Asked by tonya
A cylindrical can is to hold 20 pie mraised to the 3rd power. The material for the top and bottom costs $10/m raised to the 2nd power and material for the side costs $8/m to the 2nd power. Find the radius r and height h of the most economical can.
Answers
Answered by
drwls
Volume = pi r^2 h = 20 pi (m^3)
Therefore r^2 h = 20
h = 20/r^2
Cost = 2 pi r^2 * 10 + 2 pi r h * 8
= 20 pi r^2 + 16 pi r(20/r^2)
= 20 pi r^2 + 320 pi/r
Set the derivative with respect to r equal to zero.
40 pi r = 320 pi r^-2
r^3 = 80
r = 4.31 m
Check my math
Therefore r^2 h = 20
h = 20/r^2
Cost = 2 pi r^2 * 10 + 2 pi r h * 8
= 20 pi r^2 + 16 pi r(20/r^2)
= 20 pi r^2 + 320 pi/r
Set the derivative with respect to r equal to zero.
40 pi r = 320 pi r^-2
r^3 = 80
r = 4.31 m
Check my math
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