A thin rectangular sheet of metal is 6 inches wide and 10 inches long. The sheet of metal is to...

1. Translate the problem requirements: We have a \(6 \times 10\) inch rectangle that will form a cylinder. One dimension becomes the circumference (wraps around), the other becomes height. We need to find which orientation gives the maximum volume.
2. Set up the two possible cylinder configurations: Compare Case 1 (6-inch circumference, 10-inch height) versus Case 2 (10-inch circumference, 6-inch height).
3. Calculate radius for each case using circumference formula: Since \(\mathrm{circumference} = 2\pi r\), we can find the radius for each configuration.
4. Apply volume formula and compare: Use \(V = \pi r^2 h\) to calculate volume for both cases and identify the larger one.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what's happening here. We have a flat rectangular piece of metal that's 6 inches wide and 10 inches long. Think of it like a piece of paper that we're going to roll up to make a tube (cylinder).

When we roll up a rectangle to make a cylinder, one dimension of the rectangle wraps around to form the circumference (the distance around the circular base), and the other dimension becomes the height of the cylinder.

So we have two choices:

- Roll it so the 6-inch side wraps around (circumference = 6 inches, height = 10 inches)
- Roll it so the 10-inch side wraps around (circumference = 10 inches, height = 6 inches)

Our goal is to figure out which way gives us the bigger volume.

Process Skill: TRANSLATE - Converting the physical description into mathematical scenarios

2. Set up the two possible cylinder configurations

Let's clearly define our two cases:

Case 1: Roll along the width

- The 6-inch dimension becomes the circumference
- The 10-inch dimension becomes the height
- So: Circumference = 6 inches, Height = 10 inches

Case 2: Roll along the length

- The 10-inch dimension becomes the circumference
- The 6-inch dimension becomes the height
- So: Circumference = 10 inches, Height = 6 inches

Now we need to find the volume for each case and see which is larger.

3. Calculate radius for each case using circumference formula

To find the volume of a cylinder, we need the radius. We can get the radius from the circumference because when you go around a circle, the distance around (circumference) relates to the radius.

The relationship is: \(\mathrm{Circumference} = 2\pi \times \mathrm{radius}\)
So: \(\mathrm{radius} = \mathrm{Circumference} \div (2\pi)\)

For Case 1:
\(\mathrm{radius}_1 = 6 \div (2\pi) = \frac{3}{\pi}\) inches

For Case 2:
\(\mathrm{radius}_2 = 10 \div (2\pi) = \frac{5}{\pi}\) inches

4. Apply volume formula and compare

The volume of a cylinder is the area of the circular base times the height.
\(\mathrm{Volume} = \pi \times \mathrm{radius}^2 \times \mathrm{height}\)

Case 1 Volume:
\(V_1 = \pi \times \left(\frac{3}{\pi}\right)^2 \times 10\)
\(V_1 = \pi \times \frac{9}{\pi^2} \times 10\)
\(V_1 = \frac{9\pi}{\pi^2} \times 10\)
\(V_1 = \frac{9}{\pi} \times 10\)
\(V_1 = \frac{90}{\pi}\) cubic inches

Case 2 Volume:
\(V_2 = \pi \times \left(\frac{5}{\pi}\right)^2 \times 6\)
\(V_2 = \pi \times \frac{25}{\pi^2} \times 6\)
\(V_2 = \frac{25\pi}{\pi^2} \times 6\)
\(V_2 = \frac{25}{\pi} \times 6\)
\(V_2 = \frac{150}{\pi}\) cubic inches

Since \(\frac{150}{\pi} > \frac{90}{\pi}\), Case 2 gives us the larger volume.

Process Skill: CONSIDER ALL CASES - Systematically comparing both possible configurations

4. Final Answer

The maximum volume occurs when we roll the rectangle so that the 10-inch side becomes the circumference and the 6-inch side becomes the height.

The largest possible volume is \(\frac{150}{\pi}\) cubic inches.

This matches answer choice C) \(\frac{150}{\pi}\).

Common Faltering Points

Errors while devising the approach

Students often get confused about how rolling a rectangle creates a cylinder. They might think that the width always becomes the radius or that the length always becomes the height, without realizing that either dimension can wrap around to form the circumference. This leads them to only consider one configuration instead of comparing both possibilities.

Some students see the phrase "largest possible cylinder" but fail to understand that they need to compare two different scenarios. They might calculate the volume for just one configuration (often the first one that comes to mind) and assume that's the answer, missing the critical step of comparing both rolling orientations.

Errors while executing the approach

A very common error is directly using the circumference value as the radius in the volume formula. For example, if the circumference is 6 inches, students might use \(r = 6\) instead of correctly calculating \(r = \frac{6}{2\pi} = \frac{3}{\pi}\). This leads to dramatically incorrect volume calculations.

When calculating \(\left(\frac{3}{\pi}\right)^2\) or \(\left(\frac{5}{\pi}\right)^2\), students often make mistakes like writing \(\left(\frac{3}{\pi}\right)^2 = \frac{9}{\pi}\) instead of the correct \(\frac{9}{\pi^2}\). Similarly, when simplifying expressions like \(\pi \times \frac{9}{\pi^2} \times 10\), they might incorrectly cancel the π terms or make other fraction manipulation errors.

Students sometimes correctly identify both configurations but then mix up which height goes with which circumference. For instance, they might use circumference = 6 with height = 6, or circumference = 10 with height = 10, instead of the correct pairings.

Errors while selecting the answer

Even after correctly calculating both volumes as \(\frac{90}{\pi}\) and \(\frac{150}{\pi}\), some students accidentally select the smaller value \(\left(\frac{90}{\pi}\right)\) as their final answer, perhaps because they calculated it first or because they misread their comparison. They fail to carefully identify which volume is actually larger.