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.

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