the radius r is given by
r^2+h^2 = s^2
r^2+h^2 = 25
h = √(25-r^2)
v = π/3 r^2 h = π/3 r^2 √(25-r^2)
So, for maximum v, find r when dv/dr = 0
Find the dimensions of the right circular cone of maximum volume having a slant height of 5 ft. See the figure.
2 answers
well, if it is right circular cone, max volume will occur when the axis of the cons is perpendicular to the base.
V=1/3 areabase*height.
but height^2=slant^2-r^2
V=1/3 PI r^2*sqrt(25-r^2)
let z be V^2
z=1/9 PI^2 (r^4*25-r^6(
dz/dr=1/9 PI^2 (4r^3*25-6r^5)
setting dz/dr=0 for max
4r^3*25=6r^5
100 =6r^2
r=10/sqrt6
and h=sqrt(25-100/6)
h=2.89
r=4.08
It seemed odd to me (I have no confidence, so I checked it here, they got identical results: http://mathcentral.uregina.ca/qq/database/qq.09.06/s/christina1.html
V=1/3 areabase*height.
but height^2=slant^2-r^2
V=1/3 PI r^2*sqrt(25-r^2)
let z be V^2
z=1/9 PI^2 (r^4*25-r^6(
dz/dr=1/9 PI^2 (4r^3*25-6r^5)
setting dz/dr=0 for max
4r^3*25=6r^5
100 =6r^2
r=10/sqrt6
and h=sqrt(25-100/6)
h=2.89
r=4.08
It seemed odd to me (I have no confidence, so I checked it here, they got identical results: http://mathcentral.uregina.ca/qq/database/qq.09.06/s/christina1.html
3 answers