SOLUTION: A square piece of tin is made into an open box by cutting a 6 cm square from each corner. The volume of the box is 1536 cm^2. What was the area, in cm^2, of the original piece of t

Algebra ->  Geometry-proofs -> SOLUTION: A square piece of tin is made into an open box by cutting a 6 cm square from each corner. The volume of the box is 1536 cm^2. What was the area, in cm^2, of the original piece of t      Log On


   

Question 1102502: A square piece of tin is made into an open box by cutting a 6 cm square from each corner. The volume of the box is 1536 cm^2. What was the area, in cm^2, of the original piece of tin?
Found 3 solutions by josgarithmetic, ikleyn, greenestamps:
Answer by josgarithmetic(39626) About Me  (Show Source):
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Let x be the edge length of each square piece removed from a corner.

volume of the open box: cross%28x%28x-2%2A6%29%5E2=1536%29 6%28x%5E2-2%2A6%29%5E2=1536
6%28x-12%29%5E2=1536

(.....), %28x-12%29%5E2=1536%2F6

%28x-12%29%5E2=highlight%28256%29------the area of the base.

x-12=sqrt%28256%29
x-12=16
x=16%2B12
x=28
x%5E2=28%5E2=highlight_green%28784%29------area of original piece of tin




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A graphing tool is showing only a single root at x=20.63.----unneededandwrong


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(.....)

Answer by ikleyn(52858) About Me  (Show Source):
You can put this solution on YOUR website!
.
This problem is solved in four easy steps: 

    1.  The bottom area of the open box = Volume%2FHeight = 1536%2F6 = 256 cm^2.


    2.  The side measure of the box bottom = sqrt%28256%29 = 16 cm.


    3.  The side dimension of the original square piece of tin is 16 + 2*6 = 28 cm.


    4.  The area of the original piece of tin is 28^2 = 784 cm^2.

Answer. The area of the original piece of tin is 784 cm^2.


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Be aware !!   The approach and the solution by @josgarithmetic are   I N A D E Q U A T E   AND     I R R E L E V A N T !!



Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


The approach used by tutor josgarithmetic is in fact valid; she just didn't use it correctly.

The volume of the box, the way she set up the problem, should have been
6%28x-12%29%5E2

She mistakenly wrote it as
x%28x-12%29%5E2

Finishing the problem using this approach....

6%28x-12%29%5E2+=+1536
x-12%29%5E2+=+256
x-12+=+16 [we don't need to consider negative values, since we are talking about measurements of a piece of tin]
x+=+28

So the length of a side of the piece of tin is 28cm; the area is 28^2=784 cm^2.