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

Answer:

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

Step-by-step 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:

$4\cdot l\cdot h = A_{s} -l^{2}$

$h = \frac{A_{s}-l^{2}}{4\cdot l}$

And expand (1) by this result:

$V = l^{2}\cdot \left(\frac{A_{s}-l^{2}}{4\cdot l } \right)$

$V = \frac{A_{s}\cdot l -l^{3}}{4}$

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

First Derivative Test

$\frac{A_{s}-3\cdot l^{2}}{4} = 0$

$A_{s} = 3\cdot l^{2}$

$l =\sqrt{\frac{A_{s}}{3} }$ (1)

Second Derivative Test

$V''=-\frac{6\cdot l}{4}$

$V'' = -\frac{3\cdot l}{2}$

$V'' = -\frac{3}{2}\cdot \sqrt{\frac{A_{s}}{3} }$ (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:

$l = \sqrt{\frac{120\,cm^{2}}{3} }$

$l \approx 6.325\,cm$

$h = \frac{120\,cm^{2}-(6.325\,cm)^{2}}{4\cdot (6.325\,cm)}$

$h \approx 3.162\,cm$

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