Red Bookstore wants to ship books from its warehouses in Brooklyn and Queens to its stores, one on Long Island and one in Manhattan. Its warehouse in Brooklyn has 1,000 books and its warehouse in Queens has 2,000. Each store orders 1,500 books. It costs $5 to ship each book from Brooklyn to Long Island and $1 to ship each book from Brooklyn to Manhattan. It costs $4 to ship each book from Queens to Long Island and $2 to ship each book from Queens to Manhattan.

(a) If Red has a transportation budget of $9,000 and is willing to spend all of it, how many books should Red ship from each warehouse to each store in order to fill all the orders? (Assume Red spends the entire transportation budget.)

Brooklyn to Long Island books

Brooklyn to Manhattan books

Queens to Long Island books

Queens to Manhattan books

(b) Is there a way of doing this for less money?

A. Yes

B. No

a)

Brooklyn to Long Island = 0 books

Brooklyn to Manhattan = 1000 books

Queens to Long Island books = 1500 books

Queens to Manhattan books = 500 books

b)

YES

Step-by-step explanation:

This is a equation's system with 4 unknown variables (the number of the 4 kinds of shippings performed), lets call the 4 variables like this:

a: Brooklyn to Long Island books b: Brooklyn to Manhattan books c: Queens to Long Island books d: Queens to Manhattan books

Now, let's see what we know:

1) First we know that each bookstore has asked for 1500 books so:

1500 = b+d = > b = 1500-d (1) 1500 = a+c=> a = 1500-c (2)

2) We also know that the total amount spent in transportation is 9000$ at the given individual trasportattion prices

9000$ = 5$*a+1$*b+4$*c+2$*d

=> 9000 = 5a+b+4c+2d

=> c = (9000-5a-b-2d) / 4 (3)

3) Finally, we know that the since we have a total number of books in the two warehouses of 3000 books, the same quantity that is demanded by the two bookstores, all the books are delivered.

So, in Brooklyn there are 1000 books that can be delivered to LI or Manhattan:

1000 = a+b = > a = 1000-b (4)

and in Queens there are 2000 books:

2000 = c+d = > d = 2000-c (5)

Now, formulas (2) and (4) are equal to a, so they are equal to each other

1000-b = 1500-c = > b = c-500 (6)

Now lets solve by substitution in formula (3):

c = (9000-5a-b-2d) / 4

=> 4c = 9000-5a-b-2d

=> 4c = 9000-5 (1500-c) - (c-500) - 2 * (2000-c)

=> 4c = - 3000+6c = > c=1500

Then:

d = 2000-c = 2000-1500=> d=500. b = c-500 = > b = 1000 a = 1500-c = > c = 0

b) If we look at the trsnportation prices the cheaper is from Brooklyn to Manhattan, so lets try to maximize this tranportation.

In Brooklyn we have 1000 books, and we need 1500 in Manhattan, so lets do all the the shipping from brooklyn to Manhattan.:

1000 books * 1 $ / book = 1000$

We will need 500 additional books from Queens at 2$ / book:

500 books * 2 $/book = 1000$

Then the 1500 books left in Queens will be shipped to Long Island at a price of 4$ / book:

1500 books * 4 $/book = 6000$

Total amount spent = 1000$+1000$+6000$ = 8000$