Question

1)Find the dimensions of the box with the minimum total surface area. 2)Find the minimum total surface area of the box.

Answer 1

Let l, x, and h be the length, width, and height of the cuboid.

Therefore, its surface is given by the formula,

`A_{\text{surface}}=lx+2(hl)+2(hx)`

(Notice that the box does not have a lid.)

Furthermore,

`\begin{gathered} l=4x \\ \text{and} \\ V=\text{lxh} \\ V=400 \\ \Rightarrow\text{lxh}=400 \end{gathered}`

1) Notice that we have two conditions,

`\begin{gathered} l=4x, \\ \text{lxh}=400 \end{gathered}`

Therefore,

`\begin{gathered} \Rightarrow h=(400)/(lx) \\ \text{and} \\ l=4x \\ \Rightarrow h=(400)/(4x\cdot x)=(100)/(x^2) \\ \Rightarrow h=(100)/(x^2) \end{gathered}`

Therefore, the surface area as a function of the width is

`\begin{gathered} \Rightarrow A_{\text{surface}}=4x\cdot x+2(4x\cdot(100)/(x^2))+2(x\cdot(100)/(x^2)) \\ \Rightarrow A_{\text{surface}}=4x^2+(800)/(x)+(200)/(x) \\ \Rightarrow A_{\text{surface}}=4x^2+(1000)/(x) \end{gathered}`

We need to find the minimum of the function above; first, derivate A_surface and solve A'(x)=0, as shown below

`A^(\prime)_{\text{surface}}(x)=8x+1000(-(1)/(x^2))=8x-(1000)/(x^2)`

Then,

`\begin{gathered} A^(\prime)_{\text{surface}}(x)=0 \\ \Rightarrow8x-(1000)/(x^2)=0 \\ \Rightarrow8x^3-1000=0 \\ \Rightarrow x^3=(1000)/(8) \\ \Rightarrow x=\sqrt[3]{(1000)/(8)}=(10)/(2)=5 \\ \Rightarrow x=5 \end{gathered}`

Therefore, the width that minimizes the total surface area is x=5cm.

Finding l and h,

`\begin{gathered} \Rightarrow l=4\cdot5=20 \\ \text{and} \\ h=(100)/(25)=4 \\ \Rightarrow h=4 \end{gathered}`

The answers to part a) are l=20cm, x=5cm, h=4cm.

b)

Finally, set l=20, x=5, and h=4cm in the A_surface function, as shown below,

`\begin{gathered} A_{\text{surface}}=20\cdot5+2(20\cdot4)+2(5\cdot4)=300 \\ \Rightarrow A_{\text{surface}}=300 \end{gathered}`

The minimum total surface area is 300cm^2