Answer:
The dimensions of the box that minimize the materials used is 
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. 

The equation form when top is open,

Substitute the value of h,



Derivate w.r.t 'w',

For critical point put it to zero,





Derivate the function again w.r.t 'w',

For w=3, 
So, it is minimum at w=3.
Now, the dimensions of the box is
Width = 3 ft.
Length = 2(3)= 6 ft
Height = 
Therefore, the dimensions of the box that minimize the materials used is 