Question

A rectangular box with a square base and no top is to be made of a total of 120 cm2 of cardboard. Find the dimensions of the box of maximum volume.

Answer 1

The dimensions of the box of maximum volume are: Length: 6.325 centimeters, height: 3.162 centimeters.

Explanation:

The volume (
`V`), measured in cubic centimeters, and the surface area of the rectangular box (
`A_(s)`), measured in square centimeters, with a square base are represented by the following formulas:

`V = l^(2)\cdot h` (1)

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

Where:

`l` - Length of the side of the square base, measured in centimeters.

`h` - Height of the box, measured in centimeters.

By (2) we clear the height:

`4\cdot l\cdot h = A_(s) -l^(2)`

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

And expand (1) by this result:

`V = l^(2)\cdot \left((A_(s)-l^(2))/(4\cdot l ) \right)`

`V = (A_(s)\cdot l -l^(3))/(4)`

Lastly, we proceed to perform First and Second Derivative Test to find critical values associated with maximum volume:

First Derivative Test

`(A_(s)-3\cdot l^(2))/(4) = 0`

`A_(s) = 3\cdot l^(2)`

`l =\sqrt{(A_(s))/(3) }` (1)

Second Derivative Test

`V''=-(6\cdot l)/(4)`

`V'' = -(3\cdot l)/(2)`

`V'' = -(3)/(2)\cdot \sqrt{(A_(s))/(3) }` (2)

We conclude that critical value leads to a maximum volume.

If we know that
`A_(s) = 120\,cm^(2)`, then dimensions of the box are, respectively:

`l = \sqrt{(120\,cm^(2))/(3) }`

`l \approx 6.325\,cm`

`h = (120\,cm^(2)-(6.325\,cm)^(2))/(4\cdot (6.325\,cm))`

`h \approx 3.162\,cm`

The dimensions of the box of maximum volume are: Length: 6.325 centimeters, height: 3.162 centimeters.