Story Problem need help solving

Question

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 $8,400 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.)

Answer 1

First let's establish our variables:

 

Let B₁ = # of books shipping from Brooklyn to Long Island

 

Let B₂ = # of books shipping from Brooklyn to Manhattan

Let Q₁ = # of books shipping from Queens to Long Island

Let Q₂ = # of books shipping from Queens to Manhattan

 

 

So then, with a total of 1,000 books available to be shipped from Brooklyn, we therefore know that:

 

B1 + B2 = 1,000

 

 

And since, a total of 2,000 books are available to be shipped from Queens, we also know that:

 

Q1 + Q2 = 2,000

 

 

We also know that the total books shipped TO each location must be 1,500 each; therefore we have 2 more equations:

 

B1 + Q1 = 1,500     and     B2 + Q2 = 1,500

 

 

Now we just have one additional equation to write, so we can figure in the cost aspect here.  We know already that the total cost for shipping all of the books is going to be $8,400.

 

We also know that for each B1 book, the shipping costs $5 - so the total cost of all B1 books, to ship them all (however many that is going to wind up being) - will be 5 times the number of books shipped, or 5B1.

 

Applying that same idea to the other 3 variables, we get each of their respective total shipping costs as follows:

 

5B1

B2

4Q1

2Q2

 

(Note: "B2" in the above listing is simply written as "B2" because it is multiplied by $1.       1*B2 simply remains "B2" because multiplying by the 1 does not change the value.)

 

 

Adding all of these together, and knowing what the total must be, we get the following equation for the total cost of shipping:

 

5B1 + B2 + 4Q1 + 2Q2 = 8400

 

We also know that the total number of books shipped will be 3000, right?  So that will give us this equation:

 

B1 + B2 + Q1 + Q2 = 3000

 

 

We now find ourselves with a system of 6 equations with only 4 variables, so we ought to be more than able to solve the problem!  :)

 

 

First, let's solve our very first equation, B1 + B2 = 1000 in terms of B2.  Doing so, we arrive at: B1 = 1000 - B2.

 

Next, let us solve another of our equations in terms of B2 as well!  So, solving B2 + Q2 = 1500 (the only other of the small equations that contains B2), we get: Q2 = 1500 - B2.

 

 

Now we need a nice equation to begin to plug these into.  The idea is to start whittling down the number of variables in our equations until we can solve for one of the variables because once we have just one of these variables solved for, this will be a piece of cake!!

 

 

So, let's use our "total shipping" equation: B1 + B2 + Q1 + Q2 = 3000.

 

Substituting for B1 and Q2 as we just figured out, we will get:  (1000 - B2) + B2 + Q1 + (1500 - B2) = 3000.

 

Which simplified is: Q1 - B2 = 500

 

Now, let's substitute the same way into our "cost" equation: 5B1 + B2 + 4Q1 + 2Q2 = 8400 and see what happens!

 

Okay, substituting, we now get: 5(1000 - B2) + B2 + 4Q1 + 2(1500 - B2) = 8400.  

 

Distribute the constants and we get: 5000 - 5B2 + B2 + 4Q1 + 3000 - 2B2 = 8400.

 

Now simplify to get:  -6B2 + 4Q1 = 400

 

 

Alright!  Now we just have a system of 2 equations, in 2 variables:

 

Q1 - B2 = 500  

 

-6B2 + 4Q1 = 400 

 

So let's solve!

 

There are several ways we could do this.  (In fact there, are many ways to solve this problem, I'm quite sure!)

 

But, how about if we rewrite the first of these 2 equations in terms of B2, like this: Q1 = B2 + 500?

 

Then we are able to substitute again, putting "B2 + 500" IN the second equation in place of Q1, like this:

 

-6B2 + 4(B2 + 500) = 400

 

And all of a sudden, voila!!  We have a simple one-variable equation to solve!  :D

 

Distribute: -6B2 + 4B2 + 2000 = 400

 

Annnd, simplify to get: -2B2 = -1600

 

Divide by -2 for the value of B2: 800!

 

 

Now that we have the value of B2, it becomes so easy to solve the rest of the problem.

 

Plugging 800 into B1 + B2 = 1000 for B2 (I feel like I'm taking my vitamins), we get 200 for B1!

 

Plugging 800 into B2 + Q2 = 1500 for B2 again, we get 700 for Q2.

 

And finally, plugging 200 into B1 + Q1 = 1500 for B1, we find that Q1 = 1300!

 

 

So, we think we have found that:

B1 = 200

B2 = 800

Q1 = 1300 and

Q2 = 700

 

 

We have to check our work though!!  :)

 

Let's try out these values to make sure they meet all of our criteria:

 

*Total books shipped from Brooklyn should equal 1000, and B1 + B2 = 200 + 800 = 1000!  Check.  :)

 

*Total books shipped from Queens should equal 2000, and Q1 + Q2 = 1300 + 700 = 2000.  Yay!

 

*Total books shipped TO Long Island should equal 1500, and B1 + Q1 = 200 + 1300 = 1500.  Nice!

 

*Total books shipped TO Manhattan should equal 1500 also, and B2 + Q2 = 800 + 700, which also = 1500!  Awesome.

 

Now for our final criterion - the whole point almost of this whole word problem - they MUST add up to a total cost of $8,400!!!

 

So, we remember our cost equation, let's plug the values in!

 

5(200) + 800 + 4(1300) + 2(700) = 1000 + 800 + 5200 + 1400 (I REALLY hope this works!!  Lol)

 

Which equals...

8400

 

HOORAY!!  and hallelujah!  Hope that helps, and if anyone has a shorter or faster way, please do share!  I just offered the first method that came to  mind.