Tutorial letter 201/2/2018

Similar documents
Study Unit 3 : Linear algebra

Study Unit 2 : Linear functions Chapter 2 : Sections and 2.6

Systems of Linear Equations in Two Variables. Break Even. Example. 240x x This is when total cost equals total revenue.

MAT1300 Final Review. Pieter Hofstra. December 4, 2009

Study Guide. Quantitative Modelling 1 DSC1520. Semesters 1 and 2. Department of Decision Sciences. Bar code

Econ 110: Introduction to Economic Theory. 8th Class 2/7/11

Chapter 4 Differentiation

UNIVERSITY OF KWA-ZULU NATAL

Final Exam Review. MATH Intuitive Calculus Fall 2013 Circle lab day: Mon / Fri. Name:. Show all your work.

1 Numbers, Sets, Algebraic Expressions

NON-CALCULATOR: I. Decide whether or not the following information defines a function. Explain/support your answer x y

y = F (x) = x n + c dy/dx = F`(x) = f(x) = n x n-1 Given the derivative f(x), what is F(x)? (Integral, Anti-derivative or the Primitive function).

SECTION 5.1: Polynomials

MATH150-E01 Test #2 Summer 2016 Show all work. Name 1. Find an equation in slope-intercept form for the line through (4, 2) and (1, 3).

Math 1314 Final Exam Review. Year Profits (in millions of dollars)

Practice Questions for Math 131 Exam # 1

Online Math 1314 Final Exam Review

Mathematics 2 for Business Schools Topic 7: Application of Integration to Economics. Building Competence. Crossing Borders.

1.2 Graphs and Lines. Cartesian Coordinate System

Answer Key: Problem Set 1

Homework 1 Solutions

Bi-Variate Functions - ACTIVITES

*** Sorry...no solutions will be posted*** University of Toronto at Scarborough Department of Computer and Mathematical Sciences

Chapter 1 Linear Equations and Graphs

FACULTY OF ARTS AND SCIENCE University of Toronto FINAL EXAMINATIONS, APRIL 2012 MAT 133Y1Y Calculus and Linear Algebra for Commerce

Math 1314 Test 4 Review Lesson 16 Lesson Use Riemann sums with midpoints and 6 subdivisions to approximate the area between

Quadratic function and equations Quadratic function/equations, supply, demand, market equilibrium

Graphing Systems of Linear Equations

Finite Mathematics Chapter 1

Industrial Organization, Fall 2011: Midterm Exam Solutions and Comments Date: Wednesday October

Geometry 263 Prerequisites

LHS Algebra Pre-Test

MATH 2070 Mixed Practice KEY Sections (25) 900(.95 )

Given the table of values, determine the equation

3. (1.2.13, 19, 31) Find the given limit. If necessary, state that the limit does not exist.

Algebra 1. Predicting Patterns & Examining Experiments. Unit 5: Changing on a Plane Section 4: Try Without Angles

Consistent and Dependent

Advanced Microeconomic Analysis, Lecture 6

Simon Fraser University, Department of Economics, Econ 201, Prof. Karaivanov FINAL EXAM Answer key

Firms and returns to scale -1- Firms and returns to scale

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

Example 2: The demand function for a popular make of 12-speed bicycle is given by

Firms and returns to scale -1- John Riley

Systems of Equations and Inequalities. College Algebra

SCHOOL OF DISTANCE EDUCATION

Tutorial Letter 204/1/2016 DSC2606. Semester 2. Nonlinear mathematical programming. Department of Decision Sciences. Solutions: Assignment 04

Math 1325 Final Exam Review

Math 205 Final Exam 6:20p.m. 8:10p.m., Wednesday, Dec. 14, 2011

Contents CONTENTS 1. 1 Straight Lines and Linear Equations 1. 2 Systems of Equations 6. 3 Applications to Business Analysis 11.

Lecture 1: Introduction to IO Tom Holden

Sect 2.4 Linear Functions

Math Final Solutions - Spring Jaimos F Skriletz 1

Math 1101 Chapter 2 Review Solve the equation. 1) (y - 7) - (y + 2) = 4y A) B) D) C) ) 2 5 x x = 5

Statistics, Data Analysis, and Probability

Math Exam 3 Review

MAT 111 Final Exam Fall 2013 Name: If solving graphically, sketch a graph and label the solution.

