Question 22292: 1) In the past, the students who lived on the east side of the river went to East High, and those living on the west side went to West High. Since the city was spread out, some students needed to be bused to their schools.
Two things led to a need to change the way the students are assigned to schools:
West High has become overcrowded, while East has one extra room
Recently a federal court ordered River City to integrate the two schools. In particular they mandated that at least half of the students at East High school should come from the west side of the river
Here are some facts about the situation:
There are 300 high school students living on the east side of the river and 250 living on the west side
East High School can hold up to 350 students and West High can handle up to 225
The average cost for bussing per day is:
1. $1.20 for each east side student going to East High
2. $2.00 for each east side student going to West High
3. $3.00 for each west side student going to East High
4. $1.50 for each west side student going to West High
The problem facing the River City school board (and you) is to find out how many students to send to each school so that the bussing costs are minimized.
2) Mr. Goodfellow died and left his 300 acre farm to the city. Then the U.S. Army closed its base on the edge of town, giving 100 acres to the people. Finally there was 150 acres of mining land that was given back to the city which no resections on its use.
That was 550 acres of land that the city could use in any way it decided. The problem was that the city isn’t an “it”. A city contains many people who don’t always agree. The people in River City did not agree on how to use the 550 acres.
Essentially, there were two sides to the controversy, One side wanted as much of the land as possible for development; that is, for stores, businesses, and housing. The other side wanted to use as much of the land as possible for recreation. That is, they wanted park land, hiking trails, a wildlife preserve, and picnic areas.
The chamber of Commerce won an initial victory by getting the city counsel to agree that at least 300 acres would go for development. The chamber of Commerce also thought that the more attractive property of the army base and mining land should go to development, while any recreation land could come from Mr. Goodfellow’s property. But the Sierra Club felt that some of the more attractive land should go for recreation. The two groups finally came up with a two-part compromise.
o At most 200 acres of the army base and mining land could go for recreation
o The amount of army base land used for recreations plus the amount of land from Mr. Goodfellow used for development had to add up to exactly 100 acres
The city manager made a chart below to show for each parcel, how much each type of land use would cost the city
Parcel
Improvement costs
per acre for
recreation land Improvement costs
per acre for
development land
Goodfellow's $50 $500
Army Base $200 $2,000
Mining Land $100 $1,000
Everyone agreed that they wanted to keep the cost to River City at a minimum, while satisfying their needs. So the matter was turned over to the city manager. She was directed to decide how to slip the land use between development and recreation in a way that would minimize the cost to the city of the necessary improvements, at the same time making sure that at least 300 acres went for development and that the two-part compromise was followed. She turned the matter over to a consulting firm of city planners. Your group is to function as that consulting firm.
These problems, i have worked on them for two week and i still dont know how to get the set up equations. Please help me!!!!!!
Answer by wuwei96815(245)   (Show Source):
You can put this solution on YOUR website! 1) I don't think that you can minimize costs because you say that the number of students going to each school is mandated. However, you can figure out the costs.
350 students will attend East High. Half of them or 175 must come from the west. 200 students will attend West High. 125 will come from the east. These numbers should account for the total number of students which is 550.
$1.20 x 175 = $210 for East High students attending East High
$2.00 x 125 = $250 for East High students going to West High
$3.00 x 175 = $525 for West High students going to East High
$1.50 x 75 = $112.50 for West High students going to West High
Total cost for transporting all students is $210 + 250 + 525 + 112.50 = $1,097.50
I hope that I understood the problem correctly. Let me know.
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