Assignment 12

Write-up for Problem 6

by

Christina Reid

Maximize the volume of a lidless box formed from 5x8 sheet with a square (with a length of x) removed from each corner.

Using the figure above, the dimensions of the box are

The formula for the volume of this box would be

Use a spreadsheet to display possible values of x, (8-2x), (5-2x), and V.

 x 8-2x 5-2x V=x(8-2x)(5-2x) 0 8 5 0 0.1 7.8 4.8 3.744 0.2 7.6 4.6 6.992 0.3 7.4 4.4 9.768 0.4 7.2 4.2 12.096 0.5 7 4 14 0.6 6.8 3.8 15.504 0.7 6.6 3.6 16.632 0.8 6.4 3.4 17.408 0.9 6.2 3.2 17.856 1 6 3 18 1.1 5.8 2.8 17.864 1.2 5.6 2.6 17.472 1.3 5.4 2.4 16.848 1.4 5.2 2.2 16.016 1.5 5 2 15 1.6 4.8 1.8 13.824 1.7 4.6 1.6 12.512 1.8 4.4 1.4 11.088 1.9 4.2 1.2 9.57599999999999 2 4 0.999999999999999 7.99999999999999 2.1 3.8 0.799999999999999 6.38399999999999 2.2 3.6 0.599999999999999 4.75199999999999 2.3 3.4 0.399999999999999 3.12799999999999 2.4 3.2 0.199999999999998 1.53599999999999 2.5 3 0 0

We only need to use values of x<2.5 since the width is 5 and using a larger value of x would give us a negative number and the sheet would disappear.

Looking at the Volume column shows that the volume appears to reach a maximum volume of 18 when x=1.

This is supported using the first derivative of V.

The first derivative of V gives the critical points of 1 and 10/3. The value of 1 is the only value between 0 and 2.5.

Using the second derivative of V,

the value is positive when x=1 and that means V is a maximum at that value.

This can also be verified by the graph of the equation

At x=1, the graph shows a relative maximum point.