Given a rectangular sheet of cardboard 15 in. by 25 in. If a small square of the same size is cut from each corner and each side folded up along the cuts to from a lidless box:

What is the maximum volume of the box?

What size(s) of the square would produce a box of volume equal to 400 cubic inches?

**Figure 1** shows an illustration of the box that we need
to form.

The first thing to realize from this figure is that the size of the square that is to be cut from each corner must be less than 7.5 in by 7.5 in. This is true because if the size of the square was 7.5 in by 7.5 in. no box could be formed because the width would be zero and the size of the square cannot be greater than 7.5 in by 7.5 in. because one would not be able to cut an equal size square out of each corner because the width of the cardboard that we are given is only 15 in.

There are a couple of different ways one can go about solving the two
problems asked above. One could answer them graphically, with the help of
a spreadsheet, using hand written mathematics and solving equations, etc.
The graphical and spreadsheet methods will be used here. In order to get
a graphical representation of this problem we need an equation for the volume
of the box. Looking back to **Figure 1** to determine the length,
width, and height, the equation for the volume of the box that needs to
be solved is **V=****(25-2x)(15-2x)x**. Simplified this
is **V = 4x^3 - 80x^2 + 375x**. Now graphing this equation
(remembering that **x** must be less than 7.5 in.) we get a
nice curve that shows that there is a maximum.** Figure 2**
shows this graph.

From this figure, notice that the maximum volume is a little over 500
cubic inches and this occurs when the size of the square cut out of each
corner is about 3 in. by 3 in. Now we have narrowed down the range for which
the size of the square cut from each corner can be. It is in the range between
2.9 in. and 3.25 in. square. This is seen in **Figure 3**.
The lines for **x=2.9 in.** and **x=3.25 in.**
are shown and the maximum volume is between these two **x**-values.

Now with the range narrowed down, we can go to a spreadsheet program
and let the spreadsheet do the calculations for many different possible
values of the size of the square to be cut out and determine the size of
the square that when cut out creates a box a greatest volume. One way to
do this is to put in one column values of the width of the square to be
cut out, starting with 2.9 in. and increasing this by some increment up
to the 3.25 in. Then in another column determine the volume for each of
these values. Then look at the different volumes that were just calculated
and find the values of the width of the square where the volume is going
up, then levels off and then starts going back down. Now use the smallest
of these three widths and create another column similar to the first column
you created but you only need to let the widths of the square range between
the the smallest and largest of the three widths you just determine to be
the ones that zeroed in on the maximum volume. With multiple repetitions
of this one can find a value for the width of the square that will give
the greatest volume. With each repetition, you gain one more decimal place
of accuracy. **Table 1** below illustrates the initial calculation
of the volume of the box for values of the width of the square between 2.9
in. and 3.25 in.

Side of Square(in.) |
Volume in Cubic Inches |
Side of Square(in.) |
Volume in Cubic Inches |

2.900 | 512.256 | 3.080 | 512.960 |

2.910 | 512.370 | 3.090 | 512.917 |

2.920 | 512.476 | 3.100 | 512.864 |

2.930 | 512.573 | 3.110 | 512.803 |

2.940 | 512.661 | 3.120 | 512.733 |

2.950 | 512.740 | 3.130 | 512.655 |

2.960 | 512.809 | 3.140 | 512.569 |

2.970 | 512.870 | 3.150 | 512.474 |

2.980 | 512.922 | 3.160 | 512.370 |

2.990 | 512.966 | 3.170 | 512.258 |

3.000 | 513.000 | 3.180 | 512.138 |

3.010 | 513.026 | 3.190 | 512.009 |

3.020 | 513.042 | 3.200 | 511.872 |

3.030 | 513.051 | 3.210 | 511.727 |

3.040 | 513.050 | 3.220 | 511.573 |

3.050 | 513.041 | 3.230 | 511.411 |

3.060 | 513.022 | 3.240 | 511.241 |

3.070 | 512.996 | 3.250 | 511.063 |

The three rows highlighted in green in **Table 1** correspond
to the three values of the width of the square where the volume is increasing,
levels off and starts to decrease. These values were then broken up again
like the ones in **Table 1** but with one more decimal place
of accuracy. After repeating **Table 1** a couple more times,
a value of **x**(the width of the square to be cut from each
corner) equal to **3.03425** was determined to give the greatest
volume of the box. With more iterations of **Table 1**, one
could get a value for **x** that has more decimal places. This
value of **x=3.03425 in. **makes the volume of the box become
**V=513.0512959596 cubic inches**.

Now to answer the other question about, what size(s) of the square would
produce a box of volume equal to 400 cubic inches, we can turn again to
the graph in **Figure 2**. Since we want a volume of 400 cubic
inches, if we draw the line corresponding with **V=400** on
the graph in **Figure 2** we will end up with a graph that
looks like the one in **Figure 4**.

The intersection points of the curve and line in **Figure 4**
represent the two values of the width of the square when cut from the corners
of the piece of cardboard will generate a volume equal to 400 cubic inches.
These intersections look as though they occur at about **x=1.5**
and **x=4.75**. Now we will need to go through the steps done
above to find the value of **x** that gave the largest volume
to find the values of **x** that give a volume of 400 cubic
inches. These steps will need to be performed on **x=1.5**
and then on **x=4.75**.

After going through the same process as above to find the value of **x**,
the two values of **x** that make the volume of the box equal
400 cubic inches are **x=1.5249288 in. **which gives a volume
of **V=399.999998 cubic inches** and **x=4.7928497 in.**
which gives a volume of **V=400.0000063 cubic inches**. These
values for the volumes are not exactly equal to 400 cubic inches but are
very close. One could get the volumes even closer to 400 cubic inches simply
by repeating the steps used in** Table 1** more times on this
data.

The methods used here to determine the values of **x **that
were required to satisfy the two questions asked were just two of the many
different ways of determining them. Explore other methods as your mind contemplates
the ones shown here.