Well u must divide the area by the width this will give you the length, 5.8/2.25=2.5777778 which is around 2 3/5
1x18 2x24
2x9. 3x14
3x6. 6x7
6x3
GCF 6
The volume of such a can with base radius
and height
would be
We desire the base's circumference and the can's height to add to 120, i.e.
Substituting this into
allows us to reduce the volume to a function of a single variable
:
Taking the derivative with respect to
yields
Set this equal to 0 and find any critical points:
This suggests the can will have maximum volume when its radius is
cm, which would give a volume of about 20,371 sq. cm.