SOLUTION: Can you help me?
I am working on a problem. The problem is :
From a 12cm by12cm piece of cardboard, square corners are cut out so that the sides can be folded up to make a
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Question 87115: Can you help me?
I am working on a problem. The problem is :
From a 12cm by12cm piece of cardboard, square corners are cut out so that the sides can be folded up to make a box. Express the volume of the box as a function of the length, x, in centimeters, of a cut-out- square and determine a reasonable domain for this function.
So far I have a table that looks like : ( the lines separate the columns
length - volume
20 - 0
10 - 100
8 - 128
6 - 108
4 - 64
2 -20
0 -0
The instructor says that I am on the right track, but I need to find out how the length and the volume relate to each other. I know that the domain is {10,8, 6, 4,2}. If you could help me figure out the equation it would be helpful, I don't see patterns well.
THANK YOU
Amanda
PS I need to turn it in by tomorrow thank you again
Found 2 solutions by scott8148, stanbon:
Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
I'm not sure how you generated the table but you should reread the problem and check your calculations
if x is the length (and width) of the cut-out squares, then the length (and width) of the resulting folded box is 12-2x ... the height of the folded box is x
the volume of a box is length times width times height ... v=(12-2x)(12-2x)x ... v=144x-48x^2+4x^3
the cardboard is 12x12 so the largest the cut-out squares can be is 6 ... so the domain is 0<=x<=6
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
From a 12cm by 12cm piece of cardboard, square corners are cut out so that the sides can be folded up to make a box. Express the volume of the box as a function of the length, x, in centimeters, of a cut-out- square and determine a reasonable domain for this function.
------------
Let an edge of the cutout square be x.
After sides are folded up the bottom of box is a 12-2x by 12-2x square.
The height of the box is x.
Volume = length*width*height
Volume = (12-2x)(12-2x)x
Volume = 4x(6-x)^2
Volume = 4x[36-12x+x^2)
Volume = 4x^3-48x+144x
Domain: Since the base is 12-2x by 12-2x a reasonable domain would be 0 < x < 12
=============
Cheers,
Stan H.
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