Review of Key Algebraic Concepts and Applications to Economic Issues AGEC 317 Review of Key Algebraic Concepts and Applications to Economic Issues

Readings Review of fundamental algebra concepts (Consult any math textbook) Chapter 2, pp. 23-44, Managerial Economics

Topics Mathematical operations with algebraic expressions Addition, subtraction, multiplication, division Solving equations Linear functions Graphical analysis Slope Intercept Nonlinear functions Rate of change (marginal effects) Optimal points (minima, maxima)

Addition and Subtraction Express in simplistic terms 1. (2ac) + (-6ac) + (9ac) 2. 3x + (-7x) 3. (-8a) – (-3a) – (2a) 4. (5x) – (6x) – (7x) Add the two expressions in each problem 5. x + 2y – 8 6. 11m – 7n + 13 3x – 4y + 9 3m – 8n – 21 Subtract the two expressions in each problem 7. x + 2y – 8 8. 11m – 7n + 13 9. 10a – 17b + 24c; 13a + 14b – 16c

Addition, Subtraction, Multiplication Add the three expressions in each of the problems: 10. 7a – 3b + 11c; -14a + 10b + 10c; 8a + 8b + 13c Combine like terms: 11. 3x + 7y – 3z + 6xc – 8y – 7z + 5 -1 12. 9xy + 3x + 4a – 5ax + 10a – 7x + 3yx + 6xa Remove the symbols of grouping and simplify by combining terms: 13. 3a – (b + c) + (a + b – c) 14. -[x + (3 – x) – (4 + 3x)] 15. -{5a – b – [3b –(c - 2b + a) – 4a] + c} Carry out these multiplicative operations: 16. (5)(-4)(-2) 17. (3ab)(2a) 18. (6a2b)(3ab2) 19. (3xy2)(5x2y)(xy) 20. 3x2y(2xy2+ y) 21. -4mn(3-5m + 6mn + 3n) 22. (a + 3b)(3a2 + 6ab + 4b2) 23. (3a + 2)(a – 2)(2a + 1)

Multiplication and Division Simplify each expression: 24. x2/x2 25. -y3/y3 26. a8/a5 27. acz/ac 28. 12a4/6a 29. -93d4/3c5d3 30. 34a3b2/17a2b 31. (9x2 – 6x3 – 3x4)/(-3x2) Perform the indicated operations, giving the results in simplistic terms: 32. 33. 34. 35. 36. 37. 38. 39.

Division, Solving Equation(s) Simplify the complex fractions and other expressions: 40. 41. 42. 43. Perform the indicated operations, giving the results in simplistic terms: 44. 8x-15=3x 45. x – 4=5 + 2x 46. 47. (3x-1)(x+1)=3x2 Solve the following systems: 48. -4x + 5y + 14 = 0 49. 4xy – 2y – 11 = 0 -2x + 2y – 7 = 0 2xy – 3y – 4 = 0

Solving Equation(s), Complex fractions 50. 3x + y – 6 = 0 51. 4x – 5y = 9 4x – 3y – 21 = 0 3x – 4y = 8 Simplify the complex fractions: 52. 53. 54. 55.

Derivatives Y=f(x) The value of the ratio of for extremely small change in X. Derivative of Y with respect to X at point A is the slope of a line that is tangent to the curve at the point A.

Illustration of Derivatives

Rules of Derivatives

Linear Function: Y=aX+b Slope = dY/dX = a Interpretation: a one unit increase in X leads to an increase in Y of a units. Intercepts On x-axis; the value of X if y = 0: aX + b = 0, the x intercept is -b/a; put another way (-b/a, 0) On y-axis; the value of Y if x = 0: The y intercept is b; put another way (0,b) Graph Y = -2X + 2, x intercept is 1 or (1,0); y intercept is 2 or (0,2) Y = 2X + 4, x intercept is -2 or (-2,0); y intercept if 4 or (0,4)

Application of Linear Function: Revenue & Output Total Revenue Output $1.50 1 3.00 2 4.50 3 6.00 4 7.50 5 9.00 6 Questions: Slope Intercepts

Nonlinear function: Total, Marginal, and Average Profits Curvilinear expression of profit and output

Locating Maximum and Minimum Values of a Function Step 1: Find the derivative of the function with respect to the “independent” variable. For example, suppose that profit ( π ) = a – bQ + cQ2 Then the derivative (marginal profit) = -b + 2cQ The “independent” variable is Q and the “dependent” variable is π Step 2: Set the derivative expression from step 1 to 0 (first-order condition) so, -b + 2cQ = 0 Step 3: Find the value of the “independent” variable that solves the derivative expression -b + 2cQ = 0 2cQ = b Q = b/2c

Locating Maximum and Minimum Values of a Function (Con’t) Step 4: How to discern whether the value(s) from step 3 correspond to minimum values of the function or maximum values of the function: Calculate the second derivative with respect to the “independent” variable first derivative: -b + 2cQ second derivative: 2c If the second derivative at the value of the “independent” variable that solves the first derivative expression (step 3) is positive, then that value of the “independent” variable corresponds to a minimum. If the second derivative is negative at this point, then that value of the “independent” variable corresponds to a maximum.

Locating Maximum and Minimum Values of a Function (con’t) Step 5: Finding the Maximum of Minimum Value of the Function Simply replace the “optimum” value of the “independent” variable into the function = a – bQ + b/2c from step 3, Q = b/2c from step 4, if c > 0, then Q = b/2c corresponds to a minimum value if c < 0, then Q = b/2c corresponds to a maximum value. The minimum (maximum) value of then is

Locating Maximum and Minimum Values of a Function Profit Locating Maximum and Minimum Values of a Function

Summary of Algebraic Review Mathematical operations with algebraic expressions Solving equations Linear functions Nonlinear functions Applications to revenue and profit functions