Question 767651: A rectangular piece of tin is 3 cm longer than its width. From each corner, 3 cm^2 is cut out. The sides are turned up to form a box whose volume is 120 cm^3. Find the dimensions of the box.
Found 2 solutions by Cromlix, solver91311:
Answer by Cromlix(4381)   (Show Source):
You can put this solution on YOUR website! Width = x
Length = 3 + x
Removal of 3cm^2 from each corner.
Width = x - 6
Length = 3 + x - 6 = x - 3
Volume = length*width*height
120 = (x - 3)*(x - 6)*(3)
120 = 3x^2 - 27x + 54
Combine
3x^2 - 27x + 54 - 120 = 0
3x^2 - 27x - 66 = 0
3(x^2 - 9x - 22) = 0
3(x - 11)(x + 2) = 0
x + 2 = 0
x = -2 (no answer x<0)
x - 11 = 0
x = 11
Width = 5cm
Length = 8cm
Height = 3cm
Hope this helps.
:-)
Answer by solver91311(24713)  (Show Source):
You can put this solution on YOUR website!
I'm going to do this problem just like you stated it, even though I'm almost certain that you wrote it incorrectly.
The only way this works is if squares were cut out of the corners, so if each of the squares has an area of  , then the sides of the squares must be  .
Then, when you bend up the sides, the depth of the box is then  . Dividing the given volume by we get:
Then if the length is 3 cm greater than the width for the original piece of tin then the dimensions of the bottom of the box must be:
and
So the area of the bottom of the box as a function of the original width is:
(w\ +\ 3\ -\ 2\sqrt{3}\ =\ w^2\ +\ (3\ -\ 4\sqrt{3})w\ + (12\ -\ 6\sqrt{3}))
Setting this equal to the value for the area of the bottom of the box that we calculated earlier, and arranging it into standard quadratic equation form:
w\ + (12\ -\ 6\sqrt{3})\ -\ 40\sqrt{3}\ =\ 0)
Then using the quadratic formula and discarding the negative root, in the end you get:
and then the length is just 3 cm longer so:
Verification of the use of the quadratic formula is left as an exercise for the student.
Now what I suspect you meant in the beginning was that the squares cut out of the corners measure 3 cm on each side. That is an entirely different kettle of fish, but since it is indeed not the question you asked, I'm not going to answer that one. The lesson here is to make sure that you are precise in your communication -- say what you mean and mean what you say.
John
Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it
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