a school district needs to arrange transportation for at least 360 students for a field trip. the district…

a school district needs to arrange transportation for at least 360 students for a field trip. the district can use large buses or small buses. the large buses hold 60 students each and have an operating expense of $300 for the day. the small buses hold 45 students each and have an operating expense of $200 for the day. the district only has chaperones for, at most, seven buses. what is the minimum cost for the transportation for the field trip if each bus has \n$1,400\n$1,600\n$1,700\n$1,800

a school district needs to arrange transportation for at least 360 students for a field trip. the district can use large buses or small buses. the large buses hold 60 students each and have an operating expense of $300 for the day. the small buses hold 45 students each and have an operating expense of $200 for the day. the district only has chaperones for, at most, seven buses. what is the minimum cost for the transportation for the field trip if each bus has \n$1,400\n$1,600\n$1,700\n$1,800

Answer

Explanation:

Step1: Define variables

Let $x$ = number of large buses, $y$ = number of small buses.

Step2: List constraints

  1. Student capacity: $60x + 45y \geq 360$ (simplify to $4x + 3y \geq 24$)
  2. Bus limit: $x \leq 7$
  3. Non-negative integers: $x,y \in \mathbb{Z}_{\geq0}$

Step3: Define cost function

Cost $C = 300x + 200y$ (minimize this)

Step4: Test feasible bus combinations

  • Case 1: $x=6$ (max students: $660=360$) $606=360$ meets student need. Cost: $3006 + 2000 = 1800$
  • Case 2: $x=5$ $605=300$, need $360-300=60$ more students. Small buses needed: $\lceil\frac{60}{45}\rceil=2$ Cost: $3005 + 200*2 = 1500 + 400 = 1900$
  • Case 3: $x=4$ $604=240$, need $360-240=120$ more students. Small buses needed: $\lceil\frac{120}{45}\rceil=3$ Cost: $3004 + 200*3 = 1200 + 600 = 1800$
  • Case 4: $x=3$ $603=180$, need $360-180=180$ more students. Small buses needed: $\frac{180}{45}=4$ Cost: $3003 + 200*4 = 900 + 800 = 1700$
  • Case 5: $x=2$ $602=120$, need $360-120=240$ more students. Small buses needed: $\lceil\frac{240}{45}\rceil=6$ Cost: $3002 + 200*6 = 600 + 1200 = 1800$
  • Case 6: $x=1$ $601=60$, need $360-60=300$ more students. Small buses needed: $\lceil\frac{300}{45}\rceil=7$ Cost: $3001 + 200*7 = 300 + 1400 = 1700$
  • Case 7: $x=0$ Need $\lceil\frac{360}{45}\rceil=8$ small buses, but no chaperone limit violation, cost: $2008=1600$ Wait, recheck student capacity: $845=360$, which meets the requirement. No bus limit (limit is max 7 large buses, small buses have no stated limit).

Step5: Verify minimum cost

Compare all feasible costs: 1600, 1700, 1800, 1900. The lowest is 1600.

Answer:

$1,600$