Question

An auto manufacturer sends cars from two plants, I and II, to dealerships A and B, located in a mid-western city. Plant I has a total of 28 cars to send, and plant II has 8. Dealer A needs 20 cars, and dealer B needs 16. Transportation costs based on the distance of each dealership from each plant are $220 from I to A, $300 from I to B, $400 from II to A, and $180 from II to B. The manufacturer wants to limit transportation costs to $10,640. How many cars should be sent from each plant to each of the two dealerships? (Hints: Define each unknown value with a variable. There should be four variables in all

Answer 1

Plant I has a total of 28 cars
Plant II has 8

A needs 20 cars
B needs 16.

****************
Transportation costs (is this per car or just for the trip?)
220 from I to A,
300 from I to B,

400 from II to A,
180 from II to B.

The manufacturer wants to limit transportation costs to $10,640.

Answer 2

yes.can you help @amistre64

Answer 3

can you tell me what math this is for? linear algebra and matrixes by chance? or something else

Answer 4

linear equation

Answer 5

systems of equation

Answer 6

im thinking ......

if we divvy up the cars to the dealers we get

A = 20 = mi + nii
B = 16 = (28-m)i + (8-n)ii

if we calculate the cost per car

A = 220m + 400n
B = 300(28-m) + 180(8-n)

such that

220m + 400n + 300(28-m) + 180(8-n) <= 10640

Answer 7

220m + 400n + 300(28) -300m + 180(8) -180n <= 10640

320n -80m <= 10640 - 180(8) - 300(28)

4n - m <= 10

4n - 10 <= m

20 = (4n-10) + n

30 = 5n

n = 6

A = 20 = 14i + 6ii
B = 16 = 14i + 2ii

but this is just some rough thinking and prolly should be modified

Answer 8

i thought the question said there need to be four variables in all

Answer 9

thats what the hint said ... if youve got another way to figure this out, then by all means .....

Answer 10

please what is the answer now.is the answer 6?

Answer 11

lets see how this applies to price

220(14) from I to A,
300(6) from I to B,

400(14) from II to A,
180(2) from II to B.

gives me 10840 so its a little high, but close

Answer 12

i dont know what the answer is ... but i have an idea on how to go about it now

Answer 13

ok then please do help :-) waiting

Answer 14

to reduce the cost; lets adjust some values

220(15) from I to A,
300(6) from I to B,

400(13) from II to A,
180(2) from II to B. gets me to 10660 still a wee bit too high for 10640
---------------------------------

220(15) from I to A,
300(5) from I to B,

400(13) from II to A,
180(3) from II to B. gets me to 10540 which fulfills the costs portion 10640
--------------------------------------

so the question i have is .... does the cost have to be a minimum cost? or just less than or equal to 10640?

Answer 15

just less than or equal to i guez

Answer 16

220(20) from I to A,
300(0) from I to B,

400(8) from II to A,
180(8) from II to B. gets me to 9040

Answer 17

im mixing my data up ... so thats not correct either

Answer 18

lol.ok.i want to know how u getting the numbers in the bracket?

Answer 19

we have 28 cars from i to send; it costs 220(A) + 300(B) to divvy up 28 cars between A and B

we have 8 cars from ii to send; it costs 400(A) + 180(B) to divvy up 8 cars between A and B

Answer 20

A wants 20 total
B wants 16 total

if we go with just the cheapest stuff first:

A B
28: 220(20) + 300(8)
8: 400(0) + 180(8)
-------------------
20 16

Answer 21

thats a total charge of 8240

Answer 22

yep.so how many cars should be sent from each plant of the two dealerships

Answer 23

thats what the nifty setup tells us ....

A B
from i 28: 220(20i) + 300(8i)
from ii 8: 400(0ii) + 180(8ii)
-------------------
20 16

Answer 24

oh ok

Answer 25

are you gud in sloving a 3x3 larger system of equation

Answer 26

theres prolly a way to do it with 4 variables and whatnot ... but i just cant see it that way :/

3x3 is best solved with a matrix

Answer 27

should i send it over to you and then see to it for me?

Answer 28

@amistre64

Answer 29

you can post it right here and i will be able to view it ... i got no idea of i can solve it yet or not :)

Answer 30

dollars on his Mastercard, y dollars on his Visa, and z dollars on his American Express card. These amounts satisfy the following equations.

1.18x+1.15y+1.09z=11,244.25
3.54x-.55y+.27z=3,732.75
.12x+.10y+.06z=829.50

How much did the owner borrow on each card?

Answer 31

lol.i hope u can @amistre64

Answer 32

piece of cake :)

Answer 33

hahaha

Answer 34

rref{{1.18,1.15,1.09,11244.25},{3.54,-.55,.27,3732.75},{.12,.10,.06,829.50}}

http://www.wolframalpha.com/input/?i=rref%7B%7B1.18%2C1.15%2C1.09%2C11244.25%7D%2C%7B3.54%2C-.55%2C.27%2C3732.75%7D%2C%7B.12%2C.10%2C.06%2C829.50%7D%7D

Answer 35

the last column top to bottom is the values for xyz

Answer 36

if your intent is to do it by hand .... then no, i simply dont have time to do it :)