Intermediate Algebra. 7.6 Quadratic Inequalities. Name. Problem Set 7.6 Solutions to Every Odd-Numbered Problem. Date

Maths for Economists Tutorial 0: Revision of Basic Concepts

Math 2003 Test D This part of the Exam is to be done without a calculator

MATHEMATICAL METHODS UNIT 1 CHAPTER 3 ALGEBRAIC FOUNDATIONS

EconS Vertical Integration

Chapter 2. Functions and Graphs. Section 1 Functions

1 Functions and Graphs

Linear Functions Answer Key. 3. n(t) is a linear function and n(0) = 1 and n(4) = n(0), find its equation and sketch its graph.

Functions of One Variable

Problem Set 1: Tariffs and Quotas (Solutions) Universidad Carlos III de Madrid Economics of European Integration TA: Victor Troster Fall 2012

MATH 236 ELAC FALL 2017 CA 9 NAME: SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

3. Find the slope of the tangent line to the curve given by 3x y e x+y = 1 + ln x at (1, 1).

Doug Clark The Learning Center 100 Student Success Center learningcenter.missouri.edu Overview

Math 141:512. Practice Exam 1 (extra credit) Due: February 6, 2019

Examiners: R. Grinnell Date: April 19, 2013 E. Moore Time: 9:00 am Duration: 3 hours. Read these instructions:

1,3. f x x f x x. Lim. Lim. Lim. Lim Lim. y 13x b b 10 b So the equation of the tangent line is y 13x

UNIVERSITY OF CALICUT SCHOOL OF DISTANCE EDUCATION B Sc. Mathematics (2011 Admission Onwards) II SEMESTER Complementary Course

Final Exam Review Packet

Final Exam Review Packet

Chapter 1. Functions, Graphs, and Limits

You solve inequalities the way you solve equations: Algebra Rule Equation Inequality 2x 5 = 3 2x 5 3. Add 5 to both sides 2x = 8 2x 8.

Tutorial letter 101/3/2012

Review for Final Review

1) Find the equations of lines (in point-slope form) passing through (-1,4) having the given characteristics:

Part I: Multiple Choice Questions

Lesson 26: Characterization of Parallel Lines

Mat 210 Business Calculus Final Exam Review Spring Final on April 28 in COOR HALL 199 at 7:30 AM

Dear Algebra 2A Students and Parents,

EXAMINATION #4 ANSWER KEY. I. Multiple choice (1)a. (2)e. (3)b. (4)b. (5)d. (6)c. (7)b. (8)b. (9)c. (10)b. (11)b.

The TransPacific agreement A good thing for VietNam?

Economics 101 Homework #1 Answer Key Spring Math Review a. The drawing is as follows:

A) (-1, -1, -2) B) No solution C) Infinite solutions D) (1, 1, 2) A) (6, 5, -3) B) No solution C) Infinite solutions D) (1, -3, -7)

June To Advanced Placement Calculus Student,

Lecture 6: Sections 2.2 and 2.3 Polynomial Functions, Quadratic Models

King Fahd University of Petroleum and Minerals Prep-Year Math Program Math (001) - Term 181 Recitation (1.1)

Key Concept Solutions of a Linear-Quadratic System

Index. Cambridge University Press An Introduction to Mathematics for Economics Akihito Asano. Index.

Second Order Derivatives. Background to Topic 6 Maximisation and Minimisation

13. Valuing Impacts from Observed Behavior: Slopes, Elasticities, and Data

SOLVING SYSTEMS OF EQUATIONS WITH EXAMPLES FROM ECONOMICS, CHEMISTRY, EDUCATION AND PSYCHOLOGY. Revised: Michael Lolkus 2018

Systems and Matrices CHAPTER 7

AP Calculus BC Summer Assignment

GLOSSARY. Accumulation function A function of the form a

2. Linear Programming Problem

Transcription:

DSC1520/201/2/2018 Tutorial letter 201/2/2018 Quantitative Modelling 1 DSC1520 Semester 2 Department of Decision Sciences Solutions to Assignment 1 Bar code

