A cone is made from a sphere, proove that the volume of the cone is maximum when height of the cone = 2/3 diameter of sphere???


Let OD=x and radius of sphere be R and that of cone be r. O is the center of sphereHeight of the cone, h=AD=x+RIn OBDOB2=OD2+BD2BD=OB2-OD2r=R2-x2Volume, V=13πr2hV=13πR2-x2x+RdVdx=13πd dxR2-x2x+R=13πR2-x2d dxx+R+x+Rd dxR2-x2=13πR2-x2+x+R-2x=13πR2-x2-2x2-2xr=13π-3x2-2xR+R2For minima or maximadVdx=013π-3x2-2xR+R2=0-3x2-2xR+R2=0x=2R±4R2+12R22×-3=2R±16R22×-3x=-2R±4R6x=-2R+4R6 or x=-2R-4R6x=-R or x=13RSince x>0therefore x=R3d2Vdx2=13π-6x-2RAt x=R3,d2Vdx2=13π-6×R3-2R<0Maxima at x=R3h=x+Rh when volume is maximum=R3+R=2R3
 

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