STUDENT NAME:
STUDENT ID NUMBER:
SECTION NUMBER IN THIS COURSE:
FINAL EXAM – Fall 2020
Business Mathematics
COURSE NAME:
COURSE NUMBER:
EXAMINATION DATE:
TSF1320-001
Jan. 11-15, 2021
9:00 – 17:00
60 min / 1 h
EXAMINATION DURATION:
None
ADDITIONAL MATERIALS
ALLOWED TO USE:
SPECIAL INSTRUCTIONS:
TIME:
Qs
Q1: (30)
Q2: (30)
Q3: (20)
Q4: (20)
Total
(100)
Earned
Points:
Please do not open the examination paper until directed to do so.
___________________________________________________________________
READ INSTRUCTIONS FIRST:
Desks should be free from all unnecessary items (books, notes, technology, food, water, clothes).
Use of any electronic device (Phone, iPod, iPad, laptop) is not allowed during the examination.
Cheating, talking to fellow students, singing, turning back are not allowed.
Write your Name (capital letters), ID number and Section number in each page of your examination
paper.
Final answers must be written by only blue or black, non-erasable pen. Do not use highlighters or
correction pen.
All answers should be written in the space provided after each each question.
If you have a problem please raise your hand and wait quietly for a Proctor.
You are not allowed to leave the exam room until you submit the exam papers.
STUDENT NAME:
STUDENT ID NUMBER:
SECTION NUMBER IN THIS COURSE:
Question 1 (30 points)
A monopolist sells its product in two isolated markets/ cities with demand functions
P1 = 32 − Q1 and P2 = 40 − 2Q2 for Tashkent and Samarkand, respectively.
The total cost function is TC = 4(Q1 + Q2).
Part (a) – 5 points – Find the profit function π = TR – TC = 𝑓(𝑄1 , 𝑄2 ).
Part (b) – 5 points – Find the critical values of Q1 and Q2 which might maximise
the profit.
Part (c) – 5 points – Verify that the second-order conditions for a maximum are
satisfied. Calculate the value of the maximum profit. Calculate the values of the
prices in the two markets. In which city customers pay higher price?
Part (d) – 5 points – If the monopolist is no longer allowed to discriminate
between the two markets and must charge the same price, P, show that the total
demand, Q = Q1 + Q2, is given by Q = 52 – 3*P/2.
Part (e) – 5 points – Deduce the profit function π without price discrimination.
Part (f) – 5 points – Find the maximum profit in part (e) and hence work out how
much profit is lost by the monopolist firm when it can no longer discriminate
between the two markets. Is the price without discrimination located between the
two prices calculated in (c) for the discriminated markets?
Question 1 (Solution – HW 7, Problem 12)
Part (a)
𝜋 = 𝑇𝑅 − 𝑇𝐶 = 𝑃1 𝑄1 + 𝑃2 𝑄2 − 𝑇𝐶 =
= (32 − 𝑄1 )𝑄1 + (40 − 2𝑄2 )𝑄2 − 4(𝑄1 + 𝑄2 ) =
= −𝑄12 − 2𝑄22 + 28𝑄1 + 36𝑄2 = 𝑓(𝑄1 , 𝑄2 )
Part (b)
f1 = -2Q1 + 28 = 0, and f2 = -4Q2 + 36 = 0,
hence, (𝑄1 , 𝑄2 ) = (14, 9) is a stationary/ critical point
Part (c)
f11 = −2, f22 = −4, f12 = 0, so,
′′ ′′
′′ )2
𝑓11
𝑓22 − (𝑓12
= (−2)(−4) − 0 = 8 > 0, 𝑎𝑛𝑑 f11 = −2 < 0
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Question 1 (Solution – continue)
Therefore, (𝑄1 , 𝑄2 ) = (14, 9) yields maximum, with max profit 𝜋 = 𝑓(𝑄1 , 𝑄2 ) =
𝑓(14, 9) = -196 -2*81 + 2*196 + 4*81 = 196+162 = 358.
Furthermore, P1 = 32 − Q1 = 18 and P2 = 40 − 2Q2 = 22, hence, Samarkand’s
customers pay higher price for the same good.
Part (d)
P = 32 − 𝑄1 , and P = 40 − 2𝑄2 , hence, 2P = 64 – 2Q1, and,
3P = 104 – 2Q, so, Q = (104 – 3P)/2 = 52 - 3P/2, as well P = (104 – 2Q)/3.
Part (e)
The profit is π = (92Q – 2Q2)/3 coming from
𝜋 = 𝑃𝑄 − 𝑇𝐶 =
𝑄(104 – 2Q)
3
2
92
3
3
− 4𝑄 = − 𝑄2 +
2
𝑄 = − 𝑄(𝑄 − 46) = 𝑓(𝑄).
3
Part (f)
Maximum profit is obtained when derivative f′(Q) = 0, which implies Q=23, hence,
23
𝜋 = 2 ∗ 23 ∗ = 352.67, there is a reduction (profit lost) by 358-352.67 = 5.33.
3
The price P = (104 – 2Q)/3 = 58/3 is located between P1 = 18 and P2 = 22.
Or, without fractional decimals
232 2 ∗ 529 1058
𝜋 =2∗
=
=
= profit.
3
3
3
And, the reduction in profit is
358 −
1058 1074 − 1058
=
= 16/3.
3
3
3
STUDENT NAME:
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Question 2 (30 points)
The demand and supply functions of a good are given by
𝑃 =
8
(𝑄𝐷 +1)2
and 𝑃 = √2 + 2𝑄𝑆
where P, QD and QS denote, as usual, price, nonnegative quantity demanded and
nonnegative quantity supplied, respectively.
Part (a) – 5 points – Calculate the equilibrium price and quantity.
