Differential Calculus Applied Optimization Problem?
Guys, I need your help.
The problem is:
Prove that the right circular cylinder of greatest volume that can be inscribed in a right circular cone has volume that is (4/9) times the volume of the cone.
cone's volume = (1/3)πr²h
cylinder's volume = πr²h
I really don't know the initial process. An idea perhaps will do (at least just the process).
Any help, of course, will be of great help.
This is not an assignment, but a boardwork which i'll be presenting to my instructor.
Thank you so much!
- ?Lv 78 years agoFavorite Answer
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.
- Anonymous6 years ago