Optimising a School Fundraiser with Constraints
A student council sells two products at a fair: cupcakes and lemonade. Each cupcake earns $2 profit and each cup of lemonade earns $1.50. They have 6 hours of volunteer labour: each cupcake takes 4 minutes and each lemonade takes 2 minutes. They can store at most 150 items total, and suppliers cap cupcakes at 90 and lemonade at 120. They must sell at least 40 items overall to cover stall costs. The council wants to maximise profit and asks whether selling only the highest-profit item is best.
1. Set up the constraints and objective function, then identify the feasible region's corner points.
Read the worked answer
Let x = cupcakes, y = lemonades.
Objective: P = 2x + 1.5y (maximise).
Constraints:
- Labour: 4x + 2y ≤ 360 (6 h = 360 min) → 2x + y ≤ 180
- Storage: x + y ≤ 150
- Supply: x ≤ 90, y ≤ 120
- Minimum: x + y ≥ 40
- Non-negativity: x ≥ 0, y ≥ 0
Corner points from intersecting boundaries:
- (0, 120): supply cap y.
- (30, 120): from 2x + y = 180 with y = 120 → x = 30.
- (90, 0): x cap.
- (90, 60): x = 90 with x + y = 150.
- (60, 60): 2x + y = 180 with x + y = 150 → x = 30? Recheck: subtract gives x = 30, y = 120 - same as above. Instead use 2x + y = 180 with x = 90 → y = 0; with y = 120 → x = 30. And x + y = 150 with x = 90 → y = 60; with y = 120 → x = 30 (consistent).
Feasible corners: (40, 0), (90, 0), (90, 60), (30, 120), (0, 120), (0, 40).
2. Evaluate profit at each corner and determine whether the 'sell only the highest-profit item' strategy is optimal.
Read the worked answer
P = 2x + 1.5y:
- (40, 0): 80
- (90, 0): 180
- (90, 60): 180 + 90 = 270
- (30, 120): 60 + 180 = 240
- (0, 120): 180
- (0, 40): 60
Maximum is $270 at (90 cupcakes, 60 lemonades).
Selling only cupcakes (best single item at $2 profit) gives at most 90 cupcakes → $180, limited by supply. Selling only lemonade gives 120 → $180.
So the single-item strategy is not optimal: the mixed plan earns $270, 50% more than either single-item plan. The binding constraints at the optimum are the cupcake supply cap (x = 90) and storage cap (x + y = 150), with labour slack of 360 - (360+120) = wait: labour used = 4(90) + 2(60) = 360 + 120 = 480 min, which exceeds 360!
Recheck: 4(90) + 2(60) = 360 + 120 = 480 > 360. So (90, 60) is infeasible. Correct corner set must respect 2x + y ≤ 180: at x = 90, y ≤ 0. The feasible corners are (40, 0), (90, 0), (30, 120), (0, 120), (0, 40), plus intersection of x + y = 150 and 2x + y = 180 → x = 30, y = 120 (already listed).
Profits: (40,0)=80; (90,0)=180; (30,120)=60+180=240; (0,120)=180; (0,40)=60. Maximum $240 at 30 cupcakes and 120 lemonades. Labour used = 120 + 240 = 360 min (binding), storage = 150 (binding).
So the optimal plan sells mostly lemonade, not the higher-margin cupcake, because labour is the true bottleneck.