A rectangular peice of cardboard with a perimiter of 40ft is creased and then folded into a rectangular box with open square ends. find the length of the side of the square end for which the maximun volume of the box is realized.
2006-10-12 05:16:25 · 6 answers · asked by 3ST4X 2 in Education & Reference ➔ Homework Help
step one:
the length of the raw cardboard = l
the width of the raw cardboard = w
2l + 2w = 40
when made into a box, the length of the square end
is l/4
the volume, v = (l/4)^2 * w
substitute w=20-l into the volume
v=l^2/16 (20-l)
v=(20* l^2-l^3)/16
This can be graphed in 2-d space
Note that the point l=40/3 produces a volume of 74 ft^3
so the length will be l/4=10/3 feet
j
2006-10-12 05:18:11 · answer #1 · answered by odu83 7 · 0⤊ 0⤋
Half the perimiter is 20 that leaves 20ft for the long side and the short side. the lond side needs to be divided in 4 and as long as the end side
20/5=x
x=4
Therfore the length is 4ft. (and I have no idea what ft are, but the priciples are the same regardless of the units)
volume whould be 64ft^3
2006-10-12 12:24:01 · answer #2 · answered by enterprise17 2 · 0⤊ 0⤋
Find the derivative of the volume equation, set that equal to zero, then using that x value, do a maximization graph. :]
That's how I would do it in calculus. :]
2006-10-12 16:46:40 · answer #3 · answered by Steven Procter 2 · 0⤊ 0⤋
I think its 12 feet, but I'm not sure. good luck!
2006-10-12 12:32:02 · answer #4 · answered by Anonymous · 0⤊ 0⤋
Happiness is the way to go!
2006-10-12 12:20:44 · answer #5 · answered by Anonymous · 0⤊ 1⤋
74.07
2006-10-12 12:48:20 · answer #6 · answered by vand003 2 · 0⤊ 0⤋