Calculus
posted by Andrew on .
an open box is made by cutting out squares from the corners of a rectangular piece of cardboard and then turning up the sides. If the piece of cardboard is 12 cm by 24 cm, what are the dimensions of the box that has the largest volume made in this way?

Let each of the cutouts be x
then the base of the box is 24  2x
the width of the box is 122x
and its height is x
Volume = V = x(242x)(122x)
expand, differentiate and set equal to zero
you will be solving a quadratic. 
Would that make the dimensions x=6 and 12? or did I do that wrong?

How did you possible get these answers ?
They make no sense.
If x = 6, then you would be cutting away 12 cm from the base of width 12, so nothing is left.
x = 12 would be physically impossible, you would be cutting away 24 cm from the 12 cm width !
Did you even differentiate ?
Show me what you did. 
i got V= 4x^372x^2+288x
V=12x^2144x+288=0
Then I foiled that and my answers were x=6 and x=12. Clearly I have no idea what I'm doing, I have a test in my class on wednesday and I need to be able to make sense of this in my homework due tomorrow 
Your equation 12x^2144x+288=0 is correct
from there, divide each term by 12
x^2  12 + 24 = 0
This does not factor (FOIL is used to expand, not in factoring)
so ...
x = (12 ± √(144  4(1)(24))/2
= (12 ± √48)/2
= (12 ± 4√3)/2 = 6 ± 2√3
= aprr. 2.536 or 9.464
but clearly x < 6 or else the width would become negative, so
x = 2.536
length = 242(2.536) = 19.928
width = 12  2(2.536) = 6.928
height = 2.536 
x(242x)(182x)
(24x2x^2)(182x)
4x^384x^2+432
4x(x^221x+108)
4x(x12)(x9)
x=12,9,0
HELP! 
A machine produces open boxes using rectangular sheets of metal (12in by 20in). The machine cuts equalsized squares from each corner, and then shapes the metal into an open box by turning up the sides.
Express the volume of the box V(x) in cubic inches as a function of the length of the side of the square curs
Find the domain of the function V(x)