Show that semi-vertical angle of right circular cone of given surface area and maximum volume is \( \sin^{-1} \left(\frac{1}{3}\right)\)
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be the vertical height of a cone of semi-vertical angle αα
Surface area S=πrl+πr2S=πrl+πr2------(1)
l=S−πr2πrl=S−πr2πr
The volume of the cone V=13V=13πr2hπr2h
=13=13πr2l2−r2−−−−−√πr2l2−r2
=πr23=πr23(S−πr2)2π2r2−r2−−−−−−−−−−−−√(S−πr2)2π2r2−r2
=πr23=πr23(S−πr2)2−π2r4π2r2−−−−−−−−−−−√(S−πr2)2−π2r4π2r2
=πr23=πr23S2−2πSr2+π2r4−π2r4−−−−−−−−−−−−−−−−−−√πrS2−2πSr2+π2r4−π2r4πr
=r3=r3S2−2πSr2+π2r4−π2r4−−−−−−−−−−−−−−−−−−−−−√S2−2πSr2+π2r4−π2r4
=r3=r3S(S−2πr2)−−−−−−−−−−√S(S−2πr2)
Surface area S=πrl+πr2S=πrl+πr2------(1)
l=S−πr2πrl=S−πr2πr
The volume of the cone V=13V=13πr2hπr2h
=13=13πr2l2−r2−−−−−√πr2l2−r2
=πr23=πr23(S−πr2)2π2r2−r2−−−−−−−−−−−−√(S−πr2)2π2r2−r2
=πr23=πr23(S−πr2)2−π2r4π2r2−−−−−−−−−−−√(S−πr2)2−π2r4π2r2
=πr23=πr23S2−2πSr2+π2r4−π2r4−−−−−−−−−−−−−−−−−−√πrS2−2πSr2+π2r4−π2r4πr
=r3=r3S2−2πSr2+π2r4−π2r4−−−−−−−−−−−−−−−−−−−−−√S2−2πSr2+π2r4−π2r4
=r3=r3S(S−2πr2)−−−−−−−−−−√S(S−2πr2)
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