A box manufacturer wants to produce an open box for which the volume is 2 cubic meters and the length is twice the width. What actual dimensions of he box will require the least amount of material?
Work:
V = 2m^3
2m^3 = (w)*(2w)*(h)
Possible dimensions:
length width height surface area
2 1 1 8 units^2
1 1/2 4 12.5 units^2
1/2 1/4 16 24 1/8 units^2
To find the actual dimensions of the box that will require the least amount of material, we need to minimize the surface area of the box.
The surface area (A) of an open box can be calculated using the formula:
A = 2lw + lh + wh
Given that the volume (V) is 2 cubic meters and the length (l) is twice the width (w), we can express the dimensions as:
l = 2w
h = V/(lw) = 2/(2w*w) = 1/w
Substituting these values into the surface area formula, we get:
A = 2(2w)(w) + (2w)(1/w) + (w)(1/w)
A = 4w^2 + 2 + 1
A = 4w^2 + 3
To find the minimum surface area, we can take the derivative of A with respect to w and set it equal to zero:
dA/dw = 8w = 0
w = 0
Since w cannot be zero, there is no minimum surface area for this box. It appears that we made an error in our calculations.
Let's try another set of possible dimensions:
Length (l) Width (w) Height (h) Surface Area (A)
2m 1m 1m 8 units^2
1m 0.5m 4m 12.5 units^2
0.5m 0.25m 16m 24.125 units^2
Based on these calculations, the box with dimensions 2m x 1m x 1m requires the least amount of material, with a surface area of 8 square units.
To find the actual dimensions of the box that will require the least amount of material, we can calculate the surface area for each set of dimensions and compare them.
Let's calculate the surface area for each set of dimensions:
1) For dimensions: length=2, width=1, height=1:
Surface Area = 2lw + 2lh + 2wh
= 2(2)(1) + 2(2)(1) + 2(1)(1)
= 4 + 4 + 2
= 10 units^2
2) For dimensions: length=1, width=1/2, height=4:
Surface Area = 2lw + 2lh + 2wh
= 2(1)(1/2) + 2(1)(4) + 2(1/2)(4)
= 1 + 8 + 4
= 13 units^2
3) For dimensions: length=1/2, width=1/4, height=16:
Surface Area = 2lw + 2lh + 2wh
= 2(1/2)(1/4) + 2(1/2)(16) + 2(1/4)(16)
= 1/2 + 16 + 8
= 24 1/2 units^2
From the calculations, we can see that the set of dimensions with the least amount of material is when the length is 2, the width is 1, and the height is 1. This set of dimensions will require only 10 units^2 of material.