Do Now Exercise Solve each linear system

Daily focus Solve linear programming problems.

[3.4] Linear Programming Linear Programming has nothing to do with computer programming. It’s related to solving of systems of linear inequalities. constraints These linear inequalities are called constraints on the system. x y feasible region The intersection of the inequalities established by these constraints is called the feasible region. objective function Our concern is with optimization... either maximizing or minimizing the results which come from another simple equation, the objective function. vertices You’ll see that the optimal positions for the objective function occur at the vertices set up by the constraints.

The Paradyne Photo Corporation advertises that each of their precision cameras are “produced by human hands… not machines”. Doris is trying to make a little extra money over the summer by assembling cameras for Paradyne. restraints Here are the restraints Doris has to work under: Paradyne has granted her a $ “assembler’s budget”. She can assemble either the Excelerio ® or the Premerio ®. The Excelerio ® requires $21 from her budget and takes 5 hours to build. The Premerio ® runs her $42, but it only takes 2 hours to build. She only has 100 hours that she can spare. Her profit in building an Excelerio ® is $8. Her profit in building an Premerio ® is $10. How many of each type of camera should Doris build to maximize her profit within those 100 hours.

M = $8·e + $10·p $1260 ≥ $21·e + $42·p 100 ≥ 5·e + 2·p constraint: constraint: objective function: $1260 ≥ $21·e + $42·p e-intercept(p = 0) $1260 ≥ $21·e + $42·(0) $1260 ≥ $21·e $21 60 ≥ e p-intercept(e = 0) $1260 ≥ $21·(0) + $42·p $1260 ≥ $42·p $42 30 ≥ p 100 ≥ 5·e + 2·p e-intercept(p = 0) 100 ≥ 5·e + 2·(0) 100 ≤ 5·e 5 20 ≥ e p-intercept(e = 0) 100 ≥ 5·(0) + 2·p 100 ≥ 2·p 2 50 ≥ p e p feasible region This is the feasible region. Each point in this region satisfies all of the restraints One of these vertices will show the maximum, while another will show the minimum. But which one? Critical vertices set up as (e, p). (0, 30) (10, 25) (20, 0) (0, 0)

Test each critical vertex into the objective function. (0, 0) (Build 0 Excelerios, build 0 Premerios) M = $8·e + $10·p M = $8·(0) + $10·(0) M = M = 0 (Doris makes $0) (0, 30) (Build 0 Excelerios, build 30 Premerios) M = $8·e + $10·p M = $8·(0) + $10·(30) M = M = 300 (Doris makes $300) (10, 25) (Build 10 Excelerios, build 25 Premerios) M = $8·e + $10·p M = $8·(10) + $10·(25) M = M = 330 (Doris makes $330) (20, 0) (Build 20 Excelerios, build 0 Premerios) M = $8·e + $10·p M = $8·(20) + $10·(0) M = M = 160 (Doris makes $160) $330 is the maximum amount Doris can make. $0 is the minimum she can make.

Example: Find the maximum value and the minimum vale of C = –x + 3y subject to the following restraints. x y (2, 0) (2, 8) (5, 2) (5, 0) Try them all: (2, 0)(2, 8) (5, 2)(5, 0) minimum maximum Notice how the feasible region is completely bounded on all sides. That doesn’t always happen. Sometimes one side is left open.

Example: Find the maximum value and the minimum vale of C = x + 5y subject to the following restraints. x y Try them all: (0, 2) (1, 4) minimum maximum (0, 2) 2 (2, 4) NO… You can see from the graph that they is no minimum value. This is an unbounded region.

Do this: Find the maximum value and the minimum vale of C = 2x – y subject to the following restraints. x Max Min

What is the name of the equation that actually gives you the solution? Why do you need to find the vertices of the feasible region when using linear programming? [3.4] 9-17(odd), 21, 25, 26