Calculus
posted by Candy .
A box with an open top is to be made from a square piece of cardboard by cutting equal squares from the corners and turning up the sides. If the piece of cardboard measures 12 cm on the side, find the size of the squares that must be cut out to yield the maximum volume of the box.

Let the length of the (four) squares cut from the corner be x.
The box will then have dimensions 122x, 122x and x once the open box is made.
The volume is therefore:
V(x)=x(122x)²
Differentiate with respect to x and equate to zero to get the greatest voume:
V'(x)=(122x)²+2x(122x)(2)
=12(x²8)
Equate to zero and solve for x:
12(x²8)=0
=>
x=sqrt(8) 
Okay, so im not understanding how you went from V'(x)=(122x)²+2x(122x)(2) to V'(x)=12(x²8)... ive done this problem many times and am stuck on this part. i always end up with V'(x)= 12x²96x+144 => 12(x²8x+12)
so i end up with x=2 and x=6... which doesn't make sense because x=6 would make the base 0??? Help please!!! 
Trisha, your answer is correct; x=2 or 6. However this is where we have to use intuition and realize that x=6 is indeed extraneous (since a base of 0 will not be a box). Therefore x=2 is the answer!
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