Need an Application of Linear Equation

Question

An auto factory sends cars from plant 1 and 2 to dealership a and b. Plant 1 has 28 cars to send, Plant 2 has 8 to send. Dealer A needs 20 cars, Dealer B needs 16. Transportation costs are $220 from 1 to A, $300 from 1 to B, $400 from 2 to A, and $180 from 2 to B. Transportation costs are limited to $10640. How many cars should be sent from each plant to each of the dealerships?

Answer 1

I'll start you off with a few equations, hopefully you can go from there.

This first one is the total cost broken down by plant 1 cars to dealer a, etc:

220 p1da + 300 p1db + 400 p2da + 180 p2db = 10640

p1 has to send 28 cars:

p1da + p1db = 28

and similar for the other equations.

You could probably solve this one graphically too. As another hint, for p2db I got zero.

Answer 2

Let's let x = the number of cars sent to Dealer A from Plant 1, and
y = the number of cars sent to Dealer A from Plant 2.
Since 20 cars went to Dealer A, then we know x + y = 20

We also know that
(28 - x ) cars went to Dealer B from Plant 1, and (8 - y) cars went to Dealer B from Plant 2
Adding in the cost for transportation, we can build our second equation:

220x + (300)(28 - x) + 180(8-y) + 400y = 10640. Let's multiply a few things out

220x + 8400 - 300x + 1440 -180y + 400y = 10640

Using our first equation x+y = 20 and solving for y, y = 20-x. Substitute that into the second equation:

220x + 8400 - 300x + 144 - 180(20-x) + 400(20-x) = 10640

220x + 8400 - 300x+ 144 - 3600 + 180x + 8000 - 400x = 10640
Collect like terms
-300x + 14240 = 10640

-300x = -3600

x = 12

So there were 12 cars from Plant 1 to Dealer A, and

y = 20 - x = 8 cars from Plant 2 to dealer A

28 - x = 28 - 12 = 16 cards from Plant 1 to Dealer B, and
8 - y = 8-8 = 0 cars from Plant 2 to Dealer B.

 

Answer 3

Here's one way of solving this problem:

Plant 1 has 28 cars, x of which will be sent to dealership A, and the rest, 28-x, will be sent to dealership B.

Transportation cost for Plant 1 to Dealerships A and B will be 220(x) + 300(28-x)

Plant 2 has 8 cars, y of which will be sent to dealership A, and the rest, 8-y, will be sent to dealership B. 

Transportation cost for Plant 2 to Dealerships A and B will be 400(y) + 180(8-y)

This would make the total cost the sum of the costs for Plant 1 and Plant 2:

[220(x) + 300(28-x)] + [400(y) + 180(8-y)] = 10640

-80x + 220y +9840 = 10640

Now, since we have 2 variables, we must have 2 unique equations to solve for both of the variables. We have one equation above. Recall that Dealer A needs 20 cars. Plant 1 is sending x many cars to Dealer A, and Plant 2 is sending y many cars to Dealer A. This means x + y = 20. This is your second equation. 

Using elimination or substitution, you can now solve for x and y. Let's use substitution.

-80x + 220(20-x) +9840 = 10640

x = 12

y = 8

Plant 1 has 28 cars, 12 of which will be sent to dealership A, and the rest, 28-12, or 16, will be sent to dealership B. 

Plant 2 has 8 cars, 8 of which will be sent to dealership A, and the rest, 8-8, or 0, will be sent to dealership B.