A beverage company designs cylindrical cans so that the height of each can is 3 inches less than the diameter...
GMAT Geometry & Trigonometry : (Geo_Trig) Questions
A beverage company designs cylindrical cans so that the height of each can is 3 inches less than the diameter of its base. Let \(\mathrm{r}\) be the radius, in inches, of such a can, where \(\mathrm{r \gt 1.5}\). Which expression gives the lateral surface area (the area of the curved side only) of the can, in square inches, in terms of \(\mathrm{r}\)?
\(2\pi\mathrm{r}(2\mathrm{r} + 3)\)
\(4\pi\mathrm{r}^2 - 6\pi\mathrm{r}\)
\(\pi(2\mathrm{r})^2(2\mathrm{r} - 3)\)
\(2\pi\mathrm{r}^2 + 2\pi\mathrm{r}(2\mathrm{r} - 3)\)
1. TRANSLATE the problem information
- Given information:
- Cylindrical cans where \(\mathrm{height = diameter - 3}\) inches
- Radius = \(\mathrm{r}\) inches, where \(\mathrm{r \gt 1.5}\)
- Need to find lateral surface area (curved side only)
- What this tells us: We need to express height in terms of radius, then use the lateral surface area formula
2. TRANSLATE the height relationship
- Since \(\mathrm{diameter = 2r}\), we can write:
- \(\mathrm{Height = diameter - 3 = 2r - 3}\)
- This gives us height in terms of r, which we need for the surface area formula
3. INFER the approach
- For lateral surface area of a cylinder: \(\mathrm{L = 2\pi rh}\)
- We have r and can express h in terms of r
- Substitute \(\mathrm{h = 2r - 3}\) into the formula
4. SIMPLIFY the expression
- \(\mathrm{L = 2\pi r(2r - 3)}\)
- \(\mathrm{L = 2\pi r \times 2r - 2\pi r \times 3}\)
- \(\mathrm{L = 4\pi r^2 - 6\pi r}\)
Answer: B (\(\mathrm{4\pi r^2 - 6\pi r}\))
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak TRANSLATE reasoning: Students confuse radius and diameter relationships or misinterpret "3 inches less than the diameter."
Some students might think \(\mathrm{height = radius - 3 = r - 3}\), missing that the problem says "less than the diameter" not "less than the radius." Others might write \(\mathrm{height = 2r + 3}\) instead of \(\mathrm{2r - 3}\), getting the subtraction backwards. This leads to expressions like \(\mathrm{2\pi r(r - 3) = 2\pi r^2 - 6\pi r}\) or \(\mathrm{2\pi r(2r + 3) = 4\pi r^2 + 6\pi r}\).
This may lead them to select Choice A (\(\mathrm{2\pi r(2r + 3)}\)) if they get the sign wrong.
Second Most Common Error:
Incomplete SIMPLIFY execution: Students set up the problem correctly but make algebraic errors when expanding.
They correctly get \(\mathrm{L = 2\pi r(2r - 3)}\) but then make mistakes like forgetting to distribute the \(\mathrm{2\pi r}\) to both terms, or making sign errors during multiplication. This leads to expressions that don't match any answer choice exactly.
This leads to confusion and guessing among the remaining choices.
The Bottom Line:
This problem tests your ability to carefully translate word relationships into mathematical expressions and then execute algebraic simplification correctly. The key insight is recognizing that "diameter" must be converted to "2r" before setting up the height relationship.
\(2\pi\mathrm{r}(2\mathrm{r} + 3)\)
\(4\pi\mathrm{r}^2 - 6\pi\mathrm{r}\)
\(\pi(2\mathrm{r})^2(2\mathrm{r} - 3)\)
\(2\pi\mathrm{r}^2 + 2\pi\mathrm{r}(2\mathrm{r} - 3)\)