1 / 63

TX-1037 Mathematical Techniques for Managers

TX-1037 Mathematical Techniques for Managers . Dr Huw Owens Room B44 Sackville Street Building Telephone Number 65891 http://www.manchester.ac.uk/personal/staff/Huw.Owens. Introduction. Graph Theory Linear and quadratic equations Differentiation Integration Optimisation in management

manju
Download Presentation

TX-1037 Mathematical Techniques for Managers

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. TX-1037 Mathematical Techniques for Managers Dr Huw Owens Room B44 Sackville Street Building Telephone Number 65891 http://www.manchester.ac.uk/personal/staff/Huw.Owens

  2. Introduction • Graph Theory • Linear and quadratic equations • Differentiation • Integration • Optimisation in management • Matrix methods in management • Summation techniques

  3. Reading List • Budnick F, 1993, Applied mathematics for business, economics and social sciences, McGraw-Hill Education (ISE Editions). • Bostock and Chandler, 2000, Core A-level mathematics, Nelson Thornes. • Jacques I, 1999, Mathematics for economics and business, third edition, Addison-Wesley. • Jacques I, 2004, Mathematics for economics and business, fourth edition, Addison-Wesley. • Soper J, 2004, Mathematics for Economics and Business, An Interactive Introduction, second edition, Blackwell Publishing.

  4. Module specific learning outcomes • At the end of this module you should :- • be able to demonstrate familiarity with the basic rules of algebraic manipulations, matrix methods and applications for differentiation and integration; • have the ability to deal with unknown quantities; • have the ability to estimate order quantities, production planning skills and market forecasting etc; • have numerical skills transferable to any discipline. • Unseen exam paper worth 100% (10 credits)

  5. Module delivery • 24 hours of lectures • 76 hours of private study

  6. Lecture Outline • Monday, 30th January 2006 – Functions in Economics • Monday, 6th February 2006 – Equations in Economics • Monday, 13th February 2006 – Macroeconomic Models • Monday, 20nd February 2006 – Changes, Rates, Finance and Series • Monday, 27st February 2006 – Differentiation in Economics • Monday, 6th March 2006 – Maximum and Minimum Values • Monday, 13th March 2006 – Partial Differentiation • Monday, 20th March 2006 – Constrained Maxima and Minima • Monday, 27th March 2006 – Integration • Monday, 24th April 2006 – Integration • Monday, 1st May 2006 - Matrices • Monday, 8th May 2006 - Revision

  7. Graphs of Linear Equations - Objectives • Plot points on graph paper given their coordinates • Add, subtract, multiply and divide negative numbers • Sketch a line by finding the coordinates of two points on the line • Solve simultaneous linear equations graphically • Sketch a line by using its slope and intercept

  8. Graphs of linear equations • The horizontal solid line represents the x axis • The vertical solid line represents the y axis • O is the origin (0,0) P(x,y) y-axis or ordinate y x O or origin x-axis or abscissa

  9. Graphs of Linear Equations • How do we specify the coordinates?

  10. Graphs of Linear Equations – Rules for multiplying negative numbers • Negative * negative = positive • Negative * positive = negative • It does not matter in which order the two numbers are multiplied so • Positive * negative = negative • These rules produce • (-2)*(-3) = 6 • (-4)*5 = -20 • 7*(-5) = -35

  11. Fractions • Sometimes the functions economists use involve fractions. For example, ¼ of people’s income may be taken by the government in income tax. • Fraction: a part of a whole. • E.g. if a household spends 1/5 of its total weekly expenditure on housing, the share of housing in the household’s total weekly expenditure is 1/5. If the household’s total weekly expenditure is £250, the amount it spends on housing is one fifth of that amount. • Thus, amount spent on housing = share of housing*total weekly expenditure • =1/5*£250 = £50 • Ratio: one quantity divided by another quantity

  12. Fractions • Numerator: the value on the top of a fraction • Denominator: the value on the bottom of a fraction

  13. Fractions - cancelling • When working with fractions we can divide the top and bottom by the same amount to leave the fraction unchanged. • 10 is said to be a factor of both the numerator and the denominator, and can be cancelled.

