Problem 119 - College Algebra with Modeling and Visualization

Question

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

Step-by-Step Solution

Answer

The box's dimensions are a 12-inch square base and a 4-inch height.

Step 1: Understand the problem

We need to find the dimensions of a box with a square base, an open top, rectangular sides, a volume of 576 cubic inches, and an outside surface area of 336 square inches.

Step 2: Define the variables

Let \( x \) represent the side of the square base, and \( h \) represent the height of the box. So, the base area is \( x^2 \), the volume is \( V = x^2h = 576 \), and the surface area is \( SA = x^2 + 4xh = 336 \).

Step 3: Set up equations

From the problem, we have two equations: 1. \( x^2h = 576 \) (volume equation)2. \( x^2 + 4xh = 336 \) (surface area equation).

Step 4: Express one variable in terms of the other

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

Step 5: Substitute and solve

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

Step 6: Solve for \( x \)

Multiply through by \( x \) to clear the fraction: \( x^3 + 2304 = 336x \).Rearrange to: \( x^3 - 336x + 2304 = 0 \).Solve this cubic equation by testing logical values or using a calculator to find \( x = 12 \).

Step 7: Find \( h \) using \( x \)

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

Step 8: Verify the solution

Check the dimensions by substituting \( x = 12 \) and \( h = 4 \) into the original equations. Both the volume and surface area equations hold true for these values.

Key Concepts

Rectangular Prism

A rectangular prism is a three-dimensional shape with six faces, all of which are rectangles.
Each pair of opposite faces is congruent, which means they have the same dimensions.
In our exercise, the box with a square base is a special type of rectangular prism.

• The base of the prism, in this case, is a square, creating two equal and opposite square faces.
• There are four additional rectangular faces that are the vertical sides of the prism.

A key characteristic of a rectangular prism is its uniform cross-section when sliced along its length. Each slice parallel to the base is identical in size and shape to the base. This property is fundamental when calculating volume and surface area. It makes rectangular prisms a commonly studied shape in geometry and algebra-related problems.

Volume Calculation

The volume of a three-dimensional shape like a rectangular prism is the amount of space it occupies, measured in cubic units.
For a prism with a square base, calculating the volume becomes straightforward.
Here's the basic formula for volume: \[ V = ext{Base Area} imes ext{Height} \] For this specific problem, since the base is square:

• Base Area is given by \( x^2 \), where \( x \) is the length of a side of the square base.
• Height is \( h \), which is the vertical dimension from the base upwards.

Combining these, the volume equation becomes: \[ V = x^2 h \] For example, if \( x = 12 \) inches and \( h = 4 \) inches, then: \[ V = 12^2 imes 4 = 144 imes 4 = 576 \text{ cubic inches} \] Thus, confirming that the calculation of volume matches the given problem statement.

Surface Area Calculation

Surface area is the total area covered by the surfaces of a three-dimensional object.
For a box with an open top and a square base, the surface area calculation simplifies somewhat.
The surface area is given by: \[ SA = x^2 + 4xh \] Here, \( x^2 \) represents the area of the square base and \( 4xh \) is the combined area of the four vertical sides.

• The term \( x^2 \) indicates that there's only one square base since the box has an open top.
• Each of the four sides has an area \( x imes h \), so together they add up to \( 4xh \).

For the box with \( x = 12 \) inches and \( h = 4 \) inches:
The surface area computes as \( 12^2 + 4 imes 12 imes 4 = 144 + 192 = 336 \text{ square inches} \). This confirms the surface area as stated.

Algebraic Equations

Solving problems involving geometric shapes typically involves setting up and solving algebraic equations.
In this exercise, we use two main equations related to volume and surface area.
The equations:

• \( x^2 h = 576 \) represents the volume.
• \( x^2 + 4xh = 336 \) models the surface area.

By solving algebraically, such as expressing \( h \) in terms of \( x \) using the volume equation, we can substitute into the surface area equation to find a solvable monoculus. The substitution of \( h \) yields \( h = \frac{576}{x^2} \), providing a simpler form for equation resolution. This process involves:
- Identifying appropriate values for substitution and elimination.
- Ensuring calculations align with logical, feasible dimensions for the box. Such steps reinforce critical mathematical skills: manipulating algebraic expressions and understanding geometric properties through algebra.