Part (b) – 5 points – Specify the formula, along with the definite integral, for
consumer’s surplus: limits of integration and integrand function.
Part (c) – 5 points – Evaluate the definite integral, and the formula, and so find the
consumer’s surplus CS.
Part (d) – 5 points – Specify the formula, along with the definite integral, for
producer’s surplus: limits of integration and integrand function.
Part (e) – 5 points – Evaluate the definite integral, and the formula, and so find the
producer’s surplus PS. (Hint: you may need √2 ≈ 1.4142)
Part (f) – 5 points – Sketch the regions (areas) from the formulas for both
consumer’s surplus and producer’s surplus.
Question 2 (Solution)
Part (a)
The equilibrium quantity Q = QD = QS is a solution of the equation
8
(𝑄𝐷 +1)2
8
= 𝑃 = √2 + 2𝑄𝑆 that is, (𝑄+1)2 = √2(𝑄 + 1),
hence, (Q+1)5 = 32, so Q = 1 and P = 2.
Part (b)
By definition of CS
1
CS = ∫0
8
(𝑄+1)2
𝑑𝑄 − 𝑃 ∗ 𝑄
Part (c)
Hence,
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Question 2 (Solution – continue)
−8 𝑄=1
𝐶𝑆 =
− 2 ∗ 1 = −4 + 8 − 2 = 2
|
𝑄 + 1 𝑄=0
Part (d)
By definition of PS
1
PS = 𝑃 ∗ 𝑄 − ∫0 √2(𝑄 + 1) 𝑑𝑄
Part (e)
Hence,
2
3
2
𝑄=1
PS = 2 ∗ 1 − √2 (𝑄 + 1) ⌉
3
𝑄=0
8
2
2
= 2 − + √2 = (√2 − 1) = 0.276
3
3
3
Part (f)
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Question 3 (20 points)
Three major LCs (Logistic Companies: LC1, LC2 and LC3) are available and
running their operations in the city of Tashkent on per year contract basis. In a given
year, 40% of their customers are using only LC1, 30% are using only LC2, and 30%
are using only LC3. However, at the end of the year, some unsatisfied customers are
changing their LC provider by contracting service from another LC provider.
LC1 keeps 50% of its customers but loses 20% to LC2, and 30% to LC3.
LC2 keeps 60% of its customers but loses 10% to LC1, and 30% to LC3.
LC3 keeps 80% of its customers but loses 10% to LC1, and 10% to LC2.
Part (a) – 5 points – Find the column vector, v, representing the original customer
share for each LC provider.
Part (b) – 5 points – Find the transition matrix, T, representing the transition in
customers’ loyalty and (dis)satisfaction at the end of the given year among the three
available LC providers.
Part (c) – 5 points – Work out the matrix product, Tv, and give an interpretation of
the elements of the resulting vector.
Part (d) – 5 points – What is the percentage of customers who will use LC1 service
next year; and what is the percentage of customers who will use LC2 service next
year; and what is the “leftover” customer share for LC3 comparing to the other two
LC providers; what is the total customer share for next year?
Question 3 (Solution)
Part (a)
vT = (40, 30, 30) as it is given.
Part (b)
0.5 0.1 0.1
𝐓 = (0.2 0.6 0.1)
0.3 0.3 0.8
Part (c)
26
0.5 0.1 0.1 40
𝐓𝐯 = (0.2 0.6 0.1) (30) = (29)
45
0.3 0.3 0.8 30
Tv gives the customer’s distribution for the next year.
Part (d)
The percentage of customers who will use LC1 service next year is 26%.
The percentage of customers who will use LC2 service next year is 29%.
And, the “leftover” customer share for LC3 is actually the largest one at 45% level.
Total = 26 + 29 + 45 = 100% certainly.
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Question 4 (20 points)
At a logistic company’s depot, a big truck is processed at a cost 3 hundred thousand
UZS with a revenue of 13 hundred thousand UZS (or, cost and revenue can be seen
as 0.3 and 1.3 million UZS). The corresponding figures for a small truck are 2 and
10 hundred thousand UZS, respectively (0.2 and 1 million UZS). The maximum
number of trucks that can be processed on monthly basis at the depot’s terminals is
1200. And the company monthly budget for paying salaries to the processing crews
is 3000 hundred thousand (equivalently 300 million UZS). How many trucks of each
type should be planned for processing the depot’s terminals to maximize profit?
Part (a) – 4 points – Clearly state (describe with words) the decision variables, say
x-and-y, and – 2 points – specify (write) the correct objective function
Part (b) – 4 points – Write and explain with words the type of the of two constraints
Part (c) – 5 points – sketch the feasible region
Part (d) – 5 points – solve the linear programming model; find x =? y =? and Z =?
Question 4 (Solution)
Let x = be number of big trucks processed at the depot per month.
And y = be number of small trucks processed at the depot per month.
Then the objective function is for maximum and it is (revenue minus cost)
Z = 13x + 10y – 3x – 2y = 10x + 8y
Subject to: The budget constraint 3x + 2y ≤ 3000
and capacity constraint x + y ≤ 1200, where the intersection point, of the two straight
lines 3x + 2y = 3000 and x + y = 1200, is the solution x = y = 600, Z = 10800.
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Feasible region:
Note that in the objective function and in the first constraint
max Z = 10x + 8y
s.t.
3x + 2y ≤ 3000
x + y ≤ 1200
the measurement units for all quantities are in hundred thousand UZS, while the
second constraint compare number of trucks.
It is important to compare identical units, whatever these units are!
(units like trucks/ kilograms/ liters/ thousands/ hundred thousand/ millions etc.)
Hence, the solution “x = y = 600, Z = 10800” read as number of 600 trucks and profit
10800 hundred thousand UZS = 1080 million UZS = 1.08 billion UZS.
8