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

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

3. 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)

5. 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 – m – 7n + 13
3x – 4y m – 8n – 21
Subtract the two expressions in each problem
7. x + 2y – m – 7n + 13
9. 10a – 17b + 24c; 13a + 14b – 16c

6. 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
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) [x + (3 – x) – (4 + 3x)]
15. -{5a – b – [3b –(c - 2b + a) – 4a] + c}
Carry out these multiplicative operations:
16. (5)(-4)(-2) (3ab)(2a)
18. (6a2b)(3ab2) (3xy2)(5x2y)(xy)
20. 3x2y(2xy2+ y) mn(3-5m + 6mn + 3n)
22. (a + 3b)(3a2 + 6ab + 4b2) (3a + 2)(a – 2)(2a + 1)

7. Multiplication and Division
Simplify each expression:
24. x2/x y3/y a8/a5
27. acz/ac a4/6a d4/3c5d3
a3b2/17a2b (9x2 – 6x3 – 3x4)/(-3x2)
Perform the indicated operations, giving the results in simplistic terms:

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

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

10. 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.

11. Illustration of Derivatives

12. Rules of Derivatives

13. 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)

14. Application of Linear Function: Revenue & Output
Total Revenue Output $
Questions:
Slope
Intercepts

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

17. 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

18. 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: c
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.

19. 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

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

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