Answer:
The dimensions of the box that minimize the materials used is ![6\times 3\times 2\ ft](https://tex.z-dn.net/?f=6%5Ctimes%203%5Ctimes%202%5C%20ft)
Step-by-step explanation:
Given : An open top box is to be built with a rectangular base whose length is twice its width and with a volume of 36 ft³.
To find : The dimensions of the box that minimize the materials used ?
Solution :
An open top box is to be built with a rectangular base whose length is twice its width.
Here, width = w
Length = 2w
Height = h
The volume of the box V=36 ft³
i.e. ![w\times 2w\times h=36](https://tex.z-dn.net/?f=w%5Ctimes%202w%5Ctimes%20h%3D36)
![h=\frac{18}{w^2}](https://tex.z-dn.net/?f=h%3D%5Cfrac%7B18%7D%7Bw%5E2%7D)
The equation form when top is open,
![f(w)=2w^2+2wh+2(2w)h](https://tex.z-dn.net/?f=f%28w%29%3D2w%5E2%2B2wh%2B2%282w%29h)
Substitute the value of h,
![f(w)=2w^2+2w(\frac{18}{w^2})+2(2w)(\frac{18}{w^2})](https://tex.z-dn.net/?f=f%28w%29%3D2w%5E2%2B2w%28%5Cfrac%7B18%7D%7Bw%5E2%7D%29%2B2%282w%29%28%5Cfrac%7B18%7D%7Bw%5E2%7D%29)
![f(w)=2w^2+\frac{36}{w}+\frac{72}{w}](https://tex.z-dn.net/?f=f%28w%29%3D2w%5E2%2B%5Cfrac%7B36%7D%7Bw%7D%2B%5Cfrac%7B72%7D%7Bw%7D)
![f(w)=2w^2+\frac{108}{w}](https://tex.z-dn.net/?f=f%28w%29%3D2w%5E2%2B%5Cfrac%7B108%7D%7Bw%7D)
Derivate w.r.t 'w',
![f'(w)=4w-\frac{108}{w^2}](https://tex.z-dn.net/?f=f%27%28w%29%3D4w-%5Cfrac%7B108%7D%7Bw%5E2%7D)
For critical point put it to zero,
![4w-\frac{108}{w^2}=0](https://tex.z-dn.net/?f=4w-%5Cfrac%7B108%7D%7Bw%5E2%7D%3D0)
![4w=\frac{108}{w^2}](https://tex.z-dn.net/?f=4w%3D%5Cfrac%7B108%7D%7Bw%5E2%7D)
![w^3=27](https://tex.z-dn.net/?f=w%5E3%3D27)
![w^3=3^3](https://tex.z-dn.net/?f=w%5E3%3D3%5E3)
![w=3](https://tex.z-dn.net/?f=w%3D3)
Derivate the function again w.r.t 'w',
![f''(w)=4+\frac{216}{w^3}](https://tex.z-dn.net/?f=f%27%27%28w%29%3D4%2B%5Cfrac%7B216%7D%7Bw%5E3%7D)
For w=3, ![f''(3)=4+\frac{216}{3^3}=12 >0](https://tex.z-dn.net/?f=f%27%27%283%29%3D4%2B%5Cfrac%7B216%7D%7B3%5E3%7D%3D12%20%3E0)
So, it is minimum at w=3.
Now, the dimensions of the box is
Width = 3 ft.
Length = 2(3)= 6 ft
Height = ![\frac{18}{3^2}=2\ ft](https://tex.z-dn.net/?f=%5Cfrac%7B18%7D%7B3%5E2%7D%3D2%5C%20ft)
Therefore, the dimensions of the box that minimize the materials used is ![6\times 3\times 2\ ft](https://tex.z-dn.net/?f=6%5Ctimes%203%5Ctimes%202%5C%20ft)