The student activities department of a community college plans to rent buses and vans for a spring-break trip. Each bus has 40 regular seats and 1 handicapped seat; each van has 8 regular seats and 3 handicapped seats. The rental cost is $305 for each van and $985 for each bus. If 320 regular and 36 handicapped seats are required for the trip, how many vehicles of each type should be rented to minimize cost?
The number of buses =
The number of vans =
If there are x vans and y buses, then we want to
minimize c=305x+985y subject to
8x+40y >= 320
3x+y >= 36
use 10 vans, 6 buses
To minimize the cost, we need to find the number of buses and vans that meet the seating requirements at the lowest possible cost. Let's assume the number of buses as "x" and the number of vans as "y."
Given information:
- Each bus has 40 regular seats and 1 handicapped seat.
- Each van has 8 regular seats and 3 handicapped seats.
- The rental cost for each bus is $985.
- The rental cost for each van is $305.
- The total number of regular seats required is 320.
- The total number of handicapped seats required is 36.
Based on the seating requirements, we can write the following equations:
Equation 1: 40x + 8y = 320 (total regular seats required)
Equation 2: 1x + 3y = 36 (total handicapped seats required)
To find the number of buses (x) and vans (y) that minimize the cost, we can solve these equations.
Step 1: Rearrange Equation 1 by dividing both sides by 8 to eliminate y:
5x + y/4 = 40
Step 2: Multiply both sides of Equation 2 by 5 to eliminate x:
5x + 15y = 180
Step 3: Subtract Equation 1 from Equation 2 to isolate y:
(5x + 15y) - (5x + y/4) = 180 - 40
14y - y/4 = 140
(56y - y)/4 = 140
55y/4 = 140
Step 4: Multiply both sides by 4 and divide by 55 to solve for y:
55y/4 * 4 / 55 = 140 * 4 / 55
y = 140 * 4 / 55
y = 40/11
y ≈ 3.636 (approximate value)
Therefore, we should rent approximately 3 vans.
Step 5: Substitute the value of y into Equation 2 to find the value of x:
1x + 3(3.636) = 36
x + 10.909 = 36
x = 36 - 10.909
x = 25.091 (approximate value)
Therefore, we should rent approximately 25 buses.
So, the number of buses to be rented is approximately 25, and the number of vans to be rented is approximately 3.
To solve this problem, we need to find the number of buses and vans that should be rented to minimize the cost. We can set up a system of equations based on the given information.
Let's assume the number of buses as 'b' and the number of vans as 'v'.
Each bus has 40 regular seats, so the total number of regular seats from buses is 40b.
Each bus has 1 handicapped seat, so the total number of handicapped seats from buses is 1b.
Each van has 8 regular seats, so the total number of regular seats from vans is 8v.
Each van has 3 handicapped seats, so the total number of handicapped seats from vans is 3v.
According to the problem, the total number of regular seats required for the trip is 320, and the total number of handicapped seats required is 36.
So we can set up the following equations:
40b + 8v = 320 (equation 1)
1b + 3v = 36 (equation 2)
Now we can solve this system of equations to find the values of 'b' and 'v'.
To solve equation 1:
40b + 8v = 320
Divide by 8:
5b + v = 40 (equation 3)
To solve equation 2:
1b + 3v = 36
Multiply by 5:
5b + 15v = 180 (equation 4)
Now we have the following equations:
5b + v = 40 (equation 3)
5b + 15v = 180 (equation 4)
We can subtract equation 3 from equation 4 to eliminate 'b':
5b + 15v - (5b + v) = 180 - 40
14v = 140
Divide by 14:
v = 10
Now we can substitute the value of 'v' into equation 3 to find the value of 'b':
5b + v = 40
5b + 10 = 40
5b = 30
Divide by 5:
b = 6
Therefore, the optimal number of buses to rent is 6, and the optimal number of vans to rent is 10.