(i) p = xe^x
\({dp\over dx}=x.e^x+e^x(1)=e^x(x+1)\)
\(\eta _d={-p\over x}.{dx\over dp} = \frac{xe^x}{x}.{1\over e^x(x+1)}\)
\(\therefore\) Elasticity of supply \(=\frac{1}{x+1}\)
(ii) \(p=10 e^{-\frac{x}{3}}\)
\(\frac{d p}{d x}=10\left(-\frac{1}{3}\right) e^{\frac{-x}{3}}=\frac{-10}{3} e^{\frac{-x}{3}}\)
Elasticity of demand \(=\eta_d=\frac{-p}{x} \frac{d x}{d p}\)
\(\eta_d=-\frac{10 e^{\frac{-x}{3}}}{x} \cdot\left(\frac{-3}{10 e^{\frac{-x}{3}}}\right)=\frac{3}{x}\)

Let x be the number of passengers.
\(P=136 -\frac{40}{100}(x-100) \text { if } x \geq 100 \)
\(R=p x =136 x-\frac{2 x^2}{5}+40 x \)
\(=176 x-\frac{2 x^2}{5} \)
\(\frac{d R}{d x} =176-\frac{4 x}{5}\)
For max. revenue \(\frac{d R}{d x}=0\)
\(176 =\frac{4 x}{5} \)
\(x =\frac{176 \times 5}{4}=220 \)
\(\frac{d^2 R}{d x^2} =-\frac{4}{5}<0\)
∴ At  x = 220,
Revenue is maximum.

\(p(x)=\frac { { x }^{ 3 } }{ 3 } +{ x }^{ 2 }+x\)
p(x) = x² + 2x + 1 = (x +1)²
It is clear that P'(x)>0 for all x.
\(\therefore\) The firm is running a profitable business.

f(x) = x³ - 3x² + 4x,x \(\in\) R
f(x) = 3x² 6x+4
= 3x² - 6x +3 + 1
= 3(x- 1)² + 1
> 0, for all x \(\in\) R
\(\therefore\) the function f is strictly increasing on \((-\infty ,\infty )\)

Given p = 10e^{\(\frac { -x }{ 2 } \)}
\({dp\over dx}=10e^{-x/2}(-1/2)=-5e^{-x\over 2}\)
Elasticity of demand \((\eta_d)\)=\({-p\over x}.{dx\over dp}\) 
\(\eta_d=\frac{-10 e^{-\frac{x}{2}}}{x} \cdot \frac{1}{-5 e^{-\frac{x}{2}}}=\frac{2}{x}\)