Differential Calculus Applied Optimization Problem?

let dimensions of cone be H, R (assume const)

let dimensions of cylinder be h, r (variable)

If you draw this (2d drawing will show some rectangle in a triangle)

you will see that this involves similar triangle where:

h/(R - r) = H/R

ie h = H(R - r)/R

volume of cyclinder, v = (pi)r^2h

v = (pi)r^2H(R - r)/R = ((pi)H/R)(Rr^2 - r^3)

we want max v

This is found by dv/dr = 0 and d^v/dr^2 < 0

dv/dr = ((pi)H/R)(2Rr - 3r^2)

d^2v/dr^2 = ((pi)H/R)(2R - 6r) = (2(pi)H/R)(R - 3r)

dv/dr = 0 when

2Rr = 3r^2

one solution is r = 0 (this is a minima)

another solution is

2R = 3r or r = (2/3)R (This gives a neg d^2v/dr^2 confirming maxima)

volume of cylinder, v = ((pi)H/R)(Rr^2 - r^3) = ((pi)r^2H/R)(R - r)

v = (4/9)(pi)RH(R - (2/3)R) = (4/9)(1/3)(pi)R^2H

volume of cone, V = (1/3)(pi)R^2H

ie volume of cylinder is 4/9 volume of cone.

e23r32rf