Answer to Question #280224 in Calculus for Joshua

Question #280224

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.


1
Expert's answer
2021-12-16T17:12:28-0500

Solution:


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

Let AE=H cm,AF=R cmAE=H\ cm, AF=R\ cm

Then, CE=H5CE=H-5

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

H5H=4RRH5R=4HH=5RR4 ...(i)\Rightarrow \dfrac{H-5}{H}=\dfrac{4}{R} \\\Rightarrow RH-5R=4H \\\Rightarrow H=\dfrac{5R}{R-4}\ ...(i)


Now, volume of cone, V=13πR2HV=\dfrac13 \pi R^2H

V=13πR2(5RR4) [Using (i)]V=53π(R3R4)\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=53π((R4)3R2R3(1)(R4)2)V=53π3R312R2R3(R4)2V=53π2R2(R6)(R4)2 ...(ii)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

53π2R2(R6)(R4)2=0R6=0R=6\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=103πddR(R36R2(R4)2)=103π((R4)2(3R212R)(R36R2)2(R4)(R4)4)>0 for R=6\\ 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=5(6)64=15 cmH=\dfrac{5(6)}{6-4}=15\ cm

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


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