Resizing a Cylinder at Constant Volume
Ms. Whitfield's design class is resizing a cylindrical can. The current can has radius 4 cm and height 10 cm, giving a volume of 160π cm³ and a lateral-plus-two-bases surface area of 112π cm². Marketing wants a new can with the same volume but a radius of 5 cm, so it looks wider on the shelf. Jordan reasons: "The radius goes from 4 to 5, which is a factor of 5/4, so the height must shrink by the same factor to keep the volume constant. New height = 10 × 4/5 = 8 cm." He then argues that because the volume is unchanged, the amount of metal used in the can must also be unchanged, so the redesign has no cost impact. Ms. Whitfield needs to correct the height before the class builds physical models, and she wants the correction to expose the faulty proportional reasoning rather than just supply the right number.
1. Compute the correct new height and identify the specific flaw in Jordan's proportional reasoning.
Read the worked answer
The volume of a cylinder is $V = \pi r^2 h$, so the radius enters squared. Holding volume constant means $r^2 h$ is constant, so if the radius is multiplied by $k$, the height must be divided by $k^2$, not by $k$.
Here $k = 5/4$, so the height is multiplied by $(4/5)^2 = 16/25$: $$h_{\text{new}} = 10 \times \frac{16}{25} = 6.4\ \text{cm}.$$
Check: $\pi (5)^2 (6.4) = 25 \times 6.4 \times \pi = 160\pi$ cm³. Correct.
Jordan's error is treating the relationship as linear. A factor of 5/4 applied to the radius would give $\pi(25)(8) = 200\pi$, which is 25% too much volume - larger than the original can, not equal to it. He matched one dimension to the volume instead of accounting for the two dimensions that the radius controls.
2. Evaluate Jordan's second claim about the metal. Compute both surface areas and explain what the comparison shows.
Read the worked answer
The claim is false: constant volume does not imply constant surface area.
Original: $S = 2\pi r^2 + 2\pi r h = 2\pi(16) + 2\pi(4)(10) = 32\pi + 80\pi = 112\pi$ cm².
New: $S = 2\pi(25) + 2\pi(5)(6.4) = 50\pi + 64\pi = 114\pi$ cm².
The new can uses slightly more metal, about 1.8% more, even though the volume is identical. The reason is structural: surface area is a sum of terms that scale differently. The two circular bases scale as $r^2$ and grew from $32\pi$ to $50\pi$, while the lateral area $2\pi r h$ fell from $80\pi$ to $64\pi$. The wider can spends more area on its ends and less on its side, and the two effects do not cancel.
This is a useful generalization for students: for a fixed volume, a shape closer to a sphere or cube minimizes surface area, so making a cylinder wider and shorter increases the metal needed.
3. Ms. Whitfield wants one quick numerical check students can run to catch a linear-scaling error like Jordan's before they build. Describe the check and show it applied to Jordan's 8 cm answer.
Read the worked answer
The check is a two-line volume comparison using the proposed dimensions, done before any physical work.
"Compute $r^2 h$ for the original and for your proposed can. If the volume must stay the same, these products must be equal."
Applied to Jordan's proposal: original $r^2 h = 16 \times 10 = 160$. Proposed $r^2 h = 25 \times 8 = 200$. Since $200 \ne 160$, the proposal is wrong, and the ratio $200/160 = 1.25$ immediately reveals that he is off by exactly the factor $5/4$ he should have squared.
Solving $25h = 160$ gives $h = 6.4$ cm directly, which is faster than setting up the full volume formula with π. The same habit extends to cones and spheres: isolate the product of the non-π factors and compare, because π cancels on both sides. Students who run this check catch the error in seconds, before any cardboard is cut.