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:
[tex]V=L\cdot W\cdot H\ge1000cm^3[/tex]The material will be the sum of the areas:
[tex]A=2LW+2LH+2WH[/tex]Since the box is square-based, the width and length are equal and we can write:
[tex]L=W[/tex]Then, we can re-write the area as:
[tex]\begin{gathered} A=2L^2+2LH+2LH \\ A=2L^2+4LH \end{gathered}[/tex]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:
[tex]\begin{gathered} V=1000 \\ L\cdot W\cdot H=1000 \\ L^2\cdot H=1000 \\ H=\frac{1000}{L^2} \end{gathered}[/tex]We replace H in the expression for the area:
[tex]\begin{gathered} A=2L^2+4LH \\ A=2L^2+4L\cdot\frac{1000}{L^2} \\ A=2L^2+\frac{4000}{L} \end{gathered}[/tex]We can now optimize the area by differentiating A and then equal the result to 0:
[tex]\begin{gathered} \frac{dA}{dL}=2\frac{d(L^2)}{dL}+4000\cdot\frac{d(L^{-1})}{dL} \\ \frac{dA}{dL}=4L+4000(-1)L^{-2} \\ \frac{dA}{dL}=4L-\frac{4000}{L^2} \end{gathered}[/tex][tex]\begin{gathered} \frac{dA}{dL}=0 \\ 4L-\frac{4000}{L^2}=0 \\ 4L=\frac{4000}{L^2} \\ L\cdot L^2=\frac{4000}{4} \\ L^3=1000 \\ L=\sqrt[3]{1000} \\ L=10 \end{gathered}[/tex]We now can calculate the other dimensions as:
[tex]W=L=10[/tex][tex]H=\frac{1000}{L^2}=\frac{1000}{10^2}=\frac{1000}{100}=10[/tex]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.
