Lecture Note Differentiation
a. The elasticity of demand is
()
2
22
2
2
pdq p p
η== −=−p
qdp 300 p 300 −p
−
The demand is of unit elasticity when η = 1 , that is, when
2
2
2
2
1
300
100
10
p
p
p
p
=
−
=
=±
of which only p= 10 is in the relevent interval 03 ≤≤p 00
0
If 01 ≤<p
22
22
2210
1
300 300 10
p
p
η
×
= <=
−−
and hence the demand is inelastic.
If 10 <≤p 300
22
22
2210
1
300 300 10
p
p
η
×
= >=
−−
and hence the demand is elastic.
b. The total revenue is an increasing function of p when demand is inelastic, that
is, on the interval 0 and a decreasing function of p when demand is
elastic, that is, on the interval
≤<p 10
10 <≤p 300. At the price ൌ of unit
elasticity, the revenue function has a relative maximum.
10
c. The revenue function is ൌݍ or
R()pp= ( 300 −= −p^23 ) 300 pp
Its derivative is
R′()ppp=300 3−= − +^2 3 10( )( 10 p)
which is zero when p=± 10 , of which only ൌ 10 is in the relevant interval
0 ≤≤p 300.
On the interval 01 ≤<p 0 , R′(p)is positive and so R(p)is increasing. On the
interval 10 <≤p 300 , R′(p)is negative and so R(p) is decreasing. At the
critical value ൌ10, R()p stops increasing and starts decreasing and hence
has a relative maximum.
Exercises
1 KNnaedrIev:énGnuKmn_xageRkam rYcsRmYllT§pl[enAgaybMput (Differentiate the
following function, do as much computation as possible to simplify the results)
a.yx x=++^233 b.fx x( )=^98 +++ 5 x x 12
c. (^2)
11 1
y
xx x
=+ − d. (^) ()^33
1
fx x
x
=+
e. fx()=+ −( 2132 x)( x ) f.f(xx)=− −(^22512 )( x)
g. f() ( )( )xx=+−100 2 1 1 5x