Coffee Cart Batch Mix with an Extra-Bean Offer
A student-run coffee cart sells two cold-brew blends wholesale to campus cafes. Each batch of House Blend uses 2 kg of green coffee beans and 2 labor-hours; each batch of Espresso Blend uses 3 kg of beans and 1 labor-hour. The cart has 60 kg of beans and 40 labor-hours available for the production week. Existing wholesale orders require at least 6 batches of House Blend. Profit is $30 per House batch and $40 per Espresso batch. The student manager graphs the constraints and finds a feasible region, but must decide how many batches of each blend to make. A supplier offers an extra 6 kg of beans for $18, delivered in time for the same week, and the manager wonders whether the extra beans would pay for themselves. All batches must be whole numbers because the cart brews in fixed batch sizes.
1. Let x be House Blend batches and y be Espresso Blend batches. Write the constraint inequalities and find the corner points of the feasible region, ignoring whole-number requirements for now.
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Constraints: x ≥ 6, y ≥ 0, 2x + 3y ≤ 60 (beans), and 2x + y ≤ 40 (labor). The nonnegative and x ≥ 6 conditions give a lower-left boundary. Find intersections. At x = 6 with y = 0: (6, 0). At y = 0 with labor: 2x = 40 gives x = 20, and beans allow x ≤ 30, so (20, 0). Beans and labor intersect where 2x + 3y = 60 and 2x + y = 40. Subtract: 2y = 20, so y = 10, then 2x + 10 = 40, so x = 15. Point (15, 10). At x = 6 with beans: 12 + 3y = 60 gives y = 16; labor gives 12 + 16 = 28 ≤ 40, so (6, 16). The corner points are (6, 0), (20, 0), (15, 10), and (6, 16).
2. Determine the profit-maximizing whole-number batch plan and maximum weekly profit. Explain why testing only the corner points is sufficient here.
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Profit P = 30x + 40y. Evaluate corners: (6, 0): $180; (20, 0): $600; (15, 10): 30(15) + 40(10) = 450 + 400 = $850; (6, 16): 180 + 640 = $820. The maximum is $850 at 15 House and 10 Espresso batches. For a linear objective over a polygon, the maximum occurs at a corner unless the objective is flat along an edge, in which case endpoints share the maximum. Here the winning corner is already a whole-number point, so no integer rounding check is needed; 15 and 10 satisfy both constraints exactly at the bean and labor limits.
3. The supplier offers 6 extra kg of beans for $18. Recompute the best whole-number plan with 66 kg of beans, and decide whether the cart should buy the extra beans.
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With 66 kg, the labor constraint is unchanged. The continuous intersection is 2x + 3y = 66 and 2x + y = 40, so 2y = 26, y = 13, and 2x = 27, x = 13.5. Continuous profit would be 30(13.5) + 40(13) = 405 + 520 = $925, an increase of $75 over $850. But batches must be whole. Check nearby plans: x = 13, y = 13 uses 65 kg and 39 labor-hours, profit $390 + $520 = $910; x = 12, y = 14 uses 66 kg and 38 labor-hours, profit $360 + $560 = $920; x = 14, y = 12 uses 64 kg and 40 labor-hours, profit $420 + $480 = $900. The best whole-number plan is 12 House and 14 Espresso, profit $920. Net of the $18 bean cost, profit is $920 - $18 = $902, which is $52 better than $850. The cart should buy the extra beans.