Checking a Popcorn Booth Plan Against Constraints
The robotics team runs a popcorn booth at a regional tournament. They sell small bags for $4 and large bags for $7. Each small bag takes 5 minutes of volunteer time and $1 of supplies; each large bag takes 8 minutes and $2 of supplies. The team has 25 volunteer hours and a $150 supplies budget. The team wants revenue of at least $420. A parent proposes making 60 small bags and 50 large bags. Before ordering supplies, the team needs to know whether the proposal is even possible, which constraint it violates, and what a realistic alternative looks like. They also want to know the largest revenue they could possibly collect if they sold every bag they can afford to make, ignoring time.
1. Define variables and write a system of inequalities that models the team's time, supplies, and revenue goal. Explain what each inequality represents.
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Let x = number of small bags and y = number of large bags. Both must be whole numbers, so x >= 0 and y >= 0, with x and y integers.
Time: each small bag takes 5 minutes and each large bag takes 8 minutes. With 25 volunteer hours = 25(60) = 1,500 minutes, the constraint is 5x + 8y <= 1500.
Supplies: each small bag costs $1 and each large bag costs $2, with a $150 budget, so x + 2y <= 150.
Revenue goal: small bags earn $4 and large bags earn $7, so 4x + 7y >= 420.
The first two inequalities are resource limits; the third is a minimum requirement, not a limit. A plan is feasible only if it satisfies all of them at once with nonnegative whole numbers.
2. Test the parent's proposal of 60 small bags and 50 large bags. Which constraints does it satisfy and which does it violate? Then adjust the plan so it satisfies every constraint.
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For x = 60, y = 50: Time: 5(60) + 8(50) = 300 + 400 = 700 minutes, which is well under 1,500 minutes. Satisfied. Supplies: 60 + 2(50) = 60 + 100 = 160 dollars, which exceeds the $150 budget by $10. Violated. Revenue: 4(60) + 7(50) = 240 + 350 = $590, which is at least $420. Satisfied.
So the proposal fails only the supplies constraint. It cannot be ordered as written.
Adjustment: reduce supplies by at least $10 while keeping revenue at least $420. Cutting 10 small bags gives x = 50, y = 50: supplies = 50 + 100 = 150, time = 250 + 400 = 650 minutes, revenue = 200 + 350 = $550. That is feasible.
A better adjustment is to cut 5 large bags instead: x = 60, y = 45. Supplies = 60 + 90 = 150, time = 300 + 360 = 660 minutes, revenue = 240 + 315 = $555. Both alternatives satisfy all constraints; the second preserves more revenue because it removes fewer revenue dollars per supply dollar saved.
3. Find the largest possible revenue if every bag made can be sold, and explain whether the time constraint limits that answer.
Read the worked answer
Compare revenue per supply dollar: small bags earn $4 for $1 of supplies, or 4 dollars of revenue per supply dollar; large bags earn $7 for $2, or 3.50 per supply dollar. Small bags are the better use of the budget, so the maximum revenue occurs when the entire $150 budget goes to small bags.
With x = 150 and y = 0: supplies = 150(1) + 0 = $150, revenue = 4(150) = $600, time = 5(150) = 750 minutes. The time limit is 1,500 minutes, so 750 minutes is used and 750 minutes remain unused.
The time constraint does not bind. The maximum revenue under the supplies budget is $600, achieved by making 150 small bags. If the team made any large bags, it would trade $2 of supplies for $7 of revenue instead of using that $2 to make two small bags for $8 of revenue, so revenue would fall. Since the best plan is feasible within the time limit, $600 is the maximum.