Dear Student This tutorial letter contains the solutions to the first compulsory assignment. Please contact me if you have any questions or need any help with the next assignment. Kind regards Dr Mabe-Madisa E-mail: mabemgv@unisa.ac.za 2

DSC1520/201/2 Question 1 Let (q 1 ; C 1 ) = (10;40) and (q 2 ; C 2 ) = (20; 70) The slope m is Substituting point (10 ; 40) in equation C = q + z C = q +z m = C 2 C 1 q 2 q 1 = 70 40 20 10 = 0 10 = 40 = 10+z z = 10. The equation of the line is C = q +10 and the cost of manufacturing 5 items is y = (5)+10 = 115 [Option 1] Question 2 The slope m is m = y 2 y 1 x 2 x 1 = 2 1 4 = 1 5 = 5 giving y = 5 x+c. To determine the y-intercept c of the line we use one of the given points (x 1 ; y 1 ) and (x 2 ; y 2 ), say ( ; 1) to find 1 = 9 5 +c which results in c = 14 5. The equation of the line is therefore y = 5 x+ 14. [Option 1] 5

Question According to the textbook, the price elasticity of demand is ε d = 1 b P Q, with a and b the values of the demand function P = a bq. The given demand function is P = 70 0,5Q. It is therefore clear that a = 70 and b = 0,5. The demand function is P = 70 0,5Q. To find Q in terms of P we need to write the demand function with Q as the subject, resulting in Q = P 70 0,5. We can now substitute b and Q into the formula for elasticity of demand to find ε d = 1 0,5 P P 70 0,5 1 0,5 P P 70 0,5 1 P = P 70. Or From the demand function P = a bq, we find that bq = P a. Therefore, price elasticity of demand For our demand function with a = 70, ε d = 1 b P Q = = P bq P P a. ε d = P P 70. (See page 89, equation 2.14) [Option ] 4

DSC1520/201/2 Question 4 x 2y +z = 11 (1) 2x z = 8 (2) y +z = 10 () Substitute equation (4) into equation (5): z = 2x 8 (4) y = 10 z (5) y = 10 (2x 8) Substitute equation (2) and equation (6) into equation (1): = 18 2x (6) Substitute x = into equation (4) and equation (6): x 2 (18 2x)+(2x 8) = 11 25 x = 25 x = z = 2 8 = 2 and y = 18 2 Therefore x = ;y = 4 and z = 2. The sum is 5. = 18 6 = 4 [Option 4] Question 5 2x+ 5 6 + x ( 2x 4 x ) 2 11 4 2x+ 5 6 + x ( 4x 2x+ 2 + 20 ) 4 x 2 4x 5 5 6 x 8x 0 5 6 6 5x 25 6 6 x 5 [Option 1] 5

O N Question 6! [Option 2] Question 7 The arc price elasticity of a demand function P = a bq between two prices P 1 and P 2 is arc elasticity of demand = 1 b P 1 +P 2 Q 1 +Q 2 with b the slope of the demand function and P 1,P 2 and Q 1,Q 2 the price and quantity demanded. P = 70 0,5Q 0,5Q = 70 P Q = 140 2P We can now determine Q 1 and Q 2 by substituting P 1 = 7 and P 2 = 5 into the equation. Therefore if P 1 = 7 then Q 1 = 140 2 7 = 126 and if P 2 = 5 then Q 2 = 140 2 5 = 10. elasticity of demand = 1 b P 1 +P 2 Q 1 +Q 2 = 1 0,5 7+5 126+10 = 1 0,5 12 256 = 12 128 = 0,0975,094; inelastic. [Option 5] Question 8 The price elasticity of demand measures the responsiveness (sensitivity) of quantity demanded to changes in the good s own price i.e The sensitivity of quantity demanded to change in price. [Option 2] 6

O +, N DSC1520/201/2 Question 9 9x+2,5y 150 x+ y 100 x,y 0. [Option 2] Question 10 PS = 1 2 base height = 1 2 20 (90 50) = 800 2 = 400. [Option ] Question 11 We need to determine the maximum value of the function P = 20x+0y subject to the given constraints. The corner points of the feasible region in the graph below are the points A, B, C, D and the origin (0;0). % ) *! Point A: Point A is the point where line (2) cuts the y-axis. This coordinates of the point can be read from the graph as (0;70). Point B: 7

