1. Linear Programming

2. What is Linear Programming?

3. Say you own a 500 square acre farm. On this farm you can grow wheat, barley, corn or some combination of the 3. You have a limited supply of fertilizer and pesticide, both of which are needed (in different quantities) for each crop grown. Let’s say wheat sells at $7 a bushel, barley is $3, and corn is $3.50. So, how many of each crop should you grow to maximize your profit?

4. What is Linear Programming? A mathematical tool for maximizing or minimizing a quantity (usually profit or cost of production), subject to certain constraints. Of all computations and decisions made by management in business, 50- 90% of those involve linear programming.

5. Background on Linear Programming As a field of mathematics, LP is still a small child (in math years) Developed by Leonid Kantorovich around the time of WWII Further developed over following decades Today, easily the most commonly used field for optimization Economics, business management, transportation, technology, planning, production, … the list goes on…

6. Maximizing Profit Problem where a limited number of resources are used to produce a combination of products to maximize profit from the sale Production of… Pretty much anything

7. Maximizing Problems consist of… 1.Resources 2.Products 3.Recipes 4.Profit 5.Objective

8. Setting Up A toy manufacturer can produce skateboards and dolls. Both require the precious resource of plastic, of which there are 60 units available. Skateboards take five units of plastic and make $1 profit. Dolls take two units of plastic and make $0.55 profit. The company wants to make at least 1 doll and at least 1 skateboard. What is the number of dolls and skateboards the company can produce to maximize profit?

9. Setting Up Mixture Problems First identify components of the problem: 1. Variables (Products) 2. Constraints (Resources and Recipes) 3. Profits 4. Objective – Maximize profit

10. Make Mixture Chart or Formulas 2 Groups of Equations: - Profit Equation (profit equation) - Constraint Inequalities With these, create Feasible Region Resources Plastic (60) Profit ProductsSkateboards (x units) 5$1.00 Dolls (y units) 2$0.55

11. Feasible Region – region which consists of all possible solution choices for a particular problem Using the constraint equation we get the following graph: Constraints: 5s + 2d ≤ 60 s > 0 d > 0

12. Corner Point Principle Which point is optimal? Corner Point Principle The maximal value always corresponds to a corner point

13. Corner Point Principle Plug in corner points to profit formula:

14. Quick Practice A clothing company has 100 yards of cloth and produces shirts (x units) and vests (y units). Shirts require 10 units and have profit value of $5, while vests require 4 units and have profit value of $4. What is the optimal production solution? Steps 3 & 4: Feasible Region & Corner Points ( 0, 25 ) ( 10, 0 ) ( 0, 0 ) PointCalculation of Profit Formula $5.00x + $4.00y = P (0, 0)$5.00 (0) + $4.00 (0) = $0.00 (0, 25)$5.00 (0) + $4.00 (25) = $100.00 (10, 0)$5.00 (10) + $4.00(0) = $50.00

15. Quick Practice What if the company decides to also put a “non-zero constraint” on all production? Must produce at least 3 shirts and 10 vests. Constraints become: 10x + 4y ≤ 100 … x ≥ 3 y ≥ 10 Feasible Region becomes: ( 3, 17.5 ) ( 6, 10 ) ( 3, 10 ) Corner Points: PointCalculation of Profit Formula $5.00x + $4.00y = P (3, 10)$5.00 (3) + $4.00 (10) = $55.00 (3, 17.5)$5.00 (3) + $4.00 (17) = $83.00 (6, 10)$5.00 (6) + $4.00(10) = $70.00 Great Job!

16. Linear Programming Find the minimum and maximum values by graphing the inequalities and finding the vertices of the polygon formed. Substitute the vertices into the function and find the largest and smallest values.

17. 6 4 2 2 3 4 3 1 1 5 5 7 8 y ≤ x + 3 y ≥ 2 1 ≤ x ≤5

18. Linear Programming The vertices of the quadrilateral formed are: (1, 2) (1, 4) (5, 2) (5, 8) Plug these points into the function P = 3x - 2y

19. Linear Programming P = 3x - 2y P(1, 2) = 3(1) - 2(2) = 3 - 4 = -1 P(1, 4) = 3(1) - 2(4) = 3 - 8 = -5 P(5, 2) = 3(5) - 2(2) = 15 - 4 = 11 P(5, 8) = 3(5) - 2(8) = 15 - 16 = -1

20. Linear Programming f(1, 4) = -5 minimum f(5, 2) = 11 maximum