Budgeting for a Sari-Sari Store
Aling Nena runs a small sari-sari store. She has a starting capital of ₱5,000. She wants to buy two types of goods: rice (₱50 per kilo) and canned goods (₱30 per can). She has limited shelf space: at most 80 kilos of rice and at most 100 cans of canned goods. She also knows from experience that she can sell at most 60 kilos of rice and 80 cans of canned goods per week. She wants to maximize her weekly profit. The profit per kilo of rice is ₱10 and per can of canned goods is ₱5. She needs to decide how many kilos of rice and how many cans of canned goods to buy to maximize profit, given her capital and space constraints.
1. Formulate a linear programming model to determine the optimal number of kilos of rice and cans of canned goods Aling Nena should buy each week to maximize profit. Define variables, objective function, and constraints.
Read the worked answer
Let ( x ) = kilos of rice, ( y ) = cans of canned goods. Objective: Maximize profit ( P = 10x + 5y ) (in pesos). Constraints:
- Capital: ( 50x + 30y \leq 5000 )
- Shelf space for rice: ( x \leq 80 )
- Shelf space for canned goods: ( y \leq 100 )
- Demand: ( x \leq 60 ), ( y \leq 80 )
- Non-negativity: ( x \geq 0, y \geq 0 ) Combine space and demand: ( x \leq 60 ), ( y \leq 80 ).
2. Solve the linear programming problem graphically or by corner-point method to find the optimal quantities and maximum profit.
Read the worked answer
The feasible region is bounded by:
- ( x \leq 60 )
- ( y \leq 80 )
- ( 50x + 30y \leq 5000 )
- ( x \geq 0, y \geq 0 ) Find corner points:
- Intersection of ( x=60 ) and ( y=0 ): (60,0) → profit = 600
- Intersection of ( x=60 ) and ( 50(60)+30y=5000 ): ( 3000+30y=5000 ) → ( y=66.67 ) → (60,66.67) → profit = 10(60)+5(66.67)=600+333.35=933.35
- Intersection of ( y=80 ) and ( 50x+30(80)=5000 ): ( 50x+2400=5000 ) → ( x=52 ) → (52,80) → profit = 520+400=920
- Intersection of ( x=0 ) and ( y=80 ): (0,80) → profit = 400
- Intersection of ( x=0 ) and ( y=0 ): (0,0) → profit = 0 Maximum profit occurs at (60,66.67) with profit ≈ ₱933.35. Since cans must be whole, check integer points near: (60,66) profit=930; (60,67) profit=935 but check capital: 50(60)+30(67)=3000+2010=5010 >5000, so not feasible. (59,68): capital=2950+2040=4990 ≤5000, profit=590+340=930. So best integer solution: 60 kilos rice and 66 cans, profit ₱930.