Mom in need of help

Hi. My name is Karen. My daughter is a senior in high school taking algebra. She has a problem she wants me to check. I have been looking at it and I don't think it is completely right. I was wondering if anyone had some time, if they could show me how to do this problem right, that way I can grade my daughters work to make sure she has it right. The probelm is:

An open-topped rectangular box is formed by cutting a 6-inch square from each corner of a rectangular piece of cardboard and bending up the ends and sides. The area of the cardboard before the corners are removed is 768 inches, and the volume of the box is 1,440 cubic inches. Find the dimensions of the origina piece of cardboard.

Now she has a picture of the cardboard drawn. And she has some steps. But I haven't taken math for quite some years and I don't use this as a stay at home mother. I think the steps look mixed up or she is missing some. Thanks for anyone who is willing to take the time to help! I really appreciate it!

Karen

mathmom00 said:

An open-topped rectangular box is formed by cutting a 6-inch square from each corner of a rectangular piece of cardboard and bending up the ends and sides. The area of the cardboard before the corners are removed is 768 inches, and the volume of the box is 1,440 cubic inches. Find the dimensions of the origina piece of cardboard.

Click to expand...

It would be good to know what steps have been taken.

Name Stuff!!

x = Width of cardboard
y = Length of cardboard

*** Equation #1 *** ==> x*y = 768 in²

When you fold up the sides, it makes the width of the box

x - 2*(6")

and the length

y - 2*(6")

This means

*** Equation #2 *** ==> [x - 2*(6")]*[y - 2*(6")]*(6") = 1440 in³

Can you solve those two equations simultaneously?

I'll give this one a shot without drawing a diagram, hopefully we can get by without one. Consider what the 6 inch square cutouts means. When you fold up the four sides of the box, how tall will the box be? It should be exactly 6 inches, because thats how "tall" each of the four flaps is after you cut out the 6 inch squares. This should be pretty clear from the diagram.

Now - What is the volume of a rectangular box like this? V = L*W*H (length times width times height). You know what the height is fortunately, because we just showed that the height is 6. Therefore the volume of the box is V = L*W*6. We were given the volume of the box, so plug it in:
V = L*W*6
1440 = L*W*6

Divide each side by 6:
1440/6 = L*W
240 = L*W


Now we've got one equation. Let's go back to the original sheet of cardboard. It had an original area of 768. We removed four corner pieces, each of which is 6x6. The area of each corner piece would be 6*6 = 36. Since there are four of them, the total area we cut out is 36*4 = 144. Subtract that from the area of the whole sheet and you find that the new area is 768 - 144 = 624. Now comes the tricky part... and I think I'll have to draw a picture here.


You can see that when the four flaps fold up, the indicated arrows are the L and W of the 3-dimensional box. The height is clearly 6 inches. That confirms that the volume of the box is L*W*6. Now we are trying to find the area, however. Basically we just have to add up the individual sections since we don't have a nice rectangle any more. We do know that the total area is 624 though (from above).



This picture shows how we can find the area of the rectangle. Divide it up into three sections, with the dimensions shown. Then just find the three areas:
A1 = 6*L
A2 = W(L+12)
A3 = 6*L

Now add the three areas together, and the total must be the 624 we calculated above.

6*L + 6*L + W(L+12) = 624
12L + WL + 12W = 624.

Now remember at the beginning of this post, we found another equation?
We decided that 240=L*W

Now you have two equations with two unknowns:
12L + WL + 12W = 624
240=L*W

You can solve for one of the variables in that second equation and then plug it into the first. Can you do this part on your own?

Hope this helps!

Ted

Just to be clear here... tkhunny solved this one in the opposite way that I did. He used x and y to be the dimensions of the rectangle and then subtracted the 6 inches and solved for the volume of the box. I used L and W for the dimensions of the box and added the 6 inches later when I had to find the area.

That probably sounded complicated... but it looks like his method was definitely faster... but either one will get you there. Let us know if I just confused you by doing it the hard way!

Ted

...and that is why my Rule #1 is to write down clear definitions. It's a lot harder to be confused if you have good definitions to start.

Well, here's another slightly different approach.

x = length of box, y = width of box; since height = 6, then:
6xy = 1440
xy = 240
x = 240 / y [1]

original piece of cardboard: length = x+12, width = y+12; then:
(x+12)(y+12) = 768
xy + 12x + 12y + 144 = 768
xy + 12x = 624 - 12y
x = (624 - 12y) / (y + 12) [2]

[1][2]: 240 / y = (624 - 12y) / (y + 12)
solve for y (you'll get a nice quadratic!)

Ok Ted, I am trying your approach, but I haven't done math in awhile. I want to make sure I have the right answer before I go or even look over her work.

12L + WL + 12W = 624
240=L x W


So I am trying to solve or L and W, right?

mathmom00 said:

Ok Ted, I am trying your approach, but I haven't done math in awhile. I want to make sure I have the right answer before I go or even look over her work.
12L + WL + 12W = 624 [1]
240=L x W [2]
So I am trying to solve or L and W, right?

Click to expand...

Correct, mom.
From [2]: L = 240/W; substitute that in [1]:
12(240/W) + W(240/W) + 12W = 624
This will lead you to:
W^2 - 32W + 240 = 0
(W - 20)(W - 12) = 0

All yours :wink:

Ok Denis I can get that far but once I have

(W - 20)(W - 12) = 0

what do I do to find the length and width or to complete the problem.

Man, I took so much math in school and now can't even do this. I feel like an old dumb hag. I always wondered what am I going to use this stuff for.......

now I know the answer.......... TO HELP MY KIDS

mathmom00 said:

Ok Denis I can get that far but once I have
(W - 20)(W - 12) = 0
what do I do to find the length and width or to complete the problem.

Click to expand...

That means:
W - 20 = 0 or W - 12 = 0; so:
W = 20 or W = 12

Also, we know that LW = 240; so L = 240/W; so:
L = 240/20 or L = 240/12
L = 12 or L = 20

So L=20,W=12 or L=12,W=20 : your choice (usually Length is longer than Width!)