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
To find the dimensions of the box, we need to determine the length and width of the rectangular base.
Let's assume that the width of the base is 'x' meters.
According to the given information, the length of the rectangular base is 2 meters greater than the width. So the length would be 'x + 2' meters.
The height of the box is given as 4 meters.
To find the lateral area (L) of the box, we can use the formula:
L = 2h(w + l)
where 'h' is the height, 'w' is the width, and 'l' is the length.
Substituting the given values:
96 = 2(4)(x + (x + 2))
Now, let's simplify and solve for 'x':
96 = 2(4)(2x + 2)
96 = 8(2x + 2)
96 = 16x + 16
80 = 16x
x = 5
The width of the base (x) is 5 meters.
Now, we can find the length of the base:
Length = Width + 2
Length = 5 + 2
Length = 7 meters
So, the dimensions of the box are:
Width = 5 meters
Length = 7 meters
Height = 4 meters