For market equilibrium, P_d = p_s
⇒ 16-x² = 2x²+4
⇒ 16-4 = 2x²+x²
⇒ 3x² = 12
⇒ x² = 4
⇒ x = 土2
Since x = 2 is not possible x₀ = 2
p₀ = 16 - 2² = 16 - 4 = 12
∴ p₀x₀ = 2 x 12 = 24
Consumer's Surplus CS \(\int _{ 0 }^{ x0 }{ f(x)dx-{ p }_{ 0 }{ x }_{ 0 } } \)
\(=\int _{ 0 }^{ 2 }{ (16-{ x }^{ 2 }) } dx-24\)
\(={ \left[ 16x-\frac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 2 }-24\)
\(=16(2)-\frac { { 2 }^{ 3 } }{ 3 } -24\)
\(=32-\frac { 8 }{ 3 } -24\)
\(=8-\frac { 8 }{ 3 } -24\)
\(=\frac { 16 }{ 3 } \) units
Producer's Surplus \(PS={ p }_{ 0 }{ x }_{ 0 }-\int _{ 0 }^{ x0 }{ g(x)dx } \)
\(=24-\int _{ 0 }^{ 2 }{ (2{ x }^{ 2 }+4)dx } \)
\(=24-{ \left[ \frac { { 2x }^{ 3 } }{ 3 } +4x \right] }_{ 0 }^{ 2 }\)
\(=24-\left[ \frac { 16 }{ 3 } +8 \right] \)
\(=24-\frac { 16 }{ 3 } -8=16-\frac { 16 }{ 3 } \)
\(=\frac { 48-16 }{ 3 } =\frac { 32 }{ 3 } \)units

C(x) = 10 - 6x
⇒ ഽC(x) = ഽ10.6x dx
\(\Rightarrow C(x)=10.6\frac { { x }^{ 2 } }{ 2 } +{ k }_{ 1 }\)
=5.3x2+k1
Given fixed cost is Rs. 50
When x=s 0, C=50 ⇒ k₁ =50
∴ C(x)  5.3x² + 50 ----(1)
Total revenue =number of units sold x price per unit
∴ Pofit =R(x)-C(x)
=5x-(5.3x²+50)
[From (1)&(2)]
P=5x-5.3x²-50

Given that η_d=2
\(\Rightarrow \frac { -p }{ x } .\frac { dx }{ dp } =2\)
\(\Rightarrow \frac { dx }{ x } =-2\frac { dp }{ p } \)
Integrating both sides,
\(\int { \frac { dx }{ x } =-2\int { \frac { dp }{ p } +logk } } \)
⇒ log x+2log p=log k
log x.p²=log k[∵ a log b=log b^a]
⇒ xp²=k
When x=4, p=
4(1)²=k
⇒ k=4
∴ xp²=4
⇒ xp²=\(\frac{4}{x}\)
⇒ \(p-\frac { \sqrt { 4 } }{ x } =\frac { 2 }{ \sqrt { x } } \)
Revenue \(R=px=\frac { 2 }{ \sqrt { x } } .x\)

=2√x
\(R=2\sqrt { x } and\quad p=\frac { 2 }{ \sqrt { x } } \)

Given that MR \(=\frac { { e }^{ x } }{ 100 } +x+{ x }^{ 2 }\)
\(\int { MR } =\int { \left( \frac { { e }^{ x } }{ 100 } +x+{ x }^{ 2 } \right) } dx\)
\(R=\frac { { e }^{ x } }{ 100 } +\frac { { x }^{ 2 } }{ 2 } +\frac { { x }^{ 3 } }{ 3 } +k\)
When x=0, R=0
\(\Rightarrow 0=\frac { { e }^{ 0 } }{ 100 } +0+0+k\)
\(k=-\frac { 1 }{ 100 } [\because { e }^{ 0 }=1]\)
∴ Revenue \(R=\frac { { e }^{ x } }{ 100 } +\frac { { x }^{ 2 } }{ 2 } +\frac { { x }^{ 3 } }{ 3 } +\frac { 1 }{ 100 } \)

y = x - 5

The required area lies partially above X-axis and partially below X-axis
∴ Area \(=\int _{ 3 }^{ 5 }{ -ydx } +\int _{ 5 }^{ 7 }{ ydx } \)
\(=\int _{ 3 }^{ 5 }{ (5-x)dx+ } \int _{ 7 }^{ 5 }{ (x-5) } dx\)
\(=\left[ \because \quad y=x-5\Rightarrow -y=5-x \right] \)
\(={ \left[ 5x-\frac { { x }^{ 2 } }{ 2 } \right] }_{ 3 }^{ 5 }+{ \left[ \frac { { x }^{ 2 } }{ 2 } -5x \right] }_{ 5 }^{ 7 }\)
\(=\left( 5(5)-\frac { 25 }{ 2 } \right) -\left( 15-\frac { 9 }{ 2 } \right) +\left( \frac { 49 }{ 2 } -35 \right) -\left( \frac { 25 }{ 2 } -25 \right) \)
\(=25-\frac { 25 }{ 2 } -15+\frac { 9 }{ 2 } +\frac { 49 }{ 2 } -35-\frac { 25 }{ 2 } +25\)
\(=(25-15-35+25)+\left( \frac { 25 }{ 2 } +\frac { 9 }{ 2 } +\frac { 49 }{ 2 } -\frac { 25 }{ 2 } \right) \)
\(=0+\left( \frac { -25+9+49-25 }{ 2 } \right) \)
\(=\frac { 58-50 }{ 2 } =\frac { 8 }{ 2 } \)
A = 4 sq.units