Answer:
Side of 40 and height of 20
Step-by-step explanation:
Let s be the side of the square base and h be the height of the box. Since the box volume is restricted to 32000 cubic centimeters we have the following equation:
[tex]V = hs^2 = 32000[/tex]
[tex]h = 32000/ s^2[/tex]
Assume that we cannot change the thickness, we can minimize the weight by minimizing the surface area of the tank
Base area with open top [tex]s^2[/tex]
Side area 4hs
Total surface area [tex]A = s^2 + 4hs[/tex]
We can substitute [tex]h = 32000/ s^2[/tex]
[tex]A = s^2 + 4s\frac{32000}{s^2}[/tex]
[tex]A = s^2 + 128000/s[/tex]
To find the minimum of this function, we can take the first derivative, and set it to 0
[tex]A' = 2s - 128000/s^2 = 0[/tex]
[tex]2s = 128000/s^2[/tex]
[tex]s^3 = 64000[/tex]
[tex]s = \sqrt[3]{64000} = 40[/tex]
[tex]h = 32000/ s^2 = 32000/ 40^2 = 20[/tex]