x = number of cups the potter makes y = number of plates the potter makes

x = number of cups the potter makes y = number of plates the potter makes P = ($2/cup)(x cups) + ($1.50/plate)(y plates), so P = $2x + $1.50y Time constraint: time on cups + time on plates ≤ time available

x = number of cups the potter makes y = number of plates the potter makes P = ($2/cup)(x cups) + ($1.50/plate)(y plates), so P = $2x + $1.50y Time constraint: time on cups + time on plates ≤ time available (6 min./cup)(x cups) + (3min./plate)(y plates) ≤ (20 hrs.)(60 min./hr.) or 6x + 3y ≤ 1200 Clay constraint: clay for cups + clay for plates ≤ clay available

x = number of cups the potter makes y = number of plates the potter makes P = ($2/cup)(x cups) + ($1.50/plate)(y plates), so P = $2x + $1.50y Time constraint: time on cups + time on plates ≤ time available (6 min./cup)(x cups) + (3min./plate)(y plates) ≤ (20 hrs.)(60 min./hr.) or 6x + 3y ≤ 1200 Clay constraint: clay for cups + clay for plates ≤ clay available (3/4 lb. of clay/cup)(x cups) + (1 lb. of clay/plate)(y plates) ≤ 250 lbs. of clay.75x + y ≤ 250

x = number of cups the potter makes y = number of plates the potter makes P = ($2/cup)(x cups) + ($1.50/plate)(y plates), so P = $2x + $1.50y Time constraint: time on cups + time on plates ≤ time available (6 min./cup)(x cups) + (3min./plate)(y plates) ≤ (20 hrs.)(60 min./hr.) or 6x + 3y ≤ 1200 Clay constraint: clay for cups + clay for plates ≤ clay available (3/4 lb. of clay/cup)(x cups) + (1 lb. of clay/plate)(y plates) ≤ 250 lbs. of clay.75x + y ≤ 250 Non-negative constraints: x ≥ 0, y ≥ 0

Summary: x = number of cups the potter makes y = number of plates the potter makes The potter wants to maximize profit P = $2x + $1.50y Constraints: 6x + 3y ≤ x + y ≤ 250 x ≥ 0 y ≥ 0

400 x y y ≤ -2x y ≤ -3/4x +250 x ≥ 0 y ≥ (0, 250) (1000/3, 0) (0, 0) (?, ?)

Change from:To: 6x + 3y ≤ x + 3y = x + y ≤ 250 ¾ x + y = 250 Solving by Substitution 6x + 3y = 1200 ¾x + y = 250  y = -¾x x + 3( -¾x +250) = x + (- 9/4)x = 1200 (15/4)x = x = 1800 x = 120  6(120) + 3y = y = y = 480 y = 160

Change from:To: 6x + 3y ≤ x + 3y = x + y ≤ 250 ¾ x + y = 250 Solving by Elimination 6x + 3y = (¾x + y = 250) 6x + 3y = 1200 (-9/4)x – 3y = -750 (15/4)x + 0 = 450 (15/4)x = 450 x = 120  6(120) + 3y = y = y = 480 y = 160

400 x y y ≤ -2x y ≤ -3/4x +250 x ≥ 0 y ≥ (0, 250) (1000/3, 0) (0, 0) (120,160)

Corner Points (0, 0) : P = $2(0) + $1.50(0) = $0 + $0 = $0  P = $0 If zero cups and zero plates are made the profit will be $0. (0, 250) : P = $2(0) + $1.50(250) = $0 + $375  P = $375 If zero cups and 250 plates are made the profit will be $375. (1000/3, 0) : P = $2(333) + $1.50(0) = $666 + $0 = $666  P = $666 If 333 cups are made and 0 plates are made then the profit will be $666. (120, 160) : P = $2(120) + $1.50(160) = $240 + $240  P = $480 If 120 cups are made and 160 plates are made then the profit will be $480. For this problem the maximum should be found because we want the maximum profit. Maximum Profit!!