Maximizing the Volume of a Rectangular Box from a Given Cardboard Piece

The problem of finding the optimal dimensions of a rectangular box that can be made from a piece of cardboard is a classic optimization problem in mathematics. In this article, we will explore the detailed steps to determine the dimensions of the box that maximize the volume for a given cardboard piece measuring 24 inches by 9 inches, and provide a general analysis for any such problem.

Problem Definition

We are given a piece of cardboard measuring 24 inches by 9 inches. By cutting out identical squares from each of the four corners and turning up the sides, we can form a rectangular box. The objective is to find the dimensions of the box that will yield the maximum volume.

Steps to Solve the Problem

1. Define the Variables

Let ( x ) be the side length of the squares cut from each corner. This variable will help us determine the dimensions of the resulting box.

2. Determine the Dimensions of the Box

After cutting out squares of side length ( x ), the dimensions of the box will be:

Length: ( 24 - 2x ) Width: ( 9 - 2x ) Height: ( x )

3. Volume of the Box

The volume ( V ) of the box can be expressed as:

V  (24 - 2x) (9 - 2x) x

4. Simplify the Volume Function

Expanding and simplifying the volume function:

V  x (24 - 2x) (9 - 2x)
    x (216 - 48x - 18x   4x^2)
    x (216 - 66x   4x^2)
    216x - 66x^2   4x^3

5. Find the Critical Points

To maximize the volume, we need to take the derivative of ( V ) and set it to zero:

V'(x)  216 - 132x   12x^2  0

Divide by 12 to simplify the equation:

x^2 - 11x   18  0

6. Solve the Quadratic Equation

The quadratic equation ( x^2 - 11x 18 0 ) can be solved using the quadratic formula:

x  frac{11 pm sqrt{11^2 - 4 cdot 1 cdot 18}}{2 cdot 1}  frac{11 pm sqrt{121 - 72}}{2}  frac{11 pm sqrt{49}}{2}  frac{11 pm 7}{2}

This gives us two solutions:

( x frac{11 7}{2} 9 ) ( x frac{11 - 7}{2} 2 )

Since ( x ) represents the size of the cut squares, it must be less than half the smaller dimension of the cardboard. Therefore, the only valid solution is ( x 2 ).

7. Calculate the Dimensions of the Box

Substitute ( x 2 ) back into the dimensions:

Length: ( 24 - 2 cdot 2 20 ) inches Width: ( 9 - 2 cdot 2 5 ) inches Height: ( 2 ) inches

The dimensions of the box that maximize the volume are 20 inches long, 5 inches wide, and 2 inches high.

General Analysis

This analysis demonstrates that the optimal cut size is indeed smaller, as increasing the cut size will reduce the remaining area, leading to a lower volume. For a 9cm x 24cm piece of cardboard, if the cutouts are in inches, a 2-inch cut yields the maximum volume, as shown by the calculations.

Note that: The maximum volume for the cut size is 2 inches, resulting in a volume of 200 cubic centimeters. Any increase in cut size (1 inch or 3 inches) would reduce the volume.