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 \(MC=\frac { 100 }{ x } \)
\(\\ \int { MC } =\int { \frac { 100 }{ x } } \)
⇒ 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 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 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 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