The right circular cone of largest volume that can be enclosed by a sphere of 1m radius has a height of
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With given radius r=1m of a sphere let the inscribed cone have height h then remaining length without radius is (h-r) let R be radius of cone then there we get a right angle triangle with r as hypotanious R as adjecent and (h-r) as opposite side. Now by pythagorus theorem ((h-r)^2)+(R^2)=(r^2)
V(h)= 1/3 pi * (2rh−h2)2 * h
V′(h)=13(4rh−3h2)V′(h)=13(4rh−3h2)
V′(h)=0 when h= 4/3*r
with r=1m
the answer is h= 4/3 m
V(h)= 1/3 pi * (2rh−h2)2 * h
V′(h)=13(4rh−3h2)V′(h)=13(4rh−3h2)
V′(h)=0 when h= 4/3*r
with r=1m
the answer is h= 4/3 m
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