  14. Fractions – common denominator • Which of these fractions is larger? 3/7 or 9/20 • In order to compare these fractions we need to find a common denominator. • This is the reverse operation to cancelling and leaves the value of the fraction unchanged. • > sign: the greater than sign indicates that the value on its left is greater than the value on its right. • < sign: the less than sign indicates that the value on its left is less than the value on its right

  15. Fractions – Addition and Subtraction • If fractions have the same denominators we can immediately add or subtract them. • If the denominators are not the same we must find a common denominator for the fractions before adding or subtracting them.

  16. Fractions – Multiplication and division • To multiply two fractions we multiply the numerators and the denominators. • To divide one fraction by another we turn the divisor upside down and multiply by it. (N.B. You can check that this work by seeing that the reverse operation of multiplication gets you back to the value you started with.)

  17. Graphs of Linear Equations • Division is a similar sort of operation to multiplication (it just undoes the result of the multiplication and takes you back to where you started) and the same rules apply when one number is divided by another.

  18. Graphs of Linear Equations • Evaluate the following: • 5*(-6) • (-1)*(-1) • (-50)/10 • (-5)/-1 • 2*(-1)*(-3)*6

  19. -3 -3 -2 -2 -1 -1 0 0 1 1 2 2 3 3 4 4 -4 -4 Graphs of Linear Equations • To add or subtract negative numbers it helps to think in terms of a picture of the axis: • If b is a positive number then a-b can be thought of as an instruction to start a and to move b units to the left. E.g. 1 - 3 = -2 • Similarly, -2 - 1 = -3

  20. -3 -2 -1 0 1 2 3 4 -4 Graphs of Linear Equations • On the other hand, a-(-b) is taken to be a+b. This follows from the rule for multiplying two negative numbers since -(-b)=(-1)*(-b) = b • Consequently, to evaluate a-(-b) you start at a and move b units to the right (the positive direction). For example (-2)-(-5)= -2+5 = 3

  21. Graphs of Linear Equations • Evaluate the following without using a calculator • 1-2 • -3-4 • 1-(-4) • -1-(-1) • -72-19 • -53-(-48)

  22. Multiplication and division involving 1 and 0 • When we multiply and divide by 1 the expression is unchanged, whereas if we multiply or divide by –1 the sign of the expression changes. • For example, try • y=-(6x3-15x2+x-1) • Each term is multiplied by –1, so now we have • y = =-6x3+15x2-x+1 • When we multiply by 0, the answer is 0 • Division divides a value into parts but if there is nothing to begin with the result of the division is 0. • For example, 0/4 = 0 division of zero gives zero • Division by 0 gives an infinitely large value if it is positive or infinitely small value if it is negative

  23. Graphs of linear Equations • But, in economics we would like to be able to sketch curves represented by equations, to deduce information. • Sometimes it is more appropriate to label axes using letters other than x and y. It is convention to use Q (Quantity) and P (Price) in the analysis of supply and demand. • We will restrict our attention to graphs of straight lines at this time.

  24. Point Check (1,3) -2*1+3 = -2+3 = 1 (0,1) -2*0+1 = 0+1 = 1 (-2,-3) -2*-2-3 = 4-3 = 1 (-3,-5) -2*-3-5 = 6-5 = 1 Graphs of Linear Equations • What do you notice about the points (2,5),(1,3),(0,1), (-2,-3) and (-3,-5)? • They all lie on a straight line with the equation • -2x+y=1 • If we substitute the values for x and y into the equation for the point (2,5) • -2*2+5=1 • We can check the remaining points similarly

  25. Graphs of Linear Equations • The general equation of a straight line takes the form • A multiple of x + a multiple of y = a number • dx + ey = f • for some given numbers d,e and f. Consequently, such an equation is called a linear equation. • The numbers d and e are called coefficients. The coefficients of the linear equation –2x+y = 1, are –2 and 1. • Check the points (-2,2),(-4,4),(5,-2),(2,0) all lie on the line 2x+3y = 4 and sketch this line. • In general to sketch a line from its mathematical equation , it is sufficient to calculate the coordinates of any two distinct points lying on it. The points can be plotted on paper and a ruler used to draw the line passing through them.

