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ECON1061: Quantitative Analysis Final Assessment

QUESTION 1

James mainly sells confectionery items, newspapers, magazines and cigarettes in his convenience store. Noting his small business is not thriving, he thought of selling hot pies and rolls too. 

Suppose the total cost function for rolls and pies is, 

TC = 800 + 53𝑄, 𝑄 = 𝑄1 + 𝑄2

where 𝑄1 and 𝑄2 denote the quantities of rolls and pies respectfully. If 𝑃1 and 𝑃2 denote the corresponding prices, then the inverse demand equations are:

𝑄1 = 73 − 𝑃1 and 0.5𝑄2 = 100 − 𝑃2

REQUIRED:

  1. If James decides to charge the same price for rolls and pies per day (that is, 𝑃1 = 𝑃2), how many of rolls and pies in total should he make in order to maximize the profit of a particular day?
  1. If James decides to charge different prices as above for rolls and pies per day (that is, 𝑃1 ≠ 𝑃2), how many of rolls and pies should he make in order to maximize the profit of a particular day?
  1. Which of the above options (a) or (b) is more profitable? Provide the rationale for your answer.
  1. If James decides to make a total of 51rolls and pies per day and charges different prices as above (that is, 𝑃1 ≠ 𝑃2 ), how many of rolls and pies each should he make in order to maximize the profit of a particular day? Estimate the increase in maximum profit which results when the total number of rolls and pies per day (51) is increased to 52 [note: assume second-order conditions are satisfied].
  1. The COVID-19 pandemic saw the lockdown of many cities to reduce the spread of the virus. This unprecedented move can be viewed as a negative demand shock. Explain the impact of the lockdown of the city where James’ convenience store is located on the demand functions of rolls and pies (half a page maximum).

(2.5 +3.5 + (2+1) + (3+1) + 5 = 18 marks)

QUESTION 2

The supply and demand functions of a good are given by

𝑃𝑆 = 32 + 𝑄𝑠2

and

𝑃𝐷 = 140 − 𝑄3𝐷2

where 𝑃𝑆, 𝑃𝐷, 𝑄𝑆 and 𝑄𝐷 are the price and quantity supplied and demanded, respectively.

  1. Calculate the producer’s surplus and consumer’s surplus at the equilibrium

point.

  1. Explain the effect, if any, on producer’s surplus if the government imposes a fixed tax on this good (note: no calculation expected).

((2.5+2.5) + 2 = 7 marks) QUESTION 3

  1. A manufacturer’s marginal cost (MC) function is given by:

MC =

Find the equation of the total cost function if total costs are 5200 when 𝑄 = 8.

  1. A population of size π‘₯ is decreasing according to the law

𝑑π‘₯ = −π‘₯

𝑑𝑑 250

where 𝑑 denotes the time in days. If initially the population is of size π‘₯0 (that is, when 𝑑 = 0, π‘₯ = π‘₯0) find how long it takes for the size of the population to be halved.

(3.5 + 3.5 = 7 marks) QUESTION 4

  1. Use the inverse matrix method to solve the following system of equations (note: use the Gauss-Jordan elimination method to determine the required inverse matrix. Be sure to show your workings. If any other method, other than the Gauss-Jordan, is used or your workings not shown, you will be awarded a zero mark).

3π‘₯ + 2𝑦 − 𝑧 = 9

2𝑧 − π‘₯ + 4𝑦 = 5

−2𝑦 + π‘₯ + 𝑧 = 3

  • π‘₯ 2
  1. Let matrix A = (1 2 2 ). If |𝐴| = 0, find π‘₯
    • −1 (π‘₯ − 4)

(5 + 3 = 8 marks)  FORMULA SHEET

1. The Rules for Indices

  1. π‘Ž0 = 1
  2. π‘Ž1 = π‘Ž
  3. π‘Žπ‘š × π‘Žπ‘› = π‘Žπ‘š+𝑛
  4. π‘Žπ‘šπ‘› = π‘Žπ‘š−𝑛

π‘Ž

  1. 1𝑛 = π‘Ž−𝑛 and 𝑛 = π‘Žπ‘›

π‘Ž π‘Ž

  1. π‘Žπ‘šπ‘˜
  2. or π‘Ž

2. The Log Rules

  1. log (A) + log (B) = log (AB)
  2. log (A) – log (B) = log (A)

B

  1. log (Aπ‘˜) = π‘˜ log (A)
  2. log (1) = 0 and ln (1) = 0
  3. log (10) = 1 and ln (e) = 1
  4. ln(ek) = k and eln(k) = k
  5. log10k = k and 10logk = k

Rules for differentiation

  • Power rule:

