Algebra Word Problem help please.....

Hi Laurie! When approaching this problem, ask yourself, "What is it asking?" "What objects or shapes are we dealing with?"  

We're dealing with a piece of cardboard 10" x 18". What shape is that? That's how we talk about rectangles. Length x Width. So we start with a rectangle of cardboard 18" long and 10" wide.

Now, what's with "open top box" - what does that mean? A box is a 3-dimensional rectangular solid. An open top means what? This is poorly-worded problem because we have to guess at "open top". It probably means "no lid". Can you picture or draw a box with no lid? You're going to need to do so to solve the problem.

You are being asked to find the volume of a box. What is the formula for the volume of a box?

That's right - Box Volume = Length x Width x Height = l x w x h = V

Which box? A box made out of a piece of cardboard by cutting a square out of each corner. Can you draw this?

        18 "

|-------------------------|

|                               |  10"

|-------------------------|

(Note: I drew up this problem with MS Paint, but couldn't paste the images)

Let's keep building that box.

What is the shape of the pieces we are cutting away? Check the problem - squares!

How big are they? Check the problem - size x! What does that mean? Our rectangle takes two measurements, length and a width. How many measurements do you need to draw a square? Just one.

              ___ ?____

___x___ |               |__x____     (all the squares are x tall and x wide)

|                                         |  ?

----x-----                 ----x------                       |    

            |_________|   

What does this mean? This is our picture of what our cardboard looks like after we cut the squares off it.

It looks like a cross.

Can you imagine how to make an open top box out of this? Right, just fold the sides up. Then we'd have a box with no top.

Great, now we have a box, but what is the volume? What do we need to know about the box to find the volume? Check the earlier formula!  We need to know length, width, and height!

How long is the longer base of the box? Well, the cardboard rectangle was 18" long. Then we cut away two pieces that were x long. They were "x" inches long. If you have 18 and take away two x's, how do we write that? ..... that's right, 18 - 2x = L

How long is the shorter base of the box? Well, the cardboard rectangle was 10" wide. Then we cut away two pieces that were x long. If you have 10 and take away two x's, how do we write that? ..... that's right, 10 - 2x = W

Now, how tall is the box? Hmmm. Imagine folding up those cross pieces off the ground. They stick out a distance of x inches. When something flat and x inches long is bent straight up, how far up does it go?

That's right, x inches. The box will be x inches tall, that is, a height of x inches.

Now let's put all this together. We are looking for the volume of the folded box, and volume is length x width x height. That means the volume of the box is (18 - 2x)(10 - 2x)(x). If we multiply that out (remembering our foil method), we get V(x) = 180x - 56x^2 + 4x^3.

 The final part of the problem asks, "What is the real world domain for V(x)?" What is a domain? Remember? A domain is the set of values for which the function is defined. A real world domain is the set of values for which an expression or formula makes sense.

What values make sense for x? X is a measure of what, again? The size of the squares. We cut a square out of each corner. What are the biggest and smallest squares you can cut out of our cardboard?

Here's a question: Can a box have lengths less than zero? No! That's impossible. So we need values of x for which Length (18 - 2x), Width (10 - 2x), Height (x) are greater than zero.

What are they? Can you solve them?  1) 18 - 2x > 0    2) 10 - 2x > 0    and 3) x > 0

1) x < 9 inches.   Two nine inch squares use up all the length of the cardboard -> there's not enough board to get any longer!

2) x < 5 inches for width.   Two five inch squares use up all the width.

3) x > 0 inches.  You can't have a box without some height. x needs to be greater than 0 or we just have a flat piece of cardboard.

So, are we done? Hmmm. x should be greater than 0 but smaller than 5, or smaller than 9?

Which do we pick? Right. We have to pick x < 5, because it's impossible to go any bigger.

Thus, our real world domain of V(x) is between 0 and 5.

Hope this helps! Remember to:

1) Read the problem

2) Look up concepts and definitions

3) Write what you know.

4) Write what you are looking to solve or answer.

5) Draw a picture

6) Ask if it makes sense.

7) Are you done?

Hope this helps!

Clint, Wyzant Tutor.

You are given that the box is to be made from a piece of cardboard with the dimensions 10" X 18", which means that it is rectangular with a width of 10" and a length of 18". To make the box, you are to cut out a square shaped piece from each corner given that the size of the corner cutout is x, which I'm assuming is meant to say that each side of this square cutout is equal to x.

Start out by drawing a rectangle and label the width equal to 10" and the length equal to 18", then draw a square at each corner inside this rectangle. Since a square has equal sides, label the length and width of these square cutouts equal to x. 

After cutting out these square corners, we find that the new width (w) of the rectangle is 10" minus width of the 2 square cutouts; i.e.,  w = 10 - 2x . And the new length (l) of the rectangle is 18" minus the length of the 2 square cutouts; i.e.,  l = 18 - 2x . After folding up the sides, we find that the box made has a height (h) equal to the side of the square cutout; i.e.,  h = x .

The volume (V) of a rectangular box is given by the following formula:

          V = length · width · height = l · w · h

We found that the dimensions of the box are as follows:

          length = l = 18 - 2x

          width = w = 10 - 2x

         height = h = x

Thus, the volume of the box in terms of x is:

          V(x) = (18 - 2x)(10 - 2x)(x)

                 = (180 - 36x - 20x + 4x²)(x)

                 = (180 - 56x + 4x²)(x)

                 = 180x - 56x² + 4x³ 

The domain for V(x) is all the real values of x. It is obvious that to have a square at all, x must be greater than 0 (i.e., x > 0). And x must be less than half the width of the original rectangle; that is, x must be less than 5 (i.e., x < 5).....this is because if x were to be greater or equal to 5, then the width of the original rectangle would be less than or equal to 0 which can't be true. 

Thus, the domain of V(x) is as follows:

         0 < x < 5        

OR        

           (0, 5)