SOLUTION: Hi. I need assistance with this geometry problem, preferably within a few hours (which is when I should be able to clearly explain how to solve this problem to my professor) and I

Question 716288: Hi. I need assistance with this geometry problem, preferably within a few hours (which is when I should be able to clearly explain how to solve this problem to my professor) and I will provide an excellent review. Thank you very much!
A tin can with a circular base has a volume of 27*pi*cm^3 and the surface area of the vertical (curved) side is 18*pi*cm^2. What are the dimensions of the can?
Radius = ______cm
Height = ______cm
Answer by josgarithmetic(39617) (Show Source): You can put this solution on YOUR website!
The volume has a height and a circumference. The circular side has the same height and circumference as the volume.
and . The area equation is like stretching the circular sheet out straight, which was the circumference, and then the other dimension would be height. Circumference times height is the 18