show that the height of the cylinder of maximum volume that can be inscribed within a cone of height $h$ is...
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show that the height of the cylinder of maximum volume that can be inscribed within a cone of height $h$ is $frac{h}3$.
I have tried solving this sum but am unable to substitute the radius of the cylinder in terms of $h$.
derivatives maxima-minima
edited Nov 27 '17 at 21:02
Siong Thye Goh
94.6k1462114
asked Mar 19 '17 at 12:40
Dhruv Raghunath
151212
add a comment |
up vote
0
down vote
favorite
show that the height of the cylinder of maximum volume that can be inscribed within a cone of height $h$ is $frac{h}3$.
I have tried solving this sum but am unable to substitute the radius of the cylinder in terms of $h$.
derivatives maxima-minima
edited Nov 27 '17 at 21:02
Siong Thye Goh
94.6k1462114
asked Mar 19 '17 at 12:40
Dhruv Raghunath
151212
See this for finding the maximum volume of a cylinder in a cone of height $h$.
– heather
Mar 19 '17 at 13:03
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
show that the height of the cylinder of maximum volume that can be inscribed within a cone of height $h$ is $frac{h}3$.
I have tried solving this sum but am unable to substitute the radius of the cylinder in terms of $h$.
derivatives maxima-minima
edited Nov 27 '17 at 21:02
Siong Thye Goh
94.6k1462114
asked Mar 19 '17 at 12:40
Dhruv Raghunath
151212
show that the height of the cylinder of maximum volume that can be inscribed within a cone of height $h$ is $frac{h}3$.
I have tried solving this sum but am unable to substitute the radius of the cylinder in terms of $h$.
derivatives maxima-minima
derivatives maxima-minima
edited Nov 27 '17 at 21:02
Siong Thye Goh
94.6k1462114
asked Mar 19 '17 at 12:40
Dhruv Raghunath
151212
edited Nov 27 '17 at 21:02
Siong Thye Goh
94.6k1462114
asked Mar 19 '17 at 12:40
Dhruv Raghunath
151212
edited Nov 27 '17 at 21:02
Siong Thye Goh
94.6k1462114
edited Nov 27 '17 at 21:02
Siong Thye Goh
94.6k1462114
edited Nov 27 '17 at 21:02
Siong Thye Goh
94.6k1462114
94.6k1462114
asked Mar 19 '17 at 12:40
Dhruv Raghunath
151212
asked Mar 19 '17 at 12:40
Dhruv Raghunath
151212
asked Mar 19 '17 at 12:40
Dhruv Raghunath
151212
151212
See this for finding the maximum volume of a cylinder in a cone of height $h$.
– heather
Mar 19 '17 at 13:03
add a comment |
See this for finding the maximum volume of a cylinder in a cone of height $h$.
– heather
Mar 19 '17 at 13:03
See this for finding the maximum volume of a cylinder in a cone of height $h$.
– heather
Mar 19 '17 at 13:03
See this for finding the maximum volume of a cylinder in a cone of height $h$.
– heather
Mar 19 '17 at 13:03
add a comment |
1 Answer
1
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oldest
votes
up vote
0
down vote
Hint:
Look at the figure. With:
$$
BH=h quad HC=R quad FG=x quad HG=r
$$
from the similarity of the triangles $BHC$ and $FGC$ we have:
$$
h:x=R:(R-r)
$$
so that $r=frac{R}{h}(h-x)$
Now you can express the volume of the cylinder as $V=pi r^2x=pi frac{R^2}{h^2}(h-x)^2x $.
Maximize this function.
answered Mar 19 '17 at 13:46
Emilio Novati
50.8k43472
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Hint:
Look at the figure. With:
$$
BH=h quad HC=R quad FG=x quad HG=r
$$
from the similarity of the triangles $BHC$ and $FGC$ we have:
$$
h:x=R:(R-r)
$$
so that $r=frac{R}{h}(h-x)$
Now you can express the volume of the cylinder as $V=pi r^2x=pi frac{R^2}{h^2}(h-x)^2x $.
Maximize this function.
answered Mar 19 '17 at 13:46
Emilio Novati
50.8k43472
add a comment |
up vote
0
down vote
Hint:
Look at the figure. With:
$$
BH=h quad HC=R quad FG=x quad HG=r
$$
from the similarity of the triangles $BHC$ and $FGC$ we have:
$$
h:x=R:(R-r)
$$
so that $r=frac{R}{h}(h-x)$
Now you can express the volume of the cylinder as $V=pi r^2x=pi frac{R^2}{h^2}(h-x)^2x $.
Maximize this function.
answered Mar 19 '17 at 13:46
Emilio Novati
50.8k43472
add a comment |
up vote
0
down vote
up vote
0
down vote
Hint:
Look at the figure. With:
$$
BH=h quad HC=R quad FG=x quad HG=r
$$
from the similarity of the triangles $BHC$ and $FGC$ we have:
$$
h:x=R:(R-r)
$$
so that $r=frac{R}{h}(h-x)$
Now you can express the volume of the cylinder as $V=pi r^2x=pi frac{R^2}{h^2}(h-x)^2x $.
Maximize this function.
answered Mar 19 '17 at 13:46
Emilio Novati
50.8k43472
Hint:
Look at the figure. With:
$$
BH=h quad HC=R quad FG=x quad HG=r
$$
from the similarity of the triangles $BHC$ and $FGC$ we have:
$$
h:x=R:(R-r)
$$
so that $r=frac{R}{h}(h-x)$
Now you can express the volume of the cylinder as $V=pi r^2x=pi frac{R^2}{h^2}(h-x)^2x $.
Maximize this function.
answered Mar 19 '17 at 13:46
Emilio Novati
50.8k43472
answered Mar 19 '17 at 13:46
Emilio Novati
50.8k43472
answered Mar 19 '17 at 13:46
Emilio Novati
50.8k43472
answered Mar 19 '17 at 13:46
Emilio Novati
50.8k43472
50.8k43472
add a comment |
add a comment |
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See this for finding the maximum volume of a cylinder in a cone of height $h$.
– heather
Mar 19 '17 at 13:03