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, EAEC=AFCD
⇒HH−5=R4⇒RH−5R=4H⇒H=R−45R ...(i)
Now, volume of cone, V=31πR2H
⇒V=31πR2(R−45R) [Using (i)]⇒V=35π(R−4R3)
On differentiating w.r.t R,
V′=35π((R−4)2(R−4)3R2−R3(1))⇒V′=35π(R−4)23R3−12R2−R3⇒V′=35π(R−4)22R2(R−6) ...(ii)
Now, put V'=0
35π(R−4)22R2(R−6)=0⇒R−6=0⇒R=6
Again, differentiating (ii) w.r.t. R,
V′′=310πdRd((R−4)2R3−6R2)=310π((R−4)4(R−4)2(3R2−12R)−(R3−6R2)2(R−4))>0 for R=6
Thus, minima exists.
Now, put R=6 in (i).
H=6−45(6)=15 cm
Hence, the dimensions of the right circular cone are R=6 cm, H=15 cm.
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