Asked by skorge
find the economical proportion between the radius and height of the cylindrical can to give the least dimensions of a metal that encloses vloume of 10cu.in
Answers
Answered by
Steve
v = πr^2 h = 10
so, h = 10/(Ï€r^2)
The surface area is
a = 2Ï€r(r+h) = 2Ï€r(r + 10/(Ï€r^2))
= 2Ï€r^2 + 20/r
da/dr = 4Ï€ - 20/r^2 = (4Ï€r^2-20)/r^2
da/dr=0 when 4πr^2=20, or r=√(5/π)
so, h = 10/(Ï€*5/Ï€) = 2
so, h = 10/(Ï€r^2)
The surface area is
a = 2Ï€r(r+h) = 2Ï€r(r + 10/(Ï€r^2))
= 2Ï€r^2 + 20/r
da/dr = 4Ï€ - 20/r^2 = (4Ï€r^2-20)/r^2
da/dr=0 when 4πr^2=20, or r=√(5/π)
so, h = 10/(Ï€*5/Ï€) = 2
Answered by
Steve
Hmmm. I see that my characters for pi and sqrt have been mangled. They appeared correctly as I typed them, but on redisplay are junk.
r = sqrt(5/pi), h=2
r = sqrt(5/pi), h=2
Answered by
Calculus Student
Can you re write your solution Steve?
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