Problem 109 - Algebra and Trigonometry Real Mathematics, Real People

Question

An open box is to be made from a square piece of material 36 centimeters on a side by cutting equal squares with sides of length \(x\) from the corners and turning up the sides (see figure). (a) Verify that the volume of the box is given by the function \(V(x)=x(36-2 x)^{2}\) (b) Determine the domain of the function \(V\). (c) Use the table feature of a graphing utility to create a table that shows various box heights \(x\) and the corresponding volumes \(V\). Use the table to estimate a range of dimensions within which the maximum volume is produced. (d) Use the graphing utility to graph \(V\) and use the range of dimensions from part (c) to find the \(x\) -value for which \(V(x)\) is maximum.

Step-by-Step Solution

Answer

After graphing the function \(V(x)\), and creating a table with chosen \(x\) values from 0 to 18, one can see that the maximum volume occurs roughly around \(x = 6\). Further investigation proves that the specific value of \(x\) that maximizes \(V(x)\) within the allowable domain is in fact \(x = 6\), resulting in a maximum volume of 2160 cubic cm.

Step 1: Verifying the Volume Function

Starting with a square of side length 36 cm, squares of side length \(x\) will be cut from each corner. Therefore, each side of the resulting box (after folding up the uncut parts to form the sides of the box) will be 36 - 2x cm. The height of the box will be \(x\) cm. Therefore, the volume of the box given by length * width * height = \(V(x) = x * (36 - 2x)^2\). So, the function has been verified.

Step 2: Finding the Domain

The domain of the function \(V\) is the set of all permissible inputs. Since \(x\) represents the side of the squares that are cut for making the box, \(x\) must be greater than or equal to 0. Also, it can be at most half of the side of the square piece of material, as we are cutting squares from each corner. Hence \(x\) is less than or equal to 18. Therefore, the domain of \(V\) is [0, 18].

Step 3: Creating the Table

The graphing utility can be used to create a table with various values of \(x\) (box heights) and corresponding volumes \(V\). Make sure to choose a range of \(x\) values from 0 to 18, and at finer intervals near the peak value in order to accurately estimate the maximum volume and the corresponding value of \(x\).

Step 4: Graphing the Function and Determining the Maximum Volume

Plot the function \(V(x) = x*(36-2x)^2\). Look for the highest point on the graph, which represents the maximum volume that can be achieved with the considered range of \(x\) values. Use the estimated range found in the table to find an accurate \(x\)-value for which \(V(x)\) is maximum.

Key Concepts

Domain of a Function

Understanding the domain of a function is important because it defines the range of possible input values. In the box volume problem, we need to consider the physical constraints of the material. The side length of the original square is 36 cm. Squares are being cut from each corner to form the box, where the length of each cut is defined by the variable \( x \).

This means \( x \) must be a positive value, as you cannot cut out a negative length, and less than or equal to 18. The number 18 is derived as \( 36/2 \) because cutting squares larger than this from each corner would eliminate all material.

Hence, the domain of the function \( V(x) \) is the interval [0, 18]. This ensures \( x \) is in a reasonable range that allows the creation of a box.

Graphing Utilities

Using graphing utilities can greatly simplify understanding how a function behaves. In this exercise, graphing utilities help create a table and graph of the volume function \( V(x) = x(36-2x)^2 \).

To effectively use graphing utilities, follow these easy steps:

• Select appropriate software; many calculators or computer programs can graph functions.
• Input the volume function formula: \( V(x) = x(36-2x)^2 \).
• Set the range for input values \( x \) from 0 to 18, as defined by the domain.

Using these utilities, you can quickly create a table of \( x \) and \( V(x) \), offering insight into the function's behavior. Look closely near potential peak values, as these reveal where the function reaches a maximum. Graphing also provides a visual of the function, showcasing how the volume changes with different values of \( x \).

Quadratic Functions

Quadratic functions are a type of polynomial and appear frequently in various mathematical problems. They have the standard form \( ax^2 + bx + c \). In this exercise, the function \( V(x) = x(36-2x)^2 \) can be expanded to reveal its quadratic nature.

The basic properties of quadratic functions include:

• A parabolic graph shape, which either opens upwards or downwards.
• A vertex that represents the maximum or minimum value depending on the parabola's orientation.
• A symmetry axis that runs through the vertex, dividing the parabola into two equal mirror-like halves.

Given that our quadratic is represented with \( V(x) = x(36-2x)^2 \), our task is to determine the maximum volume, which corresponds to the vertex of the downward-opening parabola.

Box Volume Calculation

The box volume calculation is at the heart of the exercise, linking spatial dimensions to functional algebra. Calculating the volume of an open box involves multiplying its length, width, and height. In this scenario, these parameters change as \( x \) changes.

The equation given is \( V(x) = x(36-2x)^2 \), where:

• \( x \): the height of the box, and also the side length cut from each corner.
• \( 36 - 2x \): the length and width of the box, derived from reducing each original side by twice the cut square's side.

To find the volume for specific \( x \) values, plug \( x \) into the volume function. The objective is often to find the \( x \) which gives the maximum volume, pivotal when designing the most efficient box. Using graphing utilities, this critical point stands out clearly once the graph or data table of \( V(x) \) is analyzed.