Question

A company is constructing an open-top, square-based, rectangular metal tank that will have a volume of 34.5 ft3. what dimensions yield the minimum surface area? round to the nearest tenth, if necessary.

Answer 1

`l \approx 3.3\,ft`,
`x = 3.2\,ft`

Explanation:

The equations of volume and surface area are presented below:

`34.5 = l^(2)\cdot x`

`A_(s) = 2\cdot l^(2) + 4\cdot l \cdot x`

The length of the tank is:

`x = (34.5)/(l^(2))`

The expression fo the surface area is therefore simplified into an univariable form:

`A_(s) = 2\cdot l^(2) + 4\cdot \left((34.5)/(l) \right)`

`A_(s) = 2\cdot l^(2) + (138)/(l)`

The first and second derivatives of the expression are, respectively:

`A_(s)' = 4\cdot l -(138)/(l^(2))`

`A_(s)'' = 4 +(276)/(l^(3))`

The first derivative is equalized to zero and length of the square side is now found:

`4\cdot l -(138)/(l^(2)) = 0`

`4\cdot l^(3)-138 = 0`

`l^(3) = (138)/(4)`

`l = \sqrt[3]{(138)/(4) }`

`l \approx 3.3\,ft`

Now, the second derivative offers a criteria to determine if solution leads to an absolute minimum:

`A_(s)'' = 4 + (276)/((3.3\,ft)^(3))`

`A_(s)'' = 11.7` (Absolute minimum)

The depth of the tank is:

`x = (34.5\,ft^(3))/((3.3\,ft)^(2))`

`x = 3.2\,ft`