3-6: Linear Programming Linear programming is a process of finding a maximum or minimum of a function by using coordinates of the polygon formed by the graph of the constraints.

What is a constraint? A restriction... A boundary… A limitation…

What is the feasible region? The feasible region is the area of the graph in which all the constraints are met.

Example 1 A farmer has 25 days to plant cotton and soybeans. The cotton can be planted at a rate of 9 acres per day, and the soybeans can be planted at a rate of 12 acres a day. The farmer has 275 acres available. If the profit for soybeans is $18 per acre and the profit for cotton is $25 per acre, how many acres of each crop should be planted to maximize profits?

Step 1: Define the variables What are the unknown values? Let c = number of acres of cotton to plant Let s = number of acres of soybeans to plant

Step 2: Write a System of Inequalities Write the constraints. What are the limitations given in the problem? The number of acres planted in cotton must be greater than or equal to 0. The number of acres planted in soybeans must be greater than or equal to 0. The total number of acres planted must be less than or equal to 275. The time available for planting must be less than or equal to 25 days.

Step 3: Graph the Inequalities c s The purple area is the feasible region.

Step 4: Name the Vertices of the Feasible Region Find the coordinates of the vertices of the feasible region, the area inside the polygon. (0,275) (225,0) (0,0) (75,200)

Step 5: Write an Equation to be Maximized or Minimized p(c,s) = 25c + 18s Maximum profit = $25 times the number of acres of cotton planted + $18 times the number of acres of soybeans planted.

Step 6: Substitute the Coordinates into the Equation Substitute the coordinates of the vertices into the maximum profit equation.

Step 7: Find the Maximum (c,s) 25c + 18s f(c,s) (0,275) 25(0) + 18(275) 4950 (225,0) 25(225) + 18(0) 5625 (0,0) 25(0) + 18(0) (75,200) 25(75) + 18(200) 5475 225 acres of cotton and 0 acres of soybeans should be planted for a maximum profit of $5,625.

Example 2: The Bethlehem Steel Mill can convert steel into girders and rods. The mill can produce at most 100 units of steel a day. At least 20 girders and at least 60 rods are required daily by regular customers. If the profit on a girder is $8 and the profit on a rod is $6, how many units of each type of steel should the mill produce each day to maximize the profits?

Step 1: Define the Variables Let x = number of girders Let y = number of rods Step 2: Write a System of Inequalities At least 20 girders are required daily. At least 60 rods are required daily. The mill can produce at most 100 units of steel a day.

Step 3: Graph the Inequalities

Step 3: Graph the Inequalities

Step 3: Graph the Inequalities The purple region represents the feasible region.

Step 4: Name the Vertices of the Feasible Region Find the coordinates of the vertices of the feasible region, the area inside the polygon. (20, 60) (20, 80) (40, 60)

Step 5: Write an Equation to be Maximized or Minimized p(x,y) = 8x + 6y Maximum profit = $8 times the number of girders produced + $6 times the number of rods produced

Step 6: Substitute the Coordinates into the Equation Substitute the coordinates of the vertices into the maximum profit equation.

Step 7: Find the Maximum 40 girders and 60 rods of steel should be produced for a maximum profit of $680.

Explore… Suppose the mill produces 150 units of steel a day, how does this change the feasible region? 2. Can the mill increase profits with the new feasible region?

Try these: A theater, at which a drug abuse program is being presented, seats 150 people. The proceeds will be donated to a local drug information center. Admission is $2 for adults and $1 for students. Every two adults must bring at least one student. How many adults and students should attend in order to raise the maximum amount of money?

The available parking area of a parking lot is 600 square meters The available parking area of a parking lot is 600 square meters. A car requires 6 square meters of space and a bus requires 30 square meters of space. The attendant can handle no more than 60 vehicles. If a car is charged $3.00 to park and a bus is charged $8.00, how many of each should the attendant accept to maximize income?