Optimising a School Fundraiser: Profit Maximisation with Constraints
A school committee is planning a fundraiser selling two types of items: small and large. Each small item costs $2 to produce and sells for $5. Each large item costs $4 to produce and sells for $9. The committee has a budget of $400 for production and can store at most 150 items in total. They also want to sell at least 20 large items to attract customers. Due to supplier limits, they can produce at most 100 small items. The committee wants to determine how many of each item to produce to maximise profit, assuming all items sell.
1. Formulate this as a linear programming problem, clearly defining variables, objective function, and constraints. Then find the maximum profit and the number of each item that yields it.
Read the worked answer
Let (x) be the number of small items and (y) be the number of large items.
Objective: Maximise profit (P = (5-2)x + (9-4)y = 3x + 5y).
Constraints:
- Production cost: (2x + 4y \le 400) (or (x + 2y \le 200)).
- Storage: (x + y \le 150).
- At least 20 large: (y \ge 20).
- Supplier limit: (x \le 100).
- Non-negativity: (x \ge 0), (y \ge 0).
Graph the feasible region. The vertices are intersections of boundary lines:
- (x=0), (y=20): (P=100).
- (x=100), (y=20): check storage: (100+20=120\le150); cost: (2(100)+4(20)=280\le400). (P=300+100=400).
- (x=100), (y=50) (from storage: (100+y=150)): cost (200+200=400). (P=300+250=550).
- Intersection of (x+2y=200) and (x+y=150): subtract: (y=50), (x=100). Same as above.
- Intersection of (x+2y=200) and (y=20): (x=160), but (x\le100) so not feasible.
- (x=0), (y=100) (from cost: (2y=200)): (P=500).
- (x=0), (y=150) (from storage): cost (600>400) not feasible.
Check vertex (0,100): (P=5(100)=500). Vertex (100,50): (P=550). Vertex (0,20): (P=100). Vertex (100,20): (P=400).
Thus maximum profit is $550 at (x=100), (y=50).
2. Suppose the committee can negotiate a higher storage limit. Determine the minimum increase in storage capacity needed to achieve a profit of at least $600, assuming the production budget remains $400 and other constraints unchanged.
Read the worked answer
We need (P=3x+5y \ge 600) with (2x+4y \le 400) (i.e., (x+2y \le 200)), (x \le 100), (y \ge 20), and new storage (x+y \le S).
To maximise profit for given (S), we check vertices. The cost constraint (x+2y=200) and storage (x+y=S) intersect at (y=200-S), (x=2S-200). For this to be feasible, need (x \le 100) and (y \ge 20).
At that intersection, (P=3(2S-200)+5(200-S)=6S-600+1000-5S=S+400). So to get (P\ge600), need (S\ge200). But check feasibility: if (S=200), then (y=0), which violates (y\ge20). So we need another vertex.
Consider the vertex where (x=100) and (x+2y=200) gives (y=50), (P=550). To increase profit, we need to relax storage. With (x=100), storage constraint (100+y \le S) gives (y \le S-100). The cost constraint gives (y \le 50). So (y) can be at most 50 unless we also relax cost? Wait, if (x=100), cost gives (100+2y\le200 \Rightarrow y\le50). So storage does not bind at (x=100) if (S\ge150). Actually at (S=150), (y\le50) from cost, so storage not binding. To get higher profit, we need to increase (x) beyond 100? But (x\le100). So the only way to increase profit is to increase (y) while keeping (x=100), but (y) is limited by cost to 50. So profit cannot exceed 550 with (x\le100) and cost constraint? Let's check: with (x=100), (P=300+5y), and (y\le50), so max (P=550). If (x<100), can we get higher? Suppose (x=0), then (y\le100) from cost, (P=500). If (x=50), cost: (50+2y\le200 \Rightarrow y\le75), (P=150+375=525). So indeed max profit without storage constraint (only cost, (x\le100), (y\ge20)) is 550. So storage constraint is not binding at optimum. Therefore, increasing storage alone cannot achieve profit $600; we need to also increase the production budget. The question asks for minimum increase in storage capacity needed to achieve profit at least $600, assuming production budget remains $400. But with budget $400, max profit is 550, so it's impossible regardless of storage. Thus no amount of storage increase will suffice. The answer: impossible; storage increase alone cannot achieve $600 because the production budget limits profit to $550. To achieve $600, the budget must be increased. (If we misinterpreted and storage is the only change, then answer is 'no finite increase'.)