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 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 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
\(=\frac { R }{ x } =\frac { 20x{ e }^{ \frac { -x }{ 10 } } }{ x } \)
\(\Rightarrow P=20{ e }^{ \frac { -x }{ 10 } }\)

\(\frac { { E }_{ y } }{ { E }_{ x } } =\frac { x }{ x-2 } \)
\(\frac { { xd }_{ y } }{ { yd }_{ x } } =\frac { x }{ x-2 } \)
\(\frac { dy }{ y } =\frac { x }{ x-2 } .\frac { dx }{ x } \)
\(\int { \frac { dy }{ y } } =\int { \frac { dx }{ x-2 } } \)
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)