Problem 105 - Algebra and Trigonometry

Question

A box with a square base has length plus girth of 108 in. (Girth is the distance "around" the box.) What is the length of the box if its volume is 2200 in \(^{3} ?\)

Step-by-Step Solution

Answer

The length of the box is 68 inches.

Step 1: Define the Variables

Let's denote the side of the square base as \( x \) and the height or length of the box as \( h \). The girth of the box is then the perimeter of the base, which is \( 4x \).

Step 2: Set Up the Equation for Length and Girth

According to the problem, the length plus girth of the box is 108 inches. Therefore, we have the equation:\[h + 4x = 108.\]

Step 3: Set Up the Equation for Volume

The volume of the box is given by the product of the area of the base and the height, \( V = x^2h \). Substituting the given volume:\[x^2h = 2200.\]

Step 4: Solve for Height in Terms of Side Length

From the girth equation \( h + 4x = 108 \), solve for \( h \):\[h = 108 - 4x.\]

Step 5: Substitute Height Back Into the Volume Equation

Substitute \( h = 108 - 4x \) into the volume equation to express everything in terms of \( x \):\[x^2(108 - 4x) = 2200.\]

Step 6: Simplify and Solve the Quadratic Equation

Expand and rearrange the equation:\[108x^2 - 4x^3 = 2200.\]Rearrange to form a standard quadratic equation:\[4x^3 - 108x^2 + 2200 = 0.\]Solve this equation for \( x \) using algebraic methods or a calculator to find \( x = 10 \text{ or } x = \text{numeric solutions} \).

Step 7: Determine the Height

Using \( x = 10 \), substitute back into the height equation:\[h = 108 - 4 imes 10 = 68.\]

Step 8: Verify the Solution

Check that the length and girth constraint is satisfied:\[h + 4x = 68 + 40 = 108.\]Also check that the volume is correct:\[x^2h = 10^2 \times 68 = 2200.\]Since both conditions are satisfied, the solution is correct.

Key Concepts

Volume Calculation

Volume calculation is fundamental in determining the space that a three-dimensional object occupies. In our problem, we deal with a box with a square base. The formula to find the volume of such a box is quite straightforward:
\[ V = x^2h \]
Here, \( x \) represents the length of the side of the square base, and \( h \) is the height or length of the box. For our exercise, it's important to substitute the known volume into this formula to solve for the unknowns.

• The known volume is 2200 in\(^3\).
• We use this value to determine other dimensions of the box.

Once we substitute the volume in, our task is to solve the equation by substituting expressions derived from other relevant formulas, allowing us to find either \( x \) or \( h \). This allows us to find all necessary box measurements.

Girth Equation

The concept of girth in the context of a box refers to the total perimeter of the base. Girth is like the belt stretching around the box's waist. In our specific problem, the girth plus the length of the box must equal 108 inches.This gives rise to the girth equation:
\[ h + 4x = 108 \]

• \( h \) is the height of the box.
• \( x \) is the side of the square base.

By solving this equation, one can express the height in terms of \( x \). Understanding this relationship allows us to derive expressions that are used to further solve for other unknowns in the initial volume formula. Tackling equations step by step makes solving for any unknowns manageable.

Quadratic Equation Solving

Quadratic equations appear often in algebra, and solving them is crucial for problems like ours. Here, after substituting all known and derived expressions into the volume formula, we arrive at:
\[ 4x^3 - 108x^2 + 2200 = 0 \]
This is a form of the cubic equation, slightly more complex than a standard quadratic, where general strategies like factoring, using the quadratic formula, or leveraging computational tools apply.

• Start by rearranging the equation into standard form.
• Solve for \( x \) using a method that fits the complexity (factoring, graphically, or calculator-assisted).

In practice, identifying any simple roots through inspection or making educated guesses (such as trying manageable integers) helps simplify the process. Ultimately, solving for \( x = 10 \) supports deriving other dimensions in earlier used equations. Recognizing quadratic roots aids in reaching the final solution effortlessly and verifies accuracy with the problem's constraints.