The right circular cylinder's volume is where h is the height of the cylinder and r is the radius of its circular base. The quantity of aluminum used up by the can must be taken into consideration since waste . The capacity of the can that can hold the least amount of aluminum is , We can only translate A into one variable. Let's translate A into r. So have to express h in terms of r and we know that . Thus we have:
So the surface area in terms of r is
Let's now solve for r and distinguish A with respect to r by setting it to 0:
Therefore, the ratio of h to r is
(Note: Since is now larger than 2 to 1 and less than 4,)
lateral area without waste
If you use a 2r square, the base's area is
area of bottom ( for same reason )
Then Total area
Now can volume is
Considering both derivatives
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