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Largest Surface Area of Cylinder in Cone

CAT 2008 Slot 1 · Quantitative Ability · Hard · Geometry

This is a hard Quantitative Ability question from the CAT 2008 Slot 1 paper. It tests Geometry. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

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Consider a right circular cone of base radius 4 cm4\text{ cm} and height 10 cm10\text{ cm}. A cylinder is to be placed inside the cone with one of the flat surfaces resting on the base of the cone. Find the largest possible total surface area (in sq. cm) of the cylinder.

  1. A.

    100π3\frac{100\pi}{3}

  2. B.

    80π3\frac{80\pi}{3}

  3. C.

    120π7\frac{120\pi}{7}

  4. D.

    130π9\frac{130\pi}{9}

  5. E.

    110π7\frac{110\pi}{7}

Answer

A

Explanation

Let radius of cylinder be rr and height be hh. By similar triangles: 10−h10=r4  ⟹  h=10−52r=52(4−r)\frac{10 - h}{10} = \frac{r}{4} \implies h = 10 - \frac{5}{2}r = \frac{5}{2}(4 - r).

Total Surface Area A(r)=2πr2+2πrh=2π(r2+r⋅52(4−r))=2π(10r−32r2)A(r) = 2\pi r^2 + 2\pi r h = 2\pi \left(r^2 + r \cdot \frac{5}{2}(4 - r)\right) = 2\pi \left(10r - \frac{3}{2}r^2\right).

To maximize A(r)A(r), take derivative w.r.t rr: ddr(10r−32r2)=10−3r=0  ⟹  r=103\frac{d}{dr}\left(10r - \frac{3}{2}r^2\right) = 10 - 3r = 0 \implies r = \frac{10}{3}.

Max Area =2π(10(103)−32(1009))=2π(1003−503)=100π3= 2\pi \left(10\left(\frac{10}{3}\right) - \frac{3}{2}\left(\frac{100}{9}\right)\right) = 2\pi \left(\frac{100}{3} - \frac{50}{3}\right) = \frac{100\pi}{3}.

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