ECON7150 Assignment 1 - Solutions
Question 1
A firm has a demand function
100q p 
and a total cost function
3 21 7 111 50
3
C q q q   
Find the firm’s
(a) revenue function,
100p q 
2
(100 )
100
R pq
q q
q q

 
 
(b) marginal revenue function,
100 2dRMR q
dq
  
(c) marginal cost function,
2 14 111dCMC q q
dq
   
(d) profit function,
2 3 21
3
3 21
3
( )
(100 ) ( 7 111 50)
6 11 50
q R C
q q q q q
q q q
  
     
    
(e) profit maximising level of output.
Profit max when ( ) 0q  
2
1 2
( ) 12 11 0 foc
( -11)( 1) 0
1 and 11 stationary points
q q q
q q
q q
      
 
 
soc
( ) 2 12q q    
2
( 1) 2(1) 12 10q      
At q = 1, (q) is a local minimum.
( 11) 2(11) 12 10q       
At q = 11, (q) is a local maximum.
(q) is a maximum at q = 11.
Question 2
Let (, ) = ௜
ೕ
௝!
(a) Calculate (2,3)
(2,3) =
2ଷ
3!
=
4
3
(b) Calculate ∑ (2, )ଶ௝ୀ଴ , ∑ (, 3)ଷ௜ୀଵ
෍ (2, )
ଶ
௝ୀ଴
= (2,0) + (2,1) + (2,2) =
2଴
0!
+
2ଵ
1!
+
2ଶ
2!
= 5
෍ (, 3)
ଷ
௜ୀଵ
= (1,3) + (2,3) + (3,3) =
1ଷ
3!
+
2ଷ
3!
+
3ଷ
3!
= 6
(c) Calculate ∑ ∑ (, )௜௝ୀ଴ଷ௜ୀଵ
(, ) =
௝
!
1 0 (1,0) = 1଴
0!
= 1
1 1 (1,1) = 1ଵ
1!
= 1
2 0 (2,0) = 2଴
0!
= 1
2 1 (2,1) = 2ଵ
1!
= 2
2 2 (2,2) = 2ଶ
2!
= 2
3
3 0 (3,0) = 3଴
0!
= 1
3 1 (3,1) = 3ଵ
1!
= 3
3 2 (3,2) = 3ଶ
2!
=
9
2
3 3 (3,3) = 3ଷ
3!
=
9
2
෍ ෍ (, )
௜
௝ୀ଴
ଷ
௜ୀଵ
= 1 + 1 + 1 + 2 + 2 + 1 + 3 +
9
2
+
9
2
= 20
Question 3
The point (x0, y0) = (1, 0) lies on the curve
2 3 12y xxe y x e   
Show that (x1, y1) = (0, 4) lies on the tangent to the curve at (x0, y0).
Differentiate wrt x.
2 12 . 6y y xe xe y y y x e     
At (x0, y0) = (1, 0)
0 0 2 01 2.0. 6.1
1 0 6 1
4
e e y y e
y
y
    
   
 

Equation to tangent to 2 3 12y xxe y x e    at (x0, y0)
( ) ( ) ( )( )f x f a f a x a  

0 4( 1)
4 4
y x
y x
  
 
Is (x1, y1) = (0, 4) on the tangent?
4 4
4.0 4 4
y x 
   
Yes, (x1, y1) = (0, 4) is on the tangent to 2 3 12y xxe y x e    at (x0, y0).
4
Question 4
(a) Find the third-order Taylor polynomial for the function () = ଵ
√ଵି௫
about =
0.
The third-order Taylor polynomial about = 0 is
() ≈ (0) + ᇱ(0) +
1
2
ᇱᇱ(0)ଶ +
1
6
ᇱᇱ′(0)ଷ
Here we have
() =
ଵ
√ଵି௫
= (1 − )ିଵ/ଶ => (0) = 1
′() =
ଵ
ଶ
(1 − )ିଷ/ଶ => ′(0) = ଵ
ଶ
′′() =
ଷ
ସ
(1 − )ିହ/ଶ => ′′(0) = ଷ
ସ
′′′() =
ଵହ
(1 − )ି଻/ଶ => ′′′(0) = ଵହ
Thus, the Taylor polynomial is given by
() ≈ 1 +
1
2
+
3
8
ଶ +
5
16
ଷ
(b) Use your answer from part (a) to calculate the approximate value of ଵ
ଷ√ଵଵ
.
Round your answer to six decimal places.
1
3√11
=
1
√99
=
1
10√0.99
=
1
10
1
√1 − 0.01
=
1
10
(0.01)
To approximate (0.01) we plug in 0.01 to our Taylor polynomial
1
10
(0.01) ≈
1
10
൬1 +
1
2
0.01 +
3
8
0.01ଶ +
5
16
0.01ଷ൰ ≈ 0.100504
(c) Find the domain, range, and inverse of () = ଵ
√ଵି௫
. State your answer using
interval notation.
The domain is all values such that (1 − ) > 0 so (−∞, 1).
The range is (0, ∞).
Let = () = ଵ
√ଵି௫
=
1
√1 −
⇔ ଶ =
1
1 −
⇔ 1 − =
1
ଶ
⇔ = 1 −
1
ଶ
5
for ∈ (−∞, 1) and ∈ (0, ∞). Therefore, ିଵ() = 1 − ଵ
௬మ
is the inverse of () =
ଵ
√ଵି௫
.
Question 5
A firm’s price in a perfectly competitive market is 1000. Its cost function is
3 2( ) 0.01 3 1108 960C x x x x    ,
where x  0 is the number of units produced and sold.
(a) Find the profit function for x  0.
3 2
3 2
( ) 1000 ( )
1000 (0.01 3 1108 960)
0.01 3 108 960
x x C x
x x x x
x x x
  
    
    
(b) Find all stationary points and determine the profit maximising level of output.
3 2( ) 0.01 3 108 960x x x x     
foc
2( ) 0.03 6 108 0
0.03( 20)( 180)
x x x
x x
      
   
1 2( ) 0 when 20, 180x x x    
soc
( ) 0.06 6x x    
when 20 ( ) 0.06(20) 6 4.8x x      
when 180 ( ) 0.06(180) 6 4.8x x      
Stationary points are x1 = 20 and x2 = 180.
Profit is maximised when x = 180.
(c) Using a sign diagram, determine the intervals over which (x) is increasing or
decreasing.
foc
2( ) 0.03 6 108 0
0.03( 20)( 180)
x x x
x x
      
   
6
1 2( ) 0 when 20, 180x x x    
When 0 < < 20, ᇱ() < 0, profit is decreasing
When 20 < < 180, ᇱ() > 0, profit is increasing
When > 180, ᇱ() < 0, profit is decreasing
(d) Determine the intervals over which C(x) is concave and convex.
3 2( ) 0.01 3 108 960x x x x     
Profit is concave on I when
ᇱᇱ() ≤ 0 for all in I.
Profit is convex on I when
ᇱᇱ() ≥ 0 for all in I.
ᇱ() = −0.03ଶ + 6 − 108
ᇱᇱ() = −0.06 + 6 = −0.06( − 100)
When = 100 ᇱᇱ() = 0
0 < < 100 ᇱᇱ() > 0
> 100 ᇱᇱ() < 0
Profit is concave when ≥ 100
Profit is convex when ≤ 100
20 180
ℛ
−0.03( − 20)
( − 180)
−0.03( − 20)( − 180)
− − +
0
Definitions essential
7
(e) Where is the point of inflection in c (x)? Give an economic interpretation of
the point of inflection.
A point c is a point of inflection if f (c) =0 and f (c) changes sign at c.
From part (d) there is a point of inflection at x = 100.
When x  100 marginal cost is decreasing.
When x  100 marginal cost is increasing.
or x= 100 is the point at which diminishing marginal returns begin.
Question 6
Given the demand function 0aQ bP k   , where a, b and k are positive constants,
show that the price elasticity of demand is minus one when marginal revenue is zero.
Demand function 0aQ bP k   ,
1 ( )Q k bP
a
k b P
a a
 
 
( ) .p
p dqEl D p
q dp

dQ b
dP a
 
 
   
( ) .
.1
p
p dqEl D p
q dp
P b
ak bP
a
bP bP
k bP bP k



  
Revenue R PQ
2
1( ) ( ) .R Q k aQ Q
b
k aQ Q
b b
    
 
8
Marginal Revenue ( ) 2k aR Q Q
b b
  
When MR = 0
2 0
2
k a Q
b b
kQ
a
 

k aP Q
b b
 
2
.
2
2
when kQ
a
k a kP
b b a
k
b

 

 ( )
2
2
1.
2 2
1
p
bPEl D p
bP k
kb
b
kb k
b
k
k
 

   


 
Question 7
Let 1( ) , 0.xf x e x 
(a) Compute f (x) and f (x).
12
1( ) , 0.xf x e x
x
   
1 1
3 4
1
4
2 1( )
1 2
x x
x
f x e e
x x
x e
x
  

9
(b) Determine where f is concave/convex.
( ) is concave when ( ) 0
( ) is convex when ( ) 0
f x f x
f x f x
 
 
( ) 0 when (1 2 ) 0
1
2
f x x
x
   
 
There are three intervals to consider:
 1, ,0 and 0,1
2 2
,         
 
1, ( ) is concave
2
f x    
1 ,0 ( ) is convex
2
f x   
 0, ( ) is convexf x