The officers of a high school senior class are planning to...

The officers of a high school senior class are planning to rent buses and vans for a class trip. Each bus can transport 56 students, requires 5 chaperones, and costs $1,100 to rent. Each van can transport 8 students, requires 1 chaperone, and costs $120 to rent. Since there are 672 students in the senior class that may be eligible to go on the trip, the officers must plan to accommodate at least 672 students. Since only 70 parents have volunteered to serve as chaperones, the officers must plan to use at most 70 chaperones. How many vehicles of each type should the officers rent in order to minimize the transportation costs? What are the minimal transportation costs? The officers should rent buses and vans to minimize the transportation costs.

Transcript

00:01 So we're trying to decide how many vans and buses to rent.

00:04 So we're going to have a, each bus can transport 35 students.

00:09 So let's say 35 students requires three chaperones and cost a thousand dollars.

00:18 So then we're going to have each van can transport seven students requires one chaperone and cost $90.

00:25 So there are 280 students and they have to have 30 parents for chaperones.

00:37 So how many of each vehicle should they rent to maximize to minimize the cost? so we're going to let x be the buses, y be the van.

00:48 So 35 x plus seven y equals 280, three x plus one y equals 30 and we want to minimize 1000 x plus 90 y.

00:59 So let's look at the first thing.

01:01 We can see where these two equations intersect.

01:05 So i'm going to go ahead and multiply this bottom one times negative seven.

01:10 So that's going to make that negative 21 x minus seven y equals negative 210.

01:17 So then that tells me i'm going to have 14 x equals two.

01:23 What's that going to be? 70.

01:24 So x equals to five.

01:27 If x equals to five, three times five plus y equals 30.

01:31 That means that we would need a y value of 15.

01:36 Now the reason i do that is because we're going to have to look at our graph and when we graph these, we're going to need to see what our points are that we're going to have to test out.

01:46 So one of those points is going to be the point of intersection 515.

01:50 So let's go ahead and graph the other two lines.

01:53 Let's get our intercepts here.

01:55 If x is zero, y is going to be 40.

01:59 So i'm going to put four, zero 40 and if y is zero, then x is going to be whatever 280 divided by 35 is that's going to be eight.

02:09 Okay...

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