Venkat
Last Activity: 6 Years ago
I am using x for a
first equation after diff w.r.t r gives → dx/dr = -2pi*r/3x
second after diff and simplification gives →dv/dr = -2pi(xr-2r2)
again diff and simplifying d2v/dr2 = – 2pi(x-2pi*r2/3x-4r)
now dv/dr = 0 → -pi*r(x-2r) = 0 → x = 2r
at x = 2r , d2v/dr2 = 2pi(2r+pi*r/3) > 0
→ v is minium when x = 2r , that is x/2r = 1
Conclusion: when the ratio of an edge of the cube to the diameter of the sphere is 1:1, the volume is minimum...