o/t algebra help

trying to help my daughter with a homework problem. jeez i feel stupid. here it is. celia has an 18 inch by 18 inch piece of canvass. she wants to pint 3 rectangular pictures. the smallest one will be square, and the middle sized one will have twice the area of the small one. into what size pieces should she cut the canvass if she uses all of it?
we came up with the formula L squared plus 2L squared plus x = 324. where L is the length or width of the small square, 2Lsquared is the area of the middle sized one, and x is the third rectangle. i'm stumped. what am i doing wrong here????????/

You're not doing anything wrong. There's just not enough info given.

With two unknowns you need to be able to set up two equations which you can't do with the info given.

Glenn, just an idea but since they are to be rectangular in shape I figure the first one is L by L and the middle one is L by 2L. and lays along the top. That means one side of the canvas is 3L so 3L=18. That gives us L=6 so one of them is 6 by 6 and the middle one is 6 by 12. By elimination the 3rd one must be 12 by 18. When I try other numbers it does not allow the 3rd one to be the shape of a rectangle. I hope someone will add to my story if I overlooked something. Is this high School Algebra ? Barrett Jones

P.s. I really liked Algebra in school. If I didn't explain it good enough post back.

How about 6x6 for the square, 6X12 for the rectangle, the balance is 12X18. LXL+LX2L+2LX3L= 324. Lsq+2Lsq+6Lsq=324. 9Lsq=324. Lsq=324/9. Lsq=36.
Sq root of 36 is 6. L =6

If we stack the 6X12 on top of the 6X6, we end up using a space of 6X18. The balance is then 12X18.

b.jones is correct

thanks, i appreciate the help. been so long since i did algebra in school, i cant remember.

How about just go to the paint store and buy another piece of canvas.

OK, I realize it's a little late now, but here's how you solve it (as opposed to guessing).

First, consider how the canvas must be cut up. The only way that you can use all the canvas is to cut the two smaller pieces from the same side of the canvas, leaving the rest of the canvas for the same piece. Note that this means that the short side of the middle-sized piece is the same as the side of the small, square piece. So we only need to solve for the two smaller pieces.

So, let's say that:

The length of one side of the uncut canvas = a = 18
The length of the side of the small, square piece = b
The length of the long side of the middle-size piece = c
The length of the short side of the large piece = d

We know that the area of the mid-size piece is twice the area of the small piece:
2b^2 = bc

Dividing both sides of the equation by b, we have 2b - c = 0.
We also know that b + c = 18;
So we have two equations with two variables; we can solve that by adding the two equations:

2b - c = 0
b + c = 18
------------
b = 6

Solving for c, since b + c = 18, c = 12.

Checking our answers:

2b^2 = 2 * 6 * 6 = 72
bc = 6 * 12 = 72

Now to solve for d:

d = a - b = 18 - 6 = 12

So the dimensions are:

6 X 6
6 X 12
12 X 18