Problem 92 - A Graphical Approach to Precalculus with Limits

Question

Dimensions of a Box A box has an open top, rectangular sides, and a square base. The volume of the box is 576 cubic inches, and the surface area of the outside of the box is 336 square inches. Find the dimensions of the box.

Step-by-Step Solution

Answer

The box's dimensions are 12 inches by 12 inches by 4 inches.

Step 1: Define Variables

Let the side of the square base be \( x \) inches and the height of the box be \( h \) inches. Since the base is square, the area of the base is \( x^2 \).

Step 2: Write the Volume Equation

Use the formula for volume of the box, which is given by the equation \( V = x^2 h \). According to the problem, \( V = 576 \) cubic inches. So, we have the equation: \( x^2 h = 576 \).

Step 3: Write the Surface Area Equation

The surface area of the box consists of the base and the four sides (no top). The surface area \( S \) is given by \( x^2 + 4xh = 336 \).

Step 4: Solve for Height (h) from Volume Equation

From the equation \( x^2 h = 576 \), solve for \( h \) in terms of \( x \): \( h = \frac{576}{x^2} \).

Step 5: Substitute h into Surface Area Equation

Substitute \( h = \frac{576}{x^2} \) into the surface area equation: \( x^2 + 4x\left(\frac{576}{x^2}\right) = 336 \). Simplifying gives \( x^2 + \frac{2304}{x} = 336 \).

Step 6: Solve for x

Rearrange the equation to form \( x^3 + 2304 = 336x \), leading to \( x^3 - 336x + 2304 = 0 \). Solve this cubic equation by factoring or using methods such as synthetic division. After solving, we find \( x = 12 \).

Step 7: Solve for h using x value

Substitute \( x = 12 \) back into the height equation \( h = \frac{576}{12^2} = \frac{576}{144} = 4 \).

Step 8: Verify Solution

Check that both original conditions (volume and surface area) are satisfied with \( x = 12 \) and \( h = 4 \): Volume is \( 12^2 \cdot 4 = 576 \) cubic inches and surface area is \( 12^2 + 4 \cdot 12 \cdot 4 = 336 \) square inches. Both are correct.

Key Concepts

Volume of a Box

The volume of a box is a fundamental concept in precalculus, involving three-dimensional space. The formula for calculating the volume is relatively straightforward.
It involves multiplying the area of the base by the height of the box.

• For a box with a square base, the base area is given by the side length squared, so it's represented as \( x^2 \).
• The height of the box, denoted as \( h \), is then multiplied by the base area to find the total volume.
• The resulting formula is \( V = x^2 h \).

In the given exercise, we're told that this volume is specifically 576 cubic inches. Thus, we set up the equation \( x^2 h = 576 \), allowing us to solve for other dimensions when additional information is provided.

Surface Area Calculation

Surface area calculations help determine the total area covered by the outer surface of a three-dimensional object. For this box with an open top,
it consists of the bottom base and the four surrounding rectangle-like sides.

• Since the base of the box is square, the area of the base is \( x^2 \).
• Each side of the box has an area of \( x \times h \), the width \( x \), and the box's height h.
• Therefore, the four sides have a total surface area of \( 4xh \).
• The total surface area without a top is given by \( x^2 + 4xh \).

In this problem, the surface area is stated to be 336 square inches. This serves as another equation, \( x^2 + 4xh = 336 \), that needs to be solved alongside the volume equation to find the box’s dimensions.

Cubic Equations

Cubic equations are equations that contain an \( x^3 \) term, making them a bit more complex to solve than their quadratic counterparts.
For the exercise at hand, we derived a cubic equation from the surface area and volume formulas.

• By substituting the expression for \( h \) (from the volume equation \( h = \frac{576}{x^2} \)) into the surface area equation, we arrive at \( x^2 + \frac{2304}{x} = 336 \).
• Re-arranging this results in a cubic equation \( x^3 - 336x + 2304 = 0 \).
• Solving cubic equations can involve methods such as factoring, synthetic division, or using special algebraic identities.

Through these techniques, \( x \) was found to be 12, which is a key step in determining the dimensions of our box.

Rectangular and Square Geometry

Understanding the relationship between rectangles and squares is crucial when dealing with geometry problems.
In the context of the box in our exercise, the square base and rectangular sides form a clear geometric picture.

• A square is a special type of rectangle where all sides are equal, simplifying the initial computation for the base area.
• The sides of the box, being rectangles, introduce variability with one side matching the base's side length \( x \) and the other being the height \( h \).
• Geometry not only aids in visualization but, upon grasping these relationships, plays a foundational role in deriving and solving equations related to dimensions, surface areas, and volumes.

By understanding these principles, students can more easily tackle problems involving complex three-dimensional shapes like boxes.