A cylindrical container is designed so that its height is twice its radius. If the volume of the container is...

1. TRANSLATE the problem information

• Given information:
  • Volume = \(\mathrm{686π}\) cubic inches
  • Height is twice the radius → \(\mathrm{h = 2r}\)
  • Need to find height

2. INFER the solution approach

• We have one equation (volume formula) but two unknowns (r and h)
• The constraint \(\mathrm{h = 2r}\) lets us substitute and solve for one variable
• Once we find radius, we can easily find height

3. Set up the volume equation with substitution

• Start with cylinder volume formula: \(\mathrm{V = πr^2h}\)
• Substitute known values: \(\mathrm{686π = πr^2(2r)}\)
• SIMPLIFY: \(\mathrm{686π = 2πr^3}\)

4. SIMPLIFY to isolate r³

• Divide both sides by \(\mathrm{2π}\): \(\mathrm{686π ÷ 2π = r^3}\)
• This gives us: \(\mathrm{343 = r^3}\)

5. Solve for radius

• SIMPLIFY: Take cube root of both sides
• \(\mathrm{r = \sqrt[3]{343} = 7}\) inches (use calculator if needed)

6. Find the height

• Use the constraint: \(\mathrm{h = 2r = 2(7) = 14}\) inches

Answer: B (14)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students might not properly convert "height is twice its radius" into the algebraic relationship \(\mathrm{h = 2r}\), or they might set up the relationship backward as \(\mathrm{r = 2h}\).

This leads to incorrect substitution in the volume formula, producing wrong values for both radius and height. This causes them to get stuck and guess randomly.

Second Most Common Error:

Inadequate SIMPLIFY execution: Students correctly set up \(\mathrm{686π = 2πr^3}\) but make calculation errors, either when dividing by \(\mathrm{2π}\) or when computing \(\mathrm{\sqrt[3]{343}}\).

Common mistakes include forgetting that \(\mathrm{\sqrt[3]{343} = 7}\), or computational errors that lead to \(\mathrm{r = 14}\), making them incorrectly select Choice D (28) when they double this value for height.

The Bottom Line:

This problem tests whether students can handle constraint relationships in geometry formulas. The key insight is recognizing that the constraint \(\mathrm{h = 2r}\) transforms a two-variable problem into a one-variable problem through substitution.