# Managerial Economics Foundation

Topics: Marginal cost, Variable cost, Fixed cost Pages: 17 (3085 words) Published: July 24, 2013
OPEN UNIVERSITY MALAYSIA
Managerial Economics Assignment

QUESTION 1

A certain production process employs two inputs labor (L) and raw materials (R). Output (Q) is a function of these two inputs and is given by the following relationship:

Q = 6L2 R2 - 0.10L3 R3

Assume that raw materials (input R) are fixed at 10 units.

(a) Determine the total product function (TPL) for input L. (2 marks) (b) Determine the marginal product function for input L. (2 marks) (c) Determine the average product function for input L. (2 marks) (d) Find the number of units of input L that maximizes the total product function. (2 marks) (e) Find the number of units of input L that maximizes the marginal product function. (2 marks) (f) Find the number of units of input L that maximizes the average product function. (2 marks) (g) Determine the boundaries for the three stages of production. (3 marks)

[TOTAL: 15 MARKS]
Q = 6L R2 - 0.10L3 R3
Whereas : R = 10
(a) TPL For Input =6L (10) – 0.1L (10)
=6L (100) – 0.1L (1000)
=600L - 100 L

b)MP For Input=d(TPL)/dL
=600L – 100L
=1200L – 300L

c)AP For Input=TPL / L
=(600L -100L )/ L
=600L – 100L

d)Maximise Total Product Function For Input L
[ d(TPL) ]/dl=0
1200L -300L= 0
0=L / L=1200/300
L=4

e)[ d(MPL) ] / dl=0 (Condition for max point on MPL)
[ d(MPL) ] / dl=1200L – 300L
=1200 – 600L
When [ d(MPL) ] / dL= 0
1200-600L=0
-600L=-1200
L= -1200/-600
L= 2

f)[ d(APL)]/dL =0 (Condition for max point on MPL)
[ d(APL)]/dL=600L – 100L
=600 – 200L
When [ d(APL) ]/dL= 0
600 – 200L=0
L=-600/-200
L=3
g)

QUESTION 2
Consolidated Salt Company sells table salt to both retail grocery chains and commercial users (e.g., bakeries, snack food makers, etc.). The demand function for each of these markets is:

Retail grocery chains: P1 = 180 - 8Q1
Commercial users: P2 = 100 - 4Q2
where P1 and P2 are the prices charged and Q1 and Q2 are the quantities sold in the respectivemarkets. Consolidated's total cost function (which includes a "normal" return to the owners) forsalt is:

TC = 50 + 20(Q1 + Q2)
(a) Determine Consolidated's total profit function.(3 marks) (b) Assuming that Consolidated is effectively able to charge different prices in the twomarkets, what are the profit-maximizing price and output levels for the product in the twomarkets? What is Consolidated's total profit under this condition?(5 marks) (c) Assuming that Consolidated is required to charge the same price in each market, what arethe profit-maximizing price and output levels? What is Consolidated's total profit under thiscondition? (6 marks)

[TOTAL: 14 MARKS]

P1=180 – 8 Q1
P2=100 – 4 Q2
TC= 50 + 20 (Q1 + Q2)

(a) Profit=TR – Tc
=[P1 (Q1) + P2 (Q2)] – [ 50 + 20 (Q1+Q2) ]
P1= 180 – 8Q1 and P2= 100 – 4 Q2
So= [(180 – 8Q1) (Q1) + (100 – 4Q2) Q2 ] – [ 50 + 20 (Q1+Q2) ]
= [(180 – 8Q1) (Q1) + (100 – 4Q2) Q2 ] – [ 50 + 20 Q1+20 Q2) ]
= (180Q1 – 8Q1) + (100Q2-4Q2)– 50 - 20Q1- 20Q2)
= 180Q1 – 20Q1 - 8Q1 + 100Q2 – 20Q2 – 4Q2 – 50
= 160Q1 -8Q1 – 80Q2 – 4Q2 – 50

(b) Derivative over Q2
dx / dQ2=80 – 8Q2
dx / dQ2=0 (When profit is maximum)

Therfore,80 – 8Q 2=0
80=8Q2
80/8=Q2
10=Q2

Put Q1= 10 and Q2 = 10 in Profit function
Therfore, profit = 160Q1 -8Q1 + 80Q2 – 4Q2 – 50
= 160(10) -8(10) + 80(10) – 4(10) – 50
= 1600 – 800 +800 – 400 – 50
= 1150

P1=180 – 8Q1P2=100 - 4 Q2
=180 – 8((10)=100 – 4(10)
=180-80=100 – 40
=100=60

(c) When P1=P2
180 – 8Q1=100 - 4 Q2
Therfore Interm of Q2: -...