Asked by mike
Find the minimum amount of sheet that can be made into a closed clinder having a volume of 120 cu inches.
Answers
Answered by
Reiny
let the radius of the cylinder be r inches
and its height be h inches
given: V = 120 inches^3
or
π r^2 h inches^3 = 120 inches^3
h = 120/(πr^2)
Surface area
= 2πr^2 + 2πrh
= 2πr^2 + 2πr(120/πr^2)
= 2πr^2 + 240/r
d(surface area)/dr = 4πr - 240/r^2
= 0 for a min of surface area
4πr = 240/r^3
r^3 = 60/π
r = (60/π)^(1/3) = appr 2.673 inches
sub that back into
2πr^2 + 240/r to find the min surface area
and its height be h inches
given: V = 120 inches^3
or
π r^2 h inches^3 = 120 inches^3
h = 120/(πr^2)
Surface area
= 2πr^2 + 2πrh
= 2πr^2 + 2πr(120/πr^2)
= 2πr^2 + 240/r
d(surface area)/dr = 4πr - 240/r^2
= 0 for a min of surface area
4πr = 240/r^3
r^3 = 60/π
r = (60/π)^(1/3) = appr 2.673 inches
sub that back into
2πr^2 + 240/r to find the min surface area
Answered by
Damon
A = 2 pi r h + 2 *pi r^2
V = 120 = pi r^2 h
so h = 120/ (pi r^2)
A = 2 pi r 120/ (pi r^2) + 2 pi r^2
A = 240 / r + 2 pi r^2
dA/dr = -240/r^2 + 4 pi r
= 0 for max or min
4 pi r = 240/r^2
r^3 = 60 / pi
do not trust my arithmetic, working fast
V = 120 = pi r^2 h
so h = 120/ (pi r^2)
A = 2 pi r 120/ (pi r^2) + 2 pi r^2
A = 240 / r + 2 pi r^2
dA/dr = -240/r^2 + 4 pi r
= 0 for max or min
4 pi r = 240/r^2
r^3 = 60 / pi
do not trust my arithmetic, working fast
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.