If 𝑦 = π‘₯𝑛 then 𝑑𝑦 = 𝑛π‘₯𝑛−1

𝑑π‘₯

  • Exponential rule

If 𝑦 = 𝑒π‘₯ then 𝑑𝑦 = 𝑒π‘₯

𝑑π‘₯

  • Log rule If 𝑦 = ln (π‘₯) then 𝑑𝑦 = 1

𝑑π‘₯ π‘₯

  • Product or Multiplication rule

If 𝑦 = 𝑒(π‘₯) × π‘£(π‘₯) (or 𝑦 = 𝑒𝑣) then 𝑑𝑦 = {𝑒 × π‘‘π‘£} + {𝑣 × π‘‘π‘’}

𝑑π‘₯ 𝑑π‘₯ 𝑑π‘₯

  • Division or Quotient Rule if 𝑦 = 𝑒(π‘₯) (or 𝑒 𝑑𝑦 {𝑣×𝑑𝑒𝑑π‘₯} - {𝑒×𝑑π‘₯𝑑𝑣} 𝑣(π‘₯) 𝑦 = 𝑣 ) then 𝑑π‘₯ = 𝑣2
  • Chain Rule or Function of Function Rule

If 𝑦 = 𝑒(π‘₯) then 𝑑𝑦 = 𝑑𝑦 × π‘‘π‘’

𝑑π‘₯ 𝑑𝑒 𝑑π‘₯

 Rules for integration

  • Power Rule

π‘₯𝑛+1

∫ π‘₯𝑛𝑑π‘₯ = + 𝑐

𝑛+1

  • Constant rule:

∫ π‘˜π‘‘π‘₯ = π‘˜π‘₯ + 𝑐, where π‘˜ is a constant

  • Log rule

𝑐

π‘₯

  1. d) Exponential Rule

∫ 𝑒π‘₯ = 𝑒π‘₯ + 𝑐

Evaluating definite integral:

π‘₯=𝑏

∫ 𝑓(π‘₯)𝑑π‘₯ = 𝐹(𝑏) − 𝐹(π‘Ž)

π‘₯=π‘Ž

Consumer Surplus (CS) = ∫𝑄𝑄==0 𝑄0(π‘‘π‘’π‘šπ‘Žπ‘›π‘‘ π‘“π‘’π‘›π‘π‘‘π‘–π‘œπ‘›)𝑑𝑄 − 𝑃0𝑄0 Producer Surplus (PS) = 𝑃0𝑄0 (𝑠𝑒𝑝𝑝𝑙𝑦 π‘“π‘’π‘›π‘π‘‘π‘–π‘œπ‘›)𝑑𝑄 General form of linear and non-linear (quadratic) equations

Linear/straight line: 𝑦 = π‘šπ‘₯ + 𝑐 

π‘š: slope, 𝑐: intercept 

Quadratic: 𝑦 = π‘Žπ‘₯2 + 𝑏π‘₯ + 𝑐

 π‘: intercept

6. Quadratic formula

To find the roots of any quadratic equation, π‘Žπ‘₯2 + 𝑏π‘₯ + 𝑐 = 0,

−𝑏 ± √𝑏2 − 4π‘Žπ‘

 π‘₯ =

2π‘Ž

7. Arithmetic series/progression

The nth term of the arithmetic sequence is: 𝑇𝑛 = π‘Ž + (𝑛 − 1)𝑑

π‘Ž: the first term of the sequence

𝑑: common difference between numbers

The sum of the first n terms,𝑆𝑛, of an arithmetic series is given by the formula:

𝑆𝑛  = 𝑛2 [2π‘Ž + (𝑛 − 1)𝑑]

8. Geometric series/progression

The nth term of a geometric series is

𝑇𝑛 = π‘Žπ‘Ÿπ‘›−1

π‘Ž: the first term of the sequence

π‘Ÿ: common multiple/ratio between numbers 

The sum of the first 𝑛 terms of a geometric series is given by

 π‘†π‘› = π‘Ž(11π‘Ÿπ‘Ÿπ‘›)

9. Expanding Squares

  • (a + b)2 = a2 + 2ab + b2
  • (a − b)2 = a2 − 2ab + b2
  1. Difference of two squares

a2 − b2  = (a − b) × (a + b)

11. Dividing two fractions

a = (a) × (d)

d b c

12. Important economic definitions

  • Total Revenue (TR) = P×Q
  • Total Cost (TC) = Fixed cost (FC) + Variable cost (VC)
  • Average Cost (AC) = TCQ
  • Average Revenue (AR) = TRQ
  • Marginal Cost (MC) = d(TC)

dQ

  • Marginal Revenue (MR) = d(TR)

dQ

  • Profit = TR – TC
  • At the break-even, TR = TC or Profit = 0
  • At the equilibrium, Pd = Ps = P and Qd = Qs = Q
  • If price discrimination is not permitted, then P1 = P2 = P. The overall demand is the sum of the two separate demands: Q = Q1 + Q2

13. The method for finding optimum points of a function, 𝒇(𝒙)

  1. Solve the equation 𝑓(π‘₯) = 0 to find the turning point(s), π‘₯ = π‘Ž
  2. If 𝑓′′(π‘Ž) > 0, then the function has a minimum at π‘₯ = π‘Ž

If 𝑓′′(π‘Ž) < 0, then the function has a maximum at π‘₯ = π‘Ž

14. The method for finding optimum points of a function 𝒇(𝒙, π’š)

  1. Solve the simultaneous equations,

𝑓π‘₯(π‘₯, 𝑦) = 0

𝑓𝑦(π‘₯, 𝑦) = 0 to find the turning points, (π‘Ž, 𝑏).

