Aditya Gupta
Last Activity: 5 Years ago
hello praveen, we know that volume of cone V= pi*r^2*h/3
also, r^2 + h^2= l^2, where l is the slant length, r is the radius of cone (not the circular sheet), h is cone’s height. also, l= radius of circular sheet = 1
let the sector removed be of an angle x.
so, the base circumference of the cone= 2pi*r, which must be equal to the circular sheet’s sector’s length
so, 2pi*r= l*(x)= x (as l=1).
also, h= sqrt(1 – r^2), so dh/dr= – r/h
also, sectorial area removed= 1^2*x/2= x/2
now, V= pi*r^2*h/3
for max, dV/dx= 0
so, (pi/3)[d(r^2*h)/dr]*(dr/dx)= 0
but dr/dx = 1/2pi
so, d(r^2*h)/dr= 0
or 2r*h + r^2*( – r/h)= 0
or 2rh= r^3/h
or h root2 = r
or 2h^2= r^2
2 – 2r^2= r^2
or r = root(2/3)
but x= 2pi*r
so, sectorial area removed= 1^2*x/2= x/2= pi*root(2/3) sq units
kindly approve :)
