The volume of a cone = 31∗π∗r2∗h
Let the radius of the sphere = a
the radius of the cone =r
by similar triangles sin(θ)=h−aa = h2+r2r
=a∗(h2+r2)=r(h−a)
on squaring both sides
a2(h2+r2)=r2(h−a)2
=a2h2+a2r2=h2r2−2ahr2+a2r2
making r2 to be the subject of the formulae we have
r2=h−2aa2h
The volume V = 31∗π∗h−2ah2∗h
at minimum volume ;
dhdV=3πa2[(h−2a)2(h−2a)2h−h2(1)]=0
2(h−2a)h−h2=0
2h−4a−h=0
h−4a=0
h=4a
but a =8 inches
therefore h=4∗8=32
The height of the cone is 32 inches.
r2=32−2∗882∗32
r2=162048
r2=128
r=128
r=11.3137
The height of the cone is 32 inches and the radius is 11.3137inches.
Comments