Math Solution

[guingel]

ok, here's what I ended up doing, after I converted everything to inches. (note: 36^3 is 36 cubed)

36^3in=384in(3in)x
46656=1152x
x=40.5in
40.5/12=3.375ft

however. the answer given at the back of the book is 2.56ft. So I must have done something wrong somewhere. "Let me check my notes."

garden is 6 feet by 10 feet. . . once you had your width, you'd be able to stretch the border into a rectangle that would have a length of 6+6+10+10. Or 32. to convert 32 into inches, you multiply it by 12. 32x12=384. Your length is 384.
27 cubic feet of cement. Your volume is 27 cubic feet. that would be 3ftx3ftx3ft. But to convert it into inches, each of those feet needs to be multipled times 12. 3x12=36. 36inx36inx36in=46656. Your volume is 46656.
(btw, I'm going through this piece by piece of the info they give you)
the border has a depth (aka height) of 3 inches. that's already inches.

The volume formula is lengthXwidthXheight=volume. I have 3 out of four of those variables. Let me plug them in.
384(x)(3)=46656 is my formula
I simplify it by multplying 384 times 3. 384(3)=1152
1152x=46656
To get rid of the 1152, I divide both sides by 1152 (well, that's the extended kinda way of doing it). 1152x/1152--the 1152's cancel leaving me with x on that side.
x=46656/1152
I divide 46656 by 1152. I get 40.5
x=40.5 inches.
But I want feet. So I divide 40.5 by 12.
40.5/12=3.375

what is going wrong here??
I'm going to go try it on a non-graphing calculator.

edit I experimented, plugging in their width and trying to find the length. according to that, your length should be 42.1875. now I'm all confused!

Comments

[renniekins.livejournal.com]

Just for fun...here's (approximately) what I did.

(surface area of border) * (depth of 3 inches) = (volume of 37 cubic feet or 46656 inches)

But what is the surface area of the border? If you draw a picture, it would look like a rectangle inside another rectangle, right? So the surface area of the border would be the surface area of the big rectangle minus the surface area of the little one. I tried it last night, I assumed the border would have pointed corners (be a rectangle). When I re-read the problem just now though, if it's going to be of uniform width it would actually have to have rounded corners.

Anyway, the border's perimeter has to be larger than the garden's perimeter, which is what concerns me about your answer. My theory is that you have to calculate the area of a trapaziod, not a rectangle.

But I could be complicating things unnecessarily!!! (: (:

[(Anonymous)]

heheh... you're right, technically speaking my way counts the corners twice. And again, you're right, it's complicating things unnecessarily :-). And if you knew the teacher you'd agree that it's not worth it at all.

Dave

Re:

[renniekins.livejournal.com]

Well....good thing I'm not getting graded then!!! heehee!

[guingel.livejournal.com]

yeah, like david said--it doesn't matter. She grades by effort, not content. (with homework, anyway)

[guingel.livejournal.com]

Anyway, the border's perimeter has to be larger than the garden's perimeter, which is what concerns me about your answer.

I'm pretty sure that's where I went wrong, because it turns out there's another way of doing this that actually gets you the right answer. . . lol. Which I will post for everyone's edification.

Re:

[renniekins.livejournal.com]

Cool, thanks for posting it! I had the vaguely right idea...the area of the border minus the area of the garden...one of my first approaches had the formula
(72 + 2x)(120 + 2x) = 24192
which I think is the inches version of your
168 = (2x + 6)(2x + 10)
but I totally messed up trying to solve it from there...couldn't remember enough math to figure what what x was!

But then I got all worried about the rounded corners, and started to try to come up with other formulas. Ah well -- it was an interesting diversion! (:

[guingel.livejournal.com]

hehe :) glad I could help!