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The discussion focuses on optimizing the dimensions of a right circular cylinder oil can with a volume of 54π cubic inches to minimize material usage. The optimal radius is confirmed to be 3 inches. The volume formula V = πr²h is utilized alongside the surface area formula A = 2πrh + 2πr² to derive the necessary dimensions. The minimum surface area is achieved by solving for height (H) in terms of radius (R) and applying calculus to find the minimum through the first derivative.

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Got It, Thanks Guys
 
Last edited:
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Jack Jiang said:
An oil can is to be made in the form of a right circular cylinder to contain 54pi cubic inches. what dimensions of the can will require the least amt. of material.did anyone get the radius to equal 3?

Right on, the min is 3.
V=(Pi)R^2=54(Pi)
A=2(Pi)RH+2(Pi)R^2
Solve for H in volume, find minimum through first derivative.
 
Last edited:
[tex]V = \pi r^{2}h[/tex]

Thus

[tex]A = 2\pi rh + 2\pi r^{2}[/tex]
 

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