Question

The volume of a square-based rectangular cardboard box needs to be at least 1000cm^3. Determine the dimensions that require the minimum amount of material to manufacture all six faces. Assume that there will be no waste material. The Machinery available cannot fabricate material smaller than 2 cm in length.

Answer 1

We have to find the dimensions of a box with a volume that is at least 1000 cm³.

We have to find the dimensions that require the minimum amount of material.

We can draw the box as:

The volume can be expressed as:

`V=L\cdot W\cdot H\ge1000cm^3`

The material will be the sum of the areas:

`A=2LW+2LH+2WH`

Since the box is square-based, the width and length are equal and we can write:

`L=W`

Then, we can re-write the area as:

`\begin{gathered} A=2L^2+2LH+2LH \\ A=2L^2+4LH \end{gathered}`

Now, we have the area expressed in function of L and H.

We can use the volume equation to express the height H in function of L:

`\begin{gathered} V=1000 \\ L\cdot W\cdot H=1000 \\ L^2\cdot H=1000 \\ H=(1000)/(L^2) \end{gathered}`

We replace H in the expression for the area:

`\begin{gathered} A=2L^2+4LH \\ A=2L^2+4L\cdot(1000)/(L^2) \\ A=2L^2+(4000)/(L) \end{gathered}`

We can now optimize the area by differentiating A and then equal the result to 0:

`\begin{gathered} (dA)/(dL)=2(d(L^2))/(dL)+4000\cdot(d(L^(-1)))/(dL) \\ (dA)/(dL)=4L+4000(-1)L^(-2) \\ (dA)/(dL)=4L-(4000)/(L^2) \end{gathered}`
`\begin{gathered} (dA)/(dL)=0 \\ 4L-(4000)/(L^2)=0 \\ 4L=(4000)/(L^2) \\ L\cdot L^2=(4000)/(4) \\ L^3=1000 \\ L=\sqrt[3]{1000} \\ L=10 \end{gathered}`

We now can calculate the other dimensions as:

`W=L=10`
`H=(1000)/(L^2)=(1000)/(10^2)=(1000)/(100)=10`

Then, the dimensions that minimize the surface area for a fixed volume of 1000 cm³ is the length, width and height of 10 cm, which correspond to a cube (all 3 dimensions are the same).

Answer: the dimensions are length = 10 cm, width = 10 cm and height = 10 cm.