Question

A company is constructing an open-top, square-based, rectangular metal tank that will have a volume of 64 ft3. what dimensions yield the minimum surface area? round to the nearest tenth, if necessary.

Answer 1

To solve this problem you must apply the proccedure shown below:

1. You must use the formula for calculate the surface area and the volume of the tank:

$1) SA=x^{2} +4xy\\ 2) V=x^{2} y\\ 64=x^{2} y$

2. Solve for $y$ in the second equation and substitute it into the first one.Then, you have:

$y=\frac{64}{x^{2}} \\ SA=x^{2} +4x(\frac{64}{x^{2}})$

3. Now, derivate:

$SA'=2x-256x^{-2} \\ 2x-256x^{-2}=0$

4. When you solve for $x$, you obtain:

$2x^{3}-256=0\\ x=\sqrt[3]{\frac{256}{2}} \\ x=5.03$

5. Now, you must calculate $y$:

$y=\frac{64}{(5.03)^{2}} \\ y=2.52$

Therefore, the dimensions are:

The sides of the base of the tank: $5.03 feet$

The height of the tank: $2.52 feet$