1. Agricultural Economic Students and Faculty!
What: Annual Department COOKOUT!
Where: Lawn of Agriculture Administration Building (AGAD)
When: Thursday, August 23rd
Time: 4:30pm until the food is gone!
Sponsored by
Free Food & Games!

2. Announcements Do not go to scheduled lab locations.
Labs will be virtual.
Labs will start next week.
Dr. Keeney will go over lab instructions on Monday.
Key to Monday’s level exam is on course website.

3. Lecture 1: Basics of Math and Economics
AGEC 352
Fall 2012– August 22
A. Remble

4. Basic Algebra Number of unknowns Number of equations
What relationship between these two is required to solve for the unknowns?

5. Applied Algebra 20 acres of land 40 hours of labor
Planting requirements
Corn = 1 acre of land; 2 hours of labor
Soybean = 1 acre of land; 2 hours of labor
What’s an algebraic description of my situation assuming I will use all resources and plant some combination of these two crops?

6. Applied Algebra Let C = planted corn Let B = planted soybeans Land
C + B = 20
Labor
2C + 2B = 40
Can we ‘solve’ this?

7. ‘Solving’ C + B = 20 2C + 2B = 40 C = 20 – B (rewrite 1st equation)
Substitute into 2nd equation
2*(20-B) + 2B = 40
40 – 2B + 2B = 40
40 + 2B = 2B + 40???

8. ‘Solving’ C + B = 20 2C + 2B = 40 ½*(2C + 2B = 40) => C + B = 20
We can divide the second equation by 2 without changing the relationship
½*(2C + 2B = 40) => C + B = 20
The 2nd equation provides no ‘different’ information about my planting problem
Tradeoffs between the two crops
Limits I face in my planting

9. ‘Solution’? 40 + 2B = 2B + 40 E.g. : set B = 50 => C = -30
Any value works for B
Once you plug in a choice for B, then you just need to set the value of C to make sure the equation B + C = 20 holds
E.g. : set B = 50 => C = -30
But, we might not want a value of planted acres that is < 0 so we could change our problem

10. ‘Solution’? Choose values of B and C with
1) B + C = 20
II) B >=0, C>=0
These are called non-negativity conditions
Then B will be some choice on the interval [0,20]
C = 20 – B
What then is your solution to this problem?

11. Graphical ‘Solution’
Any combination that appears on the line connecting (0,20) and (20,0) is a legitimate ‘solution’.

12. Economic information Corn Net Returns/acre Soybean Net Returns/acre
$100
Soybean Net Returns/acre
$50
First, how do we represent this information mathematically?

13. Back to algebra
The equation should describe total net returns, so let’s call that R.
Every acre of corn is $100 so that gives
100*C
Every acre of soybeans is $50
50*B
R = 100C + 50B

14. Collecting our mathematical information we have…
C + B =20
R = 100C + 50B
That’s two equations for 3 variables
We’re no better off algebraically with the new information

15. But, we know the solution if…
We are willing to assume that the operator with limited land and labor wants to maximize net returns
Step 1: Compare the net returns between the 2 crops (C > B)
Step 2: Choose to produce as much of that crop as is feasible (C = 20)
Step 3: If resources remain, use those for the other crop (B = 0)

16. Optimization The optimization assumption takes care of R for us
Says, 1st find B and C that will make R bigger than any other value it can have
Then, calculate R at the end
C = 20, B =0
R = 2,000

17. Simple with two choices
When we have a large number of choices this gets more complicated
Spreadsheet modeling
Formulate the model on paper
Input it correctly into a spreadsheet
Solve
Graphical methods
Algorithm methods
Understand, interpret, and communicate the final solution

18. In general
Most of the work in this course is in developing the mathematical representation of the problem
Identifying objectives that decision makers use as optimization criteria
Identifying choices available to the decision maker that adjust their objective
Identifying the limits decision makers control in adjusting their objective

19. Next Class
Monday: review some of the stuff from the 2nd page of the level exam
Calculus, elasticity, optimization
Review virtual lab procedures