Point B is the point where line (2) and () intersect. This coordinates of the point can be read from the graph as (20;60) or you can find them by calculation. x+2y = 140 x+y = 80 () (2) and x+2y 2y = 140 2y x = 140 2y Substitute the value of x namely x = 140 2y into equation () and solve y: x+y = 80 () (140 2y)+y = 80 y = 80 140 y = 60 = 60 Substitute the value of y = 60 into equation () and solve x: x+y = 80 x+60 = 80 x = 80 60 = 20 The coordinates of Point B are (20;60). Point C: Point C is the point where line (1) and () intersect. The coordinates of the point can be read from the graph as (40;40) or you can find them by calculation. 2x+y = 120 (1) and x+y = 80 () 2x+y 2x = 120 2x y = 120 2x (4) Substituting the value of y of equation (4) into equation () and solve for x: x+y = 80 () x+(120 2x) = 80 x = 80 120 = 40 = 40. Substitute the value of x = 40 into equation () and solve y: x+y = 80 10+y = 80 y = 80 40 = 40 8

DSC1520/201/2 The coordinates of Point C are (40;40). Point D: Point D is the point where line (1) cuts the x-axis. This coordinate of the point can be read from the graph as (60;0). Substitute the corner points of the feasible region into the objective function and determine the value of the objective function for each corner point: Corner points of feasible region Value of P = 20x + 0y A : x = 0 ;y = 70 P = 20(0) +0(70) = 2100 B : x = 20;y = 60 P = 20(20) +0(60) = 2200 Maximum C : x = 40;y = 40 P = 20(40) +0(40) = 2000 D : x = 60;y = 0 P = 20(60) +0(0) = 1200 Maximum of P is at point B where x = 20, y = 60 and P = 2 200. [Option 4] Question 12 Equilibrium is the price and quantity where the demand and supply functions are equal. P d = P s 100 0,5Q = 10+0,5Q 0,5Q 0,5Q = 10 100 Q = 90 1 = 90. To calculate the price at equilibrium, we substitute the value of Q into the demand or supply function and calculate P. P = 100 0,5(90) P = 100 45 P = 55 The equilibrium price is equal to 55 and the quantity is 90. [Option 1] 9

Question 1 Revenue or Income is defined as price times quantity or R = p q or p x. Now given is quantity as x and price is given as p(x) = 5 x 1000. Thus Revenue = p x = (5 x 1 000 ) x = 5x x2 1 000 = 5x ( 1 1 000 )x2 = 5x 0,001x 2. [Option ] Question 14 Let x = number produced Total cost = variable cost + fixed cost TC = 4x+64 = 4(200) +64 = 800+64 = 864 [Option 2] Question 15 We need to find the consumer surplus for demand function when the market price P = 16. P = 60 4Q From the textbook page 128, we know that the consumer surplus is calculated as CS = Amount willing to pay Amount actually paid. This is determined by calculating the area of the triangle P 0 E 0 a in the following graph which is given to with 1 2 base height = 1 2 Q 0(a P 0 ) P 0 the market price, Q 0 the number of units demanded at price P 0, and a the y-intercept of the demand function P = a bq. 10

2 2 DSC1520/201/2 O = N E I E J A H? A F J = B 2 = > ) H A = + 5 2 = > 2 - In general we can summarise the steps of determining the consumer surplus as follows: 1. Calculate Q 0 if P 0 is given. 2. Draw a rough graph of the demand function.. Read the value of a from the demand function, that is the y-intercept of the demand function. 4. Calculate the area of CS = 1 / 2 Q 0 (a P 0 ). First we need to determine Q from the demand function P = 60 4Q if P = 16. That is 16 = 60 4Q giving Q = 11. Draw the demand function by using the point (11;16) found before and (0; a)(0; 60). = $ 2 $ " $ 11

The value of a is found by substituting Q = 0 into the demand function, that is P = 50 bq = 50. The consumer surplus is the area of the shaded triangle in the sketch, that is CS = 1 2 base height = 1 2 11 (60 16) = 482 2 = 242. The consumer surplus is equal to 242 if the price P is equal to 16. [Option 1] 12