Posted by Andrew on Monday, July 18, 2011 at 1:42pm.
Let each of the cut-outs be x
then the base of the box is 24 - 2x
the width of the box is 12-2x
and its height is x
Volume = V = x(24-2x)(12-2x)
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^3-72x^2+288x
V=12x^2-144x+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^2-144x+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 = 24-2(2.536) = 19.928
width = 12 - 2(2.536) = 6.928
height = 2.536
x(24-2x)(18-2x)
(24x-2x^2)(18-2x)
4x^3-84x^2+432
4x(x^2-21x+108)
4x(x-12)(x-9)
x=12,9,0
HELP!
A machine produces open boxes using rectangular sheets of metal (12in by 20in). The machine cuts equal-sized 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)