• Nov 2nd 2005, 05:29 PM
gw16
Ok this chapter we are using vertex form to solve story problems.

Here is the Prob

A square, which is 5cm by 5cm, is cut from each corner of a rectangular piece of cardboard snd the sides are folded up to make a box. If the bottom of the box must have a perimiter of 50cm, what would be the length, width and height for the maximum volume?

thanks

Greg
• Nov 3rd 2005, 11:16 AM
ticbol
Quote:

A square, which is 5cm by 5cm, is cut from each corner of a rectangular piece of cardboard snd the sides are folded up to make a box. If the bottom of the box must have a perimiter of 50cm, what would be the length, width and height for the maximum volume
Maximum volume?
Okay.
Box.
V = L*w*h

The cut squares at each corner of the rectangular cardboard are 5cm by 5cm each. So if the sides of the box are folded up, the h of the box is 5cm.
Then the bottom of the box is L by w, in cms.

"If the bottom of the box must have a perimiter of 50cm,..."
L+w+L+w = 50
2L +2W = 50
L +w = 25 -------(i)

Hence, V = L*w*5 = 5Lw -----(ii)
So that we will have one variable only for V, say we eliminate L.
From (i), L = 25 -w. Substitute that into (ii),
V = 5(25 -w)w
V = 125w -5w^2 ----(iii)
Differentiate both sides of (iii) with respect to w,
dV/dw = 125 -10w
Set dV/dw to zero,
0 = 125 -10w
10w = 125
w = 125/10 = 12.5 cm.
Hence,
L = 25 -w = 25 -12.5 = 12.5 cm also.

Therefore, for maximum volume, length is 12.5cm, width is 12.5cm, and height is 5cm. ------answer.