Maximising an Inscribed Cylinder in a Cone
A right circular cone has base radius R = 6.0 cm and height H = 12.0 cm. A cylinder is inscribed in it: the cylinder's axis coincides with the cone's axis, its lower base lies in the cone's base plane, and its upper rim touches the cone's curved surface. A student must maximise the cylinder's volume. He first writes the constraint as h = H − r, treating the cone's side as though it sloped at 45°, and obtains r = 4.0 cm, h = 8.0 cm. A second student uses similar triangles from the apex and gets a different relation. Because R = 6 cm and H = 12 cm happen to make the optimal cylinder's height equal to its radius, a third student concludes that h = r is the general condition for maximum volume of a cylinder inscribed in any cone. The group must settle the correct constraint, the true optimum, and whether that last generalisation survives a change of cone dimensions.
1. Derive the correct constraint and find the maximum cylinder volume for R = 6 cm, H = 12 cm.
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Let r, h be the cylinder's radius and height. A vertical section through the axis gives similar triangles between the small cone above the cylinder and the full cone:
(R − r)/R = (H − h)/H → h = H(1 − r/R) = 12(1 − r/6) = 12 − 2r
V = πr²h = π(12r² − 2r³) dV/dr = π(24r − 6r²) = 6πr(4 − r) = 0 → r = 4.0 cm, h = 4.0 cm V = π(16)(4) = 64π ≈ 201 cm³. Since V → 0 as r → 0 and as r → 6, this stationary point is the maximum.
The first student's h = 12 − r is the wrong constraint (it assumes a 45° slope); with r = 4 it predicts h = 8, but only 4 cm of height is geometrically available there.
2. Generalise to arbitrary R and H, and test the claim that h = r at the optimum.
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With h = H(1 − r/R): V = πH(r² − r³/R), so dV/dr = πH(2r − 3r²/R) = 0 → r = 2R/3, h = H/3.
V_max = π(4R²/9)(H/3) = 4πR²H/27, which is exactly 4/9 of the cone's volume πR²H/3 - a fixed fraction, independent of dimensions.
h = r only when H/3 = 2R/3, i.e. H = 2R, which happens to hold for 12 = 2 × 6. The generalisation fails immediately for another cone: with R = 6, H = 24, the optimum is r = 4, h = 8, giving V = 128π cm³ against a cone volume of 288π cm³ - again the 4/9 ratio.
3. A classmate uses r = 4, h = 8 and gets V = 128π cm³. Is that answer merely 'too large', or is the underlying reasoning internally inconsistent?
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It is internally inconsistent, not just numerically large. The pair (r = 4, h = 8) is impossible inside this cone: at radius 4 the side of the cone has already descended to height 12 − 2(4) = 4 cm, so a cylinder of height 8 would pierce the curved surface.
The error lies entirely in the constraint, not the calculus: the derivative was performed correctly on the wrong relation h = 12 − r. Good practice is to check the physical feasibility of the optimum (0 ≤ h ≤ 12 − 2r) before trusting the stationary point, and to confirm the answer lies below the cone's own volume 144π cm³.