x = 10 - p
\(\frac { dx }{ dp } \) = -1
Elasticity of demand n_d = \(\frac { -p }{ x } .\frac { dx }{ dp } \)
\(\Rightarrow\)  n_d = \(\frac { -p }{ 10-p } (-1)=\frac { p }{ 10-p } \)
At p = 6,  n_d  \(=\frac { 6 }{ 4 } =\frac { 3 }{ 2 } > 1\) = |n_d| = 1 = elastic
Hence the demand is elastic.

Given that f(x) = x² – 4x + 6
Differentiate with respect to x,
f' (x) = 2x = 4
when  f'(x) = 0 \(\Rightarrow\) 2x - 4 = 0 \(\Rightarrow\) x = 2
Then the real line is divided into two intervals namely (-\(\infty \),2) and (2,\(\infty \))

[to choose the sign of f' (x) choose any values for x from the intervals ans substitute in f' (x) and geth the sign]

given C(x) = \(\frac { { x }^{ 2 } }{ 6 } \)+ 5x + 200 .....(1)
and  p(x) = 40 - x .....(2)
Profit is maximized when, marginal revenue = marginal cost
(i..e) R'(x) = C'(x) 
= \(\frac { x }{ 3 } +5\)
R = p . x
= 40x– x²
R'(x) = 40–2x
Hence 40 - 2x = \(\frac { x }{ 3 } +5\)
x = 15
At x = 15, the profit is maximum.

\(A C =2 x-11+\frac{50}{x} \)
\(\frac{d(A C)}{d x} =2-\frac{50}{x^2} \)
\(\frac{d(A C)}{d x} =0 \Rightarrow 2-\frac{50}{x^2}=0 \Rightarrow 2=\frac{50}{x^2} \)
\(x^2 =25 \Rightarrow x=\pm 5\)
The real line is divided into 2 intervals namely  ( \(-\infty \), -5) (-5, 5) and  (5, \(\infty \))

Average cost is increasing at x > 5

Annual demand : R = 4000
Storage cost: C₁ = Rs. 50
Setup cost per production: C₃= Rs. 160
EOQ = \(\sqrt { \frac { 2R{ c }_{ 3 } }{ { c }_{ 1 } } } \) = \(\sqrt { \frac { 2\times 4000\times 160 }{ 50 } } =160\)
\(\therefore\) To minimize the production cost number of units produced per run is 160 units.