3 } \right) }_{ 0 }^{ 2 }-38\)
\(=25(2)-\frac { 4 }{ 2 } -\frac { 8 }{ 3 } -38\)
\(=50-2-\frac { 8 }{ 3 } -38\)
\(=10-\frac { 8 }{ 3 } =\frac { 30-8 }{ 3 } \)
\(CS=\frac { 22 }{ 3 } \) units

Given ηd = 1
\(\Rightarrow \frac { -p }{ x } .\frac { dx }{ dp } =1\)
\(\Rightarrow \frac { dx }{ x } =\frac { -dp }{ p } \)
⇒ log x = log p + log k
⇒log x+ log p = log k
⇒ log px = log k
⇒ px = k
Demand function P = \(\frac{k}{x}\)

Given \(MC=\frac { 100 }{ x } \)
\(\\ \int { MC } =\int { \frac { 100 }{ x } } \)
⇒ C = 100 log x + k
Given C(16) =100 ⇒ When x = 16,
C = 100
∴ 100 = 100 log 16 + k
⇒ k = 100-100 log 16
C = 100 log x + 100 - 100 log 16
= 100 (log x - log16 + 1)
\(C=100(log\left( \frac { x }{ 16 } \right) +1)\)

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

Given MC = 3x² - 2x + 8
⇒ ഽMC = ഽ(3x² - 2x + 8)dx
\(\Rightarrow C=\frac { { 3x }^{ 3 } }{ 3 } -\frac { { 2x }^{ 2 } }{ 2 } +8x+k\)
⇒ C = x³ - x² + 8x + k
Since there is no fixed cost,
when x = 0, C = 0 ⇒ k = 0
∴ C = x³ - x² + 8x

Given curve is y = 4x - X²
The limits are from x = 0 to x = 3
∴ Area \(\int _{ 0 }^{ 3 }{ y } dx=\int _{ 0 }^{ 3 }{ (4x-{ x }^{ 2 }) } dx\)
\({ \left[ \frac { 4{ x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 3 }=2(9)-\frac { 27 }{ 3 } \)
=18 - 9
Area = 9 sq.units

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

\(y=4{ - }x^{ 2 }\)
Required area = \(\int _{ 0 }^{ 2 }{ ydx } =\int _{ 0 }^{ 2 }{ (4- } { x }^{ 2 })dx\)
= \({ \left[ 4x-\frac { { x }^{ 2 } }{ 3 } \right] }_{ 0 }^{ 2 }=8-\frac { 8 }{ 3 } \)
= \(\frac { 16 }{ 3 } \) 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 functionp = 122 - 5x - 2x² and x = 6
When x₀ = 6, p₀ = 122-5(6)-2(6)²
= 122-30-72
= 122-102
P₀ = 20
p₀x₀ = 20 \(\times\) 6 = 120
Consumer's Surplus
CS \(=\int _{ 0 }^{ x }{ f(x) } dx-{ p }_{ 0 }{ x }_{ 0 }\)
\(=\int _{ 0 }^{ 6 }{ (122-5x-2{ x }^{ 2 })dx-120 } \)
\(={ \left[ 122x-\frac { 5{ x }^{ 2 } }{ 2 } -\frac { { 2x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 6 }-120\)
\(=122(6)-5\frac { \left( { 6 }^{ 2 } \right) }{ 2 } -2\frac { \left( { 6 }^{ 3 } \right) }{ 3 } -120\)
\(=732-\frac { 180 }{ 2 } -\frac { 432 }{ 3 } -120\)
= 732-90-144-120
= 732-354
C.S = 378 units

Given \(MC=\frac { 14000 }{ \sqrt { 7x+4 } } \)
\(\Rightarrow \frac { dc }{ dx } =\frac { 14000 }{ \sqrt { 7x+4 } } \)
\(\Rightarrow dC=\frac { 14000 }{ \sqrt { 7x+4 } } dx\)
\(\Rightarrow \int { dC } =14000\int { \frac { dx }{ \sqrt { 7x+4 } } } \)
\(\Rightarrow 14000\int { (7x+4)^{ -\frac { 1 }{ 2 } }dx } \)
\(\Rightarrow C=14000\frac { { (7x+4) }^{ -\frac { 1 }{ 2 } +1 } }{ \left( -\frac { 1 }{ 2 } +1 \right) 7 } +k\)
\(\Rightarrow C=14000\frac { { (7x+4) }^{ \frac { 1 }{ 2 } } }{ \frac { 7 }{ 2 } } +k\)
\(C={\not14000} \times \frac{2}{\not 7} \sqrt{7 x+4}+k\)
\(\Rightarrow C=4000\sqrt { 7x+4 } +k\)  ...(1)
Given fixed cost is 18,000
When x = 0, C = 18000
⇒ 18000 = 4000√4 + k
⇒ 18000 = 4000(2) + k
⇒ k = 18000-8000
⇒ k = 10000
∴(1) becomes,
C = 4000\(\sqrt { 7x+4 } +10000\), 
Average Cost (AC) \(=\frac{C}{x}\)
= \(\frac { 4000 }{ x } \sqrt { 7x+4 } +\frac { 10000 }{ x } \)

Given marginal revenue
MR = 9-4x²
\(\Rightarrow \frac { dR }{ dx } =9-4{ x }^{ 2 }\) 
\(\Rightarrow \int { dR } =\int { (9-4{ x }^{ 2 })dx } \)
\(\Rightarrow R=9x-\frac { { 4x }^{ 3 } }{ 3 } +k\quad ...(1)\)
When no product is sold, revenue is zero
∴ When x = 0, R = 0
⇒0 = 0 - 0 + k ⇒ k = 0
∴ (1) becomes 
\(R=9x-\frac { 4{ x }^{ 3 } }{ 3 } \)
We know that R = Px where P is the demand function
\(\Rightarrow \mathrm{P} \not x=\not x\left(9-\frac{4 x^{2}}{3}\right)\)
\(\Rightarrow P=9-\frac { 4{ x }^{ 3 } }{ 3 } \)
Hence, the demand function is \(9-\frac { 4{ x }^{ 3 } }{ 3 } \)