Geometry
posted by Jordie .
For a given box, the height measures 4 meters. If the length of the rectangular base is 2 meters greater than the width of the base and the lateral area L is 96 square meters, find the dimensions of the box.

let the width be w
then length = x+2
SA = 2lw + 2lh + 2 wh
2w(w+2) + 2(w+2)(4) + 2(w)(4) = 96
2w^2 + 4w + 8w + 16 + 8w = 96
w^2 + 10w  40 = 0
completing the square
w^2 + 10w + 25 = 40+25
(w+5)^2 = 65
w +5 = ±√65
w = 5 + √65 or x = 5 √65 , the last being negative, thus resulting in "silly answer"
w = 5+√65 = appr. 3.06 or whatever number of decimals you want
so the width = 3.06
lenth = 5.06
height = 4
check:
2(3.06)(5.06) + 2(3.06)(4) + 2( 5.06)(4) = 95.92...
(not bad) 
area = 2(lh + wh)
you know that h=4, l=w+2, area=96
96 = 2((w+2)(4) + 4(4))
48 = 4w + 8 + 16
w = 6
the box is 6x8x4 
hmm. forgot the w
96 = 2((w+2)(4) + w(4))
48 = 4w+8+4w
40 = 8w
w=5
box is 5x7x4
Lateral area is height*perimeter = 4*2(5+7) = 4*24 = 96 
I think Steve has an error
2(5x7) + 2(5x4) + 2(7x4) ≠ 96 
Ahem. Lateral area does not include the bases.

Thanks Steve, didn't know that
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