Chapter 11.8, Problem 22AYU

Spring Break 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 special seat designed to accommodate travelers with disabilities. Each van has 8 regular seats and 3 special seats. The rental cost is $\$350$ for each van and $\$975$ for each bus. If 320 regular and 36 special seats are required for the trip, how many vehicles of each type should be rented to minimize cost?

Source: www.busrates.com

To determine

To solve: The given linear programming problem.

Answer to Problem 22AYU

Solution:

The minimum cost is $\$9350$, when renting 6 Buses and 10 vans.

Explanation of Solution

Given:

• Types of Seats in a Bus - 40 Regular and 1 Special.
• Types of Seats in a Van - 8 Regular and 3 Special.
• Rental Cost in a each Bus - $\$975$.
• Rental Cost in a each Van - $\$350$.
• Number of Regular Seats - 320.
• Number of Special Seats - 36.

Calculation:

Step 1:

Begin by assigning symbols for the two variables.

$x=$ Number of buses.

$y=$ Number of vans.

If $C$ is the total cost of renting the vehicles, then

$C=975x+350y$

Step 2:

The goal is to minimize $C$ subject to certain constraints on $x$ and $y$. Because $x$ and $y$ represents the number of vehicles, the only meaningful values of $x$ and $y$ are non-negative.

Therefore, $x\ge 0,y\ge 0$.

From the given data we get

$40x+8y\ge 320 =>5x+y\ge 40$

$x+3y\ge 36$

Therefore, the linear programming problem may be stated as

Minimize $P=975x+350y$

Subject to

$x\ge 0$

$y\ge 0$

$5x+y\ge 40$

$x+3y\ge 36$

Step 3:

The graph of the constraints is illustrated in the figure below.

Corner points are	Value of objective function
$\left(0,40\right)$	$\$14,000$
$\left(36,0\right)$	$\$35,100$
$\left(6,10\right)$	$\$9,350$

The minimum cost is $\$9350$, when renting 6 Buses and 10 vans.