  26. Graphs of Linear Equations - Example • Sketch the line 4x+3y = 11 • For the first point, we could choose x=5. Substitution gives:- • 4*5+3*y=11 • 20+3y=11 • Now we need to determine y but how? • We could guess values of y using trial and error. • Actually, we only need to apply one simple rule • “You can apply whatever mathematical operation you like to an equation, provided that you do the same thing to both sides” • BUT there is one exception; never divide both sides by zero.

  27. Graphs of Linear Equations - Example • 20+3y = 11 • 20+3y –20=11-20 • 3y=-9 • 3y/3=-9/3 • y=-3 • Consequently the coordinates of one point on the line are (5,-3). • But we need two point to sketch the line. • If we choose x=-1 and substitute into the equation • 4*-1+3*y = 11 • 3y=11+4 • y=5, therefore the coordinates of the second point are (-1,5) • Usually we select x=0 and y=0

  28. Graphs of Linear Equations • Finding where two lines intersect • 4x+3y=11 • 2x+y=5 • y=1 • If y=1 • 4x+3=11 • 4x=11-3 • x=2

  29. Graphs of Linear Equations • It can be shown that provided e is non-zero any equation given by • dx+ey=f • Can be rearranged into the form • y=ax+b

  30. -3 -2 -1 0 1 2 -4 -3 -2 -1 0 1 2 3 4 4 -4 3 Graphs of Linear Equations Positive slope Zero slope Intercept Negative slope

  31. Graphs of linear Equations • Use the slope intercept approach to sketch the line • 2x + 3y = 12 • 3y=12-2x • y=4-2/3x 3 units 2 units

  32. Graphs of linear equations • Use the slope-intercept approach to sketch the lines • y=x+2 • 4x+2y=1

  33. Algebra • Algebra is boring!!!!!!! • In evaluating algebraic or arithmetic statements certain rules need to be observed about the various operations. • E.g. y = 10 + 6x2 • If x=3 • First substitute the value 3 for x and square it (9) • Multiply this by 6 (54) • Finally, add the result to the value 10 giving y = 64.

  34. The order of algebraic operation • Brackets - If there are brackets, do what is inside the brackets first • Exponentiation – exponentiation: raising to a power • Multiplication and division • Addition and subtraction • Acronymn (BEDMAS) • Remember – An expression in brackets immediately preceded or followed by a value implies that the whole expression in the brackets is to be multiplied by that value. E.g. y = (10+6)x2 • If x=3 then y = 144

  35. Order within an expression • Algebraic expressions are usually evaluated from left to right. • Addition or multiplication can occur in any order. • In subtraction and division the order is important • For example, • 8-6 = 2 but 6-8 = -2 • 8/4 = 2 but 4/8 = 1/2

  36. Algebraic solution of simultaneous linear equations - Objectives • Solve a system of two simultaneous linear equations with unknowns using elimination • Detect when a system of equations does not have a solution • Detect when a system of equations have infinitely many solutions • Solve a system of three linear equations in three unknowns using elimination

  37. The elimination method • Why use elimination? • The graphical method has several drawbacks • How do you decide suitable axes? • Accuracy of the graphical solution? • Complex problems with > three equations and > three unknowns?

  38. Example • 4x+3y = 11 (1) • 2x+y = 5 (2) • The coefficient of x in equation 1 is 4 and the coefficient of x in equation 2 is 2 • By multiplying equation 2 by 2 we get • 4x+2y = 10 (3) • Subtract equation 3 from equation 1 to get

  39. Example • If we substitute y=1 back into one of the original equations we can deduce the value of x. • If we substitute into equation 1 then • 4x+3(1)=11 • 4x=11-3 • 4x=8 • x=2 • To check this put substitute these values (2,1) back into one of the original equations • 2*2+1 = 5

  40. Summary of the method of elimination • Step 1 – Add/subtract a multiple of one equation to/from a multiple of the other to eliminate x. • Step 2- Solve the resulting equation for y. • Substitute the value of y into one of the original equations to deduce x. • Step 4 – Check that no mistakes have been made by substituting both x and y into the other original equation.

