Answer to Question #280224 in Calculus for Joshua

Question

Find the dimensions of the right circular cone of least volume which can be

circumscribed about a right circular cylinder of radius 4 cm and height of 5 cm.

Answer 1

Solution:

Given, $AB=CD=4\ cm, AC=5\ cm$

Let $AE=H\ cm, AF=R\ cm$

Then, $CE=H-5$

Since, triangle EDC and EFA are similar by AA rule, $\dfrac{EC}{EA}=\dfrac{CD}{AF}$

$\Rightarrow \dfrac{H-5}{H}=\dfrac{4}{R} \\\Rightarrow RH-5R=4H \\\Rightarrow H=\dfrac{5R}{R-4}\ ...(i)$

Now, volume of cone, $V=\dfrac13 \pi R^2H$

$\Rightarrow V=\dfrac13 \pi R^2(\dfrac{5R}{R-4})\ [Using \ (i)] \\ \Rightarrow V=\dfrac53 \pi (\dfrac{R^3}{R-4})$

On differentiating w.r.t R,

$V'=\dfrac53 \pi (\dfrac{(R-4)3R^2-R^3(1)}{(R-4)^2}) \\ \Rightarrow V'=\dfrac53 \pi \dfrac{3R^3-12R^2-R^3}{(R-4)^2} \\ \Rightarrow V'=\dfrac53 \pi \dfrac{2R^2(R-6)}{(R-4)^2}\ ...(ii)$

Now, put V'=0

$\dfrac53 \pi \dfrac{2R^2(R-6)}{(R-4)^2}=0 \\\Rightarrow R-6=0 \\\Rightarrow R=6$

Again, differentiating (ii) w.r.t. R,

$\\ V''=\dfrac{10}3 \pi \dfrac{d}{dR}(\dfrac{R^3-6R^2}{(R-4)^2}) \\=\dfrac{10}3 \pi (\dfrac{(R-4)^2(3R^2-12R)-(R^3-6R^2)2(R-4)}{(R-4)^4}) \\>0\ for \ R=6$

Thus, minima exists.

Now, put R=6 in (i).

$H=\dfrac{5(6)}{6-4}=15\ cm$

Hence, the dimensions of the right circular cone are R=6 cm, H=15 cm.