Asked by JEZRELL
A company wants to manufacture cylindrical aluminum cans with a value of 1000cm^3 (1 Liter) What radius and height of the can be to minimize the amount of aluminum used? Please help! This is very hard. Thanks.
Answers
Answered by
Steve
You know that
pi r^2 h = 1000, so
h = 1000/(pi r^2)
the surface area is
a = 2pi r^2 + 2pi r h
= 2pi r^2 + 2pi r * 1000/(pi r^2)
= 2pi r^2 + 2000/r
da/dr = 4pi r - 2000/r^2
= (4pi r^3 - 2000)/r^2
Since r > 0, da/dr=0 when
4pi r^3 = 2000
r^3 = 2000/4pi = 500/pi
r = 5 ∛(4/pi) ~= 5.42
go to wolframalpha.com and enter
plot y = 2pi r^2 + 2000/r where 3<r<8
to see where it is minimum
pi r^2 h = 1000, so
h = 1000/(pi r^2)
the surface area is
a = 2pi r^2 + 2pi r h
= 2pi r^2 + 2pi r * 1000/(pi r^2)
= 2pi r^2 + 2000/r
da/dr = 4pi r - 2000/r^2
= (4pi r^3 - 2000)/r^2
Since r > 0, da/dr=0 when
4pi r^3 = 2000
r^3 = 2000/4pi = 500/pi
r = 5 ∛(4/pi) ~= 5.42
go to wolframalpha.com and enter
plot y = 2pi r^2 + 2000/r where 3<r<8
to see where it is minimum
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