  41. Example involving fractions • Solve the system of equations • 3x+2y =1 (1) • -2x + y = 2 (2) • Solution • Step 1 - Set the x coefficients of the two equations to the same value. We can do this by multiplying the first equation by 2 and the second by 3 to give • 6x+4y = 2 (3) • -6x+3y = 6 (4) • Add equations 3 and 4 together to cancel the x coefficients • 7y = 8 • y=8/7 • Step three substitute y = 8/7 into one of the original equations • 3x+2*8/7=1

  42. Example • 3x=1-16/7 • 3x=-9/7 • x = -9/7*1/3 • x= -3/7 • The solution is therefore x= -3/7, y= 8/7 • Step 4 check using equation 2 • -2*(-3/7)+8/7 = 2 • 6/7+8/7 = 2 • 14/7 = 2 • 2=2

  43. Problems • 1) Solve the following using the elimination method • 3x-2y = 4 • x-2y =2 • 2) Solve the following using the elimination method • 3x+5y = 19 • -5x+2y = -11

  44. Special Cases • Solve the system of equations • x-2y = 1 • 2x-4y=-3 • The original system of equations does not have a solution. Why? • Solve the system of equations • 2x-4y = 1 • 5x-10y = 5/2 • This original system of equations does not have a unique solution

  45. Special Cases • There can be a unique solution, no solution or infinitely many solutions. We can detect this in Step 2. • If the equation resulting from elimination of x looks like the following then the equations have a unique solution • If the elimination of x looks like the following then the equations have no solutions Any non-zero number Any number * y = Any non-zero number * y = zero

  46. Special Cases • If the elimination of x looks like the following then the equations have infinitely many solutions zero * y = zero

  47. Elimination Strategy for three equations with three unknowns • Step 1 – Add/Subtract multiples of the first equation to/from multiples of the second and third equations to eliminate x. This produces a new system of the form • ?x + ?y + ?z = ? • ?y+?z = ? • ?y+?z =? • Step 2 – Add/subtract a multiple of the second equation to/from a multiple of the third to eliminate y. This produces a new system of the form • ?x + ?y + ?z = ? • ?y+?z = ? • ?z = ?

  48. Step 3 – Solve the last equation for z. Substitute the value of z into the second equation to deduce y. Finally, substitute the values of both y and z into the first equation to deduce x. • Step 4 – Check that no mistakes have been made by substituting the values of x,y and z into the original equations. • Example – Solve the equations • 4x+y+3z = 8 (1) • -2x+5y+z = 4 (2) • 3x+2y+4z = 9 (3) • Step 1 – To eliminate x from the second equation multiply it by 2 and then add to equation 1

  49. To eliminate x from the third equation we multiply equation 1 by 3, multiply equation 3 by 4 and subtract • Step 2 – To eliminate y from the new third equation (5) we multiply equation 4 by 5, multiply equation 5 by 11 and add • This gives us z = 1. Substitute back into equation 4. This gives us y = 1. • Finally substituting y=1 and z=1 into equation 1 gives the solution x=1, y=1, z=1 • Step 4 Check the original equations give • 4(1)+1+3(1) = 8 • -2(1)+5(1)+1=4 • 3(1)+2(1)+4(1)=9 • respectively

  50. Practice Problems • Sketch the following lines on the same diagram • 2x-3y=6 • 4x-6y=18 • x-3/2y=3 • Hence comment on the nature of the solutions of the following system of equations • A) • 2x-3y = 6 • x-3/2y=3 • B) • 4x-6y=18 • x-3/2y=3

More Related