  1. Let Δ = 𝑓π‘₯π‘₯𝑓𝑦𝑦 − 𝑓π‘₯𝑦2 .
    • if 𝑓π‘₯π‘₯ > 0 and 𝑓𝑦𝑦 > 0 and Δ > 0 at (π‘Ž, 𝑏), then the function has a minimum at (π‘Ž, 𝑏)
    • if 𝑓π‘₯π‘₯ < 0 and 𝑓𝑦𝑦 < 0 and Δ > 0 at (π‘Ž, 𝑏), then the function has a maximum at (π‘Ž, 𝑏)
    • The point is a point of inflection if both second derivatives have the same sign but Δ < 0
    • The point is a saddle point if the second derivatives have different signs and Δ < 0
    • If Δ = 0 then there is no conclusion

15. Profit maximization via MR and MC

  • Profit is maximized when MR = MC and 𝑀𝑅< 𝑀𝐢
  • Profit is minimized when MR = MC and 𝑀𝑅> 𝑀𝐢

16. Constrained Optimization and Lagrange Multipliers 

To find the optimum values of a function,𝑓(π‘₯, 𝑦), subject to a constraint, π‘Žπ‘₯ + 𝑏𝑦 = 𝑀, define the Lagrangian function, 𝐿, where

𝐿 = 𝐿(π‘₯, 𝑦, πœ†) = 𝑓(π‘₯, 𝑦) + πœ†(𝑀 – π‘Žπ‘₯ – 𝑏𝑦)

where λ is called a Lagrange multiplier.

18. Solution of differential equations of the form π’…π’š= 𝒇(𝒙)

𝒅𝒙

  1. integrate both sides of the differential equation with respect to π‘₯. This gives the general solution.
  2. If conditions are given for π‘₯ and 𝑦, substitute these values into the general solution and solve for the arbitrary constant, 𝑐.
  3. Substitute this value of 𝑐 into the general solution to find the particular solution.

19. Matrices

  • Evaluating a 2×2 determinant:

|π‘Ž 𝑏| = (π‘Ž × π‘‘) – (𝑐 × π‘) = π‘Žπ‘‘ – 𝑐𝑏

𝑐 𝑑

  • Evaluating a 3×3 determinant:

π‘Žπ‘Ž1121 π‘Žπ‘Ž1222 π‘Ž13 π‘Ž22 π‘Ž23 π‘Ž21 π‘Ž23

|π‘Ž31 π‘Ž32 π‘Ž23| =(π‘Ž11)× |π‘Ž32 π‘Ž33| - (π‘Ž12)×|π‘Ž31 π‘Ž33| +

π‘Ž33

(π‘Ž13)×|π‘Žπ‘Ž2131 π‘Žπ‘Ž2232| 

  • To write a system of equations in matrix form:

π‘Ž1π‘₯ + 𝑏1𝑦 + 𝑐1𝑧 = 𝑑1

π‘Ž2π‘₯ + 𝑏2𝑦 + 𝑐2𝑧 = 𝑑2

π‘Ž3π‘₯ + 𝑏3𝑦 + 𝑐3𝑧 = 𝑑3

π‘Ž1

(π‘Ž2 π‘Ž3

𝑏1

𝑏2

𝑏3

𝑐1 π‘₯ 𝑑1

𝑐2) ×(𝑦) = (𝑑2)

𝑐3 𝑧 𝑑3

This is known as 𝐴𝑋 = 𝐡 format, 

     

π‘Ž1

where 𝐴 = (π‘Ž2

π‘Ž3

𝑏1

𝑏2

𝑏3

𝑐1 π‘₯ 𝑑1

𝑐2), 𝑋 = (𝑦) and 𝐡 = (𝑑2)

𝑐3 𝑧 𝑑3

  • Inverse matrix method involves solving 𝑋 using 𝑋 = 𝐴−1𝐡
  • Finding the inverse of matrix A using the Gauss-Jordan elimination method:

π‘Ž1

Write down the augmented matrix as (π‘Ž2 π‘Ž3

𝑏1

𝑏2

𝑏3

𝑐1 1

𝑐2|0

𝑐3 0

0

1

0

0

0)

1

Transform the above augmented matrix as

1

(0

0

0

1

0

0 𝑝1

0|𝑝2

1 𝑝3

π‘ž1

π‘ž2

π‘ž3

π‘Ÿ1

π‘Ÿ2)

π‘Ÿ3

The original matrix 𝐴, is now reduced to the identity/unit matrix. The inverse of 𝐴 is given by the transformed unit matrix. That is,

𝑝1 π‘ž1 π‘Ÿ1

𝐴−1 = (𝑝2 π‘ž2 π‘Ÿ2)

𝑝3 π‘ž3 π‘Ÿ3

There are three elementary row operations used to achieve the row echelon form:

  1. Swap the positions of two rows
  2. Multiply (or divide) each element of a row by a nonzero constant Replace a row by the sum of itself and a constant multiple of another row of the matrix.
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