Question

anyone get this? Find the most economical dimension for a closed cylindrical can containing a quart..
his solution is:
For the question posed in the body of the post: I believe the term "most economical dimensions" means using the least possible amount of material for the can. In other words - find the minimum possible surface area for a can that has volume = 1 quart. The area of the can is equal to the area of the bottom + lid + sides:

Answer 1

....For the question posed in the body of the post: I believe the term "most economical dimensions" means using the least possible amount of material for the can. In other words - find the minimum possible surface area for a can that has volume = 1 quart. The area of the can is equal to the area of the bottom + lid + sides:



The volume must equal 1 quart, which is 57.75 cubic inches:



Rearrange to get:



You can now substitute this value for H into the first equation to get the area as a function of R alone. Then take the derivative of A with respect to R and set that to zero - this will let you determine a value for R that yields either a minimum or maximum value of A. You can tell which it is by considering the second derivative derivateive - if the second derivative is positive at this value of R then you have the minimum for A, and have found the radius that gives the minimal amount of surface area for a quart can.