Answer:
Side of 22 and height of 11
Step-by-step explanation:
Let s be the side of the square base and h be the height of the tank. Since the tank volume is restricted to 5324 ft cubed we have the following equation:
[tex]V = s^2h = 5324[/tex]
[tex]h = 5324 / s^2[/tex]
As the thickness is already defined, we can minimize the weight by minimizing the surface area of the tank
Base area with open top [tex]s^2[/tex]
Side area 4sh
Total surface area [tex]A = s^2 + 4sh[/tex]
We can substitute [tex]h = 5324 / s^2[/tex]
[tex]A = s^2 + 4s\frac{5324}{s^2}[/tex]
[tex]A = s^2 + 21296/s[/tex]
To find the minimum of this function, we can take the first derivative, and set it to 0
[tex]A' = 2s - 21296/s^2 = 0[/tex]
[tex]2s = 21296/s^2[/tex]
[tex]s^3 = 10648[/tex]
[tex]s = \sqrt[3]{10648} = 22[/tex]
[tex]h = 5324 / s^2 = 5324 / 22^2 = 11[/tex]
