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 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

\(\frac { EY }{ Ex } =\frac { -7x }{ (1-2x)(2+3x) } \)
\(\frac { 7 }{ (2x-1)(3x+2) } =\frac { A }{ 2x-1 } +\frac { B }{ 3x+2 } \)
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 \(x=2,y=\frac { 3 }{ 8 } \)
\(\Rightarrow \frac { 3 }{ 8 } =c\left( \frac { 4-1 }{ 8 } \right) \)
\(\frac { 3 }{ 8 } =c\left( \frac { 3 }{ 8 } \right) =c=1\)
\(y=\left( \frac { 2x-1 }{ 3x+2 } \right) \)
\(\Rightarrow y= \frac { 2x-1 }{ 3x+2 }\)

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

\({ y }^{ 2 }=8x\) (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.