Question

A certain cube has edges of length L inches, surface area of A square inches, and volume of \(\mathrm{B}\) cubic inches. For what value of \(\mathrm{L}\) would \(A=B ?\)

Short answer

The value of \( L \) for which \( A = B \) is 6 inches.

Step-by-step solution

Understand the Problem

We have a cube with edge length denoted by \( L \). The surface area \( A \) of the cube is given by the formula: \( A = 6L^2 \). The volume \( B \) of the cube is given by the formula: \( B = L^3 \). We need to find the value of \( L \) such that the surface area equals the volume, i.e., \( A = B \).

Set the Equations Equal

Since we want \( A = B \), substitute the formulas for surface area and volume into this equality: \( 6L^2 = L^3 \).

Simplify the Equation

Divide both sides of the equation \( 6L^2 = L^3 \) by \( L^2 \) (assuming \( L eq 0 \)), to get: \( 6 = L \).

Solution Verification

Verify that \( L = 6 \) satisfies the condition \( A = B \). Calculate \( A = 6L^2 = 6(6)^2 = 216 \) and \( B = L^3 = 6^3 = 216 \). Both the surface area and volume are 216, thus confirming \( A = B \).

Key concepts

Surface Area of a Cube

The surface area of a cube is a fundamental geometric property that quantifies the total area covered by its six faces. Each face of a cube is a square, and since a cube has six identical square faces, the formula for the surface area is simple:

• To find the surface area, use the formula \( A = 6L^2 \), where \( L \) represents the edge length of a cube.
• This formula results from multiplying the area of one face \( (L^2) \) by the six faces of the cube.

Understanding this formula helps in visualizing how the total area is just an accumulation of the areas of all the individual faces. It is crucial to the exercise because the surface area forms one part of the equation that needs to be equal to the volume for this specific problem.

Volume of a Cube

The volume of a cube measures the space contained within it. For any cube, the volume is determined by how much material it would take to fill the cube or how much space it occupies. The formula to find the volume is:

• Volume, \( B \), is calculated as \( L^3 \), where \( L \) is the edge length of the cube.
• This formula signifies that the volume is the product of the cube's length, width, and height, all of which are equal \( (L \times L \times L) \).

In this problem, the cube's volume is directly compared to its surface area. Knowing just this formula already equips you with the knowledge to derive the cube's internal capacity, given its dimensions. This is especially vital as it forms the second part of the equation in our solved problem.

Algebraic Equations

Algebraic equations are vital tools in relating different mathematical expressions to solve for unknown values, such as in this cube problem. By setting up and simplifying equations, you can efficiently explore relationships between variables. Here's how it applies here:

• The key equation from this task is \( 6L^2 = L^3 \), which arises from equating the cube's surface area to its volume.
• Simplifying involves removing common terms (in this case, dividing both sides by \( L^2 \)), provided \( L eq 0 \), which leaves us with \( 6 = L \).

This method of simplification helps in isolating \( L \), which reveals it as the only solution under the given conditions. Mastering how to manipulate such equations is crucial in solving for unknowns across various mathematical challenges.

Problem Solving

Problem solving in geometry often involves a systematic approach to understanding and finding solutions to mathematical issues, like equalizing a cube's surface area and volume in this exercise. Here’s how the steps unfold:

• First, correctly identify what information is given and what is being asked—here, finding \( L \) such that surface area equals volume.
• Then, use established formulas for surface area \( (6L^2) \) and volume \( (L^3) \) and set them equal to form an equation.
• Next, solve the equation algebraically by isolating \( L \), checking assumptions and not dividing by variables that might be zero.
• Finally, verify the solution by plugging it back into initial expressions to ensure it satisfies the problem's conditions.

This organized methodology makes tackling similar problems more manageable and is critical in solving complex problems effectively.