SOLUTION: Dear Sir/Madam, I am confronted with the following problem: "A box with an open top is to be constructed by cutting 3-in squares from the corners of a rectangular sheet of tin

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: Dear Sir/Madam, I am confronted with the following problem: "A box with an open top is to be constructed by cutting 3-in squares from the corners of a rectangular sheet of tin      Log On


   



Question 6611: Dear Sir/Madam,
I am confronted with the following problem:
"A box with an open top is to be constructed by cutting 3-in squares from the corners of a rectangular sheet of tin, whose length is twice its width, and then folding up the sides. What size sheet of tin will produce a box having a volume of 60 cubic inches?"
I did the following:
1) Volume = l * w * h
l = 2w
h = 3
2) 60 = 2w * w * 3
3) 6w^2 = 60
4) w^2 = 10
5) w = +sqrt%2810%29+
But that is not correct. The right answer for the width is 8 inches. What did I do wrong?
Thanks in advance.
Regards,
-Mike

Answer by prince_abubu(198) About Me  (Show Source):
You can put this solution on YOUR website!
Alright. We're cool with V = l * w * h, right, and we know that h = 3 and the volume is 60 in^2. So far we've got

+60+=+l+%2A+w+%2A+3+

The trick is with the L and the W. You were right when you said that L = 2W. The length of the original sheet is twice the width of the original sheet BEFORE chopping of 3-inch by 3-inch squares on all four corners. When you chop off square corners, the actual length and width of the resulting folded box will definitely be shorter than the original L and W you started with.

Let's take the width. We decided to call it w. Now, when we cut off 3 inches from all 4 corners, that would actually chop 6 inches off the width. So the width of your box will have to be w - 6.

We know that the length = 2w BEFORE cutting off squares. Like the width, 3 inches were also chopped off the length from BOTH sides, making the actual length (of your box) 6 inches shorter. So the length of the box will be 2w - 6.

Let's now plug in the box's width and length:

+60+=+%282w+-+6%29%28w+-+6%29%283%29+ <----- We'll solve for the width eventually.

+20+=+%282w+-+6%29%28w+-+6%29+ <----- Divided both sides by 3

+20+=+2w%5E2+-+18x+%2B+36+ <----- Performed "FOIL"

+0+=+2w%5E2+-+18x+%2B+16+ <----- Subtract 20 from both sides.

+0+=+w%5E2+-+9x+%2B+8+ <---- Divide both sides by 2

+0+=+%28w+-+8%29%28w+-+1%29+ <---- "UnFOILED" or factored

Just by looking at the above equation, the width is either 8 or 1. We'll have to throw out the 1 because that measure for the width won't work if you'll have to chop 6 inches off it. The 8 would, though. If 8 is the width, and the length is twice the width, then the length is 16. So you've got a sheet that was 16" x 8".