QUESTION
Curtin University
School of Economics and Finance
Assignment
Semester 1, 2012
Subject: Economic Techniques 201
SPK: 10815
Total Marks: 20
Answer all four (4) questions.
All questions are of equal value.
Due:
Friday 11 May 2012, 5:00 pm
Submit to:
Mailbox, labelled ‘Gary Madden’, Level 5, Building 402
Cover Sheet
: To be marked, a completed ‘Assignment Cover Page’ MUST be
stapled to the front page of the assignment. The special Cover
Page is available for downloading from Blackboard.
QUESTION 1
Suppose the accountant at the Independent Insurance Company thinks that an
appropriate model of the productivity gains from their training program is the logistic
function:
90
18
=
+
y
e
−
0.4
x
where
y
is the total productivity gain, and
x
is the number of weeks spent in the
program.
(a) Find the initial level of productivity.
(b) Find the derivative of this function after
8x =
weeks. Interpret this value.
(c) At the point of inflection we have:
2
0
dy
dx
=
2
After how many weeks do we have a point of inflection for this function? What can
we say about the rate of learning at this value of
x
.
QUESTION 2
The Dixie Darling Company makes the famous Southern Belle bridal doll that is sold
in L-Mart stores. Sales of their product
q
can be described by the demand function:
1
−
=
where
p
is the price of their doll,
1
0.6 0.4
1 12
100q pp Y
p
is the price of their main competitor, the Sweet
Rose doll, and
Y
is the level of income.
2
(a) Find the three partial derivatives and sign them. Use the economic theory of
demand to interpret these derivatives.
(b) Classify the elasticity of demand values for the three variables
p
and
Y
when
6.5p =
,
5.85p =
and
40Y =
.
QUESTION 3
1
2
CC’s Fabulous Swimwear Company design and manufacture designer swimsuits. The
manager Claude Cee knows that he can sell as many units as he produces. The
technology used at CC’s can be modelled using the Cobb-Douglas production
function:
15Q KL=
.
0.7 0.8
Page 2 of 3
p
,
1
2
While there is no constraints on either capital inputs
K
or labour inputs
L
, the
business requires a bank overdraft to operate. Claude knows that the cost of each unit
of these inputs is:
pp= =
10 12
KL
And the total amount that can be spent on the two types of inputs is $20,000.
(a) Find the appropriate Lagrangean if Claude is to maximize output
Q
subject to
the amount of working capital available for inputs.
(b) What value of
K
and
L
satisfy the first-order conditions for this constrained
maximization.
(c) Find and interpret the value of the Lagrangean multiplier.
QUESTION 4
A firm produces two products—cardboard and string—denoted
x
and
,y
respectively. The corresponding cost function has the form
ln(2 12 10 ).c x xy y= − +
22
Find the critical values and indicate whether the function is at a maximum or
minimum.
Page 3 of 3
SOLUTION
Solution1. The productivity function is given in the question as
a) The initial level of productivity can be find out by substituting the value of x as zero in the above equation. Thus we will get 90/ (1+8) = 10
b) The value of the derivative of the function can be find out by first differentiating the above equation and then substituting the value of x equal to 8 weeks.
The value of e is taken as 2.71. Solving the above derivative we get
(288 X (2.71)-(0.4 X8))/ ((8 X 2.71(0.4 X8)) + 1)2
=1.85/ 1.76 = 6.71
c) The point of inflection is that point where the where the second derivative of the given productivity equation is equated to zero and thus the value of x is calculated. The second derivative of the above equation is developed as shown below:
When the above equation is equated to zero the value of x can be calculated as
16e-0.2x=8e-4x+1
Solving the above equation we get the value of x.
Q2. Below is the equation given in the question
As required below are the derivatives of the given equation with respect to the different variables. Thus the three partial derivatives are
The partial derivative with respect to the price of competitor is given by differentiating with respect to p2. Thus we get
The partial derivative with respect to the level of income is given by differentiating with respect to y. Thus we get
a) The values that are given are substituted in the above equations to compute the value of the derivatives.
The value for the derivatives of the partial derivatives for the price of doll, competitor and the income is -38.1, 28 and 5.21 respectively.
The elasticity of the demand can be seen by analyzing the value of the partial derivative for each parameter. The value of three parameters is given to be p1=6.5, p2=5.85 and y=40. These have been shown.
Thus it can be said that the price of the doll has the maximum elasticity whereas for the income level the elasticity is least.
Thus the demand is most elastic for the change in price off doll and the elasticity is minimum for change in income.
Q3.
a) The given equation is as shown below
15K0.7L0.8
This equation is subject to constraint given in the problem, this has been expressed as
10K + 12L= 20000
Writing the above in Langrangian form. It has been shown below
f(K,L)= 15K0.7L0.8
g(K,L)= 10K + 12L= 20000
| Thus we get | ||||
| F(K,L,ƛ)= | 15K0.7L0.8 -ƛ(10K + 12L-20000) | |||
| The partial derivative equations for the above equation are | ||||
| dF(K,L,ƛ)/dK= | (10.5L0.8/K0.3) – 10ƛ | |||
| dF(K,L,ƛ)/dL= | (12K0.7/L0.2) – 12ƛ | |||
| b) Thus we will solve the following three equations | ||||
| (10.5L0.8/K0.3) – 10ƛ=0 | Eqn 1 | |||
| (12K0.7/L0.2) – 12ƛ=0 | Eqn 2 | |||
| 10K + 12L= 20000 | Eqn 3 | |||
| From Eqn 2 we get | ƛ = K0.7/L0.2 | |||
| putting this in Eqn 1 | 1.05L=K | |||
| From this we get | 22.5L=20000 | |||
| Thus all the values are | L=888.88 | |||
| K=933.33 | ||||
| ƛ =30.85 | ||||
| c) The above values are used to calculate the maximum values and thus this will compute the maximum value of 4011336 obtained from the given equation. | ||||
Q4. The given function is as shown below
To find the critical values the first derivative is taken and equated to zero. The partial derivative of this equation is shown below and equated to zero
,
Equating to zero we get x=3 and y=5.
L089
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