Question

A rectangular tank that is 5324 ft cubed with a square base and open top is to be constructed of sheet steel of a given thickness. Find the dimensions of the tank with minimum weight.

Answer 1

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:

$V = s^2h = 5324$

$h = 5324 / s^2$

As the thickness is already defined, we can minimize the weight by minimizing the surface area of the tank

Base area with open top $s^2$

Side area 4sh

Total surface area $A = s^2 + 4sh$

We can substitute $h = 5324 / s^2$

$A = s^2 + 4s\frac{5324}{s^2}$

$A = s^2 + 21296/s$

To find the minimum of this function, we can take the first derivative, and set it to 0

$A' = 2s - 21296/s^2 = 0$

$2s = 21296/s^2$

$s^3 = 10648$

$s = \sqrt[3]{10648} = 22$

$h = 5324 / s^2 = 5324 / 22^2 = 11$