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Number of internal frame packs (x) Number of external frame packs (y) 50x + 80y 4   8  12 16 x + 3y ≤ 18 2x + y ≤ 16 x ≥ 0 y ≥ 0 4    8   12  16
(0, 0)		50(0) + 80(0) = 0 (8, 0)		 50(8) + 80(0) = 400 (6, 4)		 50(6) + 80(4) = 620 (0, 6)		 50(0) + 80(6) = 480 6 internal frame packs, 4 external frame packs

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Alg2 lesson 3-4 day 2

  • 1. Number of internal frame packs (x) Number of external frame packs (y) 50x + 80y 4 8 12 16 x + 3y ≤ 18 2x + y ≤ 16 x ≥ 0 y ≥ 0 4 8 12 16
  • 2. (0, 0) 50(0) + 80(0) = 0 (8, 0) 50(8) + 80(0) = 400 (6, 4) 50(6) + 80(4) = 620 (0, 6) 50(0) + 80(6) = 480 6 internal frame packs, 4 external frame packs
  • 3. The pep club is selling two types of hamburgers as a fundraiser. The Junior Burger is made of two pieces of bread and one hamburger patty. The Deluxe Burger is made of two pieces of bread and two hamburger patties. They have purchased 40 hamburger patties and 50 pieces of bread to make their burgers. If they are charging $3 for a Junior Burger and $4 for a Deluxe burger, how many of each burger do they need to sell to maximize their profit?
  • 4. The pep club is selling two types of hamburgers as a fundraiser. The Junior Burger is made of two pieces of bread and one hamburger patty. The Deluxe Burger is made of two pieces of bread and two hamburger patties. They have purchased 40 hamburger patties and 50 pieces of bread to make their burgers. If they are charging $3 for a Junior Burger and $4 for a Deluxe burger, how many of each burger do they need to sell to maximize their profit? What are the variables? J = Number of Junior Burgers sold D = Number of Deluxe Burgers sold
  • 5. The pep club is selling two types of hamburgers as a fundraiser. The Junior Burger is made of two pieces of bread and one hamburger patty. The Deluxe Burger is made of two pieces of bread and two hamburger patties. They have purchased 40 hamburger patties and 50 pieces of bread to make their burgers. What are the limits on the resources? 1J + 2D ≤ 40 40 hamburger patties 50 pieces of bread 2J + 2D ≤ 50 2 pieces of bread each J = Number of Junior Burgers sold D = Number of Deluxe Burgers sold 1 hamburger patty for each 2 pieces of bread each 2 patties for each
  • 6. The pep club is selling two types of hamburgers as a fundraiser. The Junior Burger is made of two pieces of bread and one hamburger patty. The Deluxe Burger is made of two pieces of bread and two hamburger patties. They have purchased 40 hamburger patties and 50 pieces of bread to make their burgers. J ≥ 0 1J + 2D ≤ 40 D ≥ 0 2J + 2D ≤ 50 J = Number of Junior Burgers sold D = Number of Deluxe Burgers sold
  • 7. 1J + 2D ≤ 40 D 2J + 2D ≤ 50 J ≥ 0 D ≥ 0 Corner points: (0,0) (25, 0) (0, 20) (10, 15) 5 10 15 20 25 30 35 40 45 50 5 10 15 20 25 30 35 40 45 50 J
  • 8. If they are charging $3 for a Junior Burger and $4 for a Deluxe burger, how many of each burger do they need to sell to maximize their profit? Profit = $3 for each Junior burger + $4 for each Deluxe Burger profit = 3J + 4D J = Number of Junior Burgers sold D = Number of Deluxe Burgers sold
  • 9. Corner points: (0,0) (25, 0) (0, 20) (10, 15) profit= 3J + 4D J D 0 0 3(0) + 4(0)=0 0 3(25) + 4(0)=75 25 0 20 3(0) + 4(20)=80 15 3(10) + 4(15)=90 10 Maximum profit is $90 when 10 Junior Burgers and 15 Deluxe burgers are sold
  • 10. A landscaping company has crews who mow lawns and trim trees. The company schedules 1 hour for mowing jobs and 3 hours for trimming jobs. Each crew is scheduled for no more than 2 trimming jobs per day. Each crew’s schedule is set up for a maximum of 9 hours per day. On the average, the charge for mowing a lawn is $40 and the charge for tree trimming is $120. Find a combination of mowing lawns and trimming trees that will maximize the income the company receives per day from one of its crews. What are the variables? Lawns mowed and tree trimming jobs x = Lawns mowed y = Trim Jobs Example 4-3a
  • 11. A landscaping company has crews who mow lawns and trim trees. The company schedules 1 hour for mowing jobs and 3 hours for trimming jobs. Each crew is scheduled for no more than 2 trimming jobs per day. Each crew’s schedule is set up for a maximum of 9 hours per day. On the average, the charge for mowing a lawn is $40 and the charge for tree trimming is $120. Find a combination of mowing lawns and trimming trees that will maximize the income the company receives per day from one of its crews. What are the limits on resources? 9 hours work 1x + 3y ≤ 9 y ≤ 2 No more than 2 trim jobs 0 ≤ x ≥ 0 lawns mowed nonnegative 1 hour per lawn 3 hour per trim job x = Lawns mowed y = Trim Jobs Example 4-3a
  • 12. x ≥ 0 0 ≤ y ≤ 2 1x + 3y ≤ 9 Corner points: (0, 0) (9, 0) (0, 2) (3, 2) x = Lawns mowed y = Trim Jobs Example 4-3a
  • 13. A landscaping company has crews who mow lawns and trim trees. The company schedules 1 hour for mowing jobs and 3 hours for trimming jobs. Each crew is scheduled for no more than 2 trimming jobs per day. Each crew’s schedule is set up for a maximum of 9 hours per day. On the average, the charge for mowing a lawn is $40 and the charge for tree trimming is $120. Find a combination of mowing lawns and trimming trees that will maximize the income the company receives per day from one of its crews. What are we trying to maximize or minimize? Maximize the income Income = $ from mowing + $ from trimming Income = 40x + 120y $120 per trim $40 per lawn x = Lawns mowed y = Trim Jobs Example 4-3a
  • 14. x ≥ 0 y ≥ 0 y ≤ 2 1x + 3y ≤ 9 Maximize 40x + 120y x = Lawns mowed y = Trim Jobs The maximum profit will be $360 when 3 lawns are mowed and 2 tree trim jobs are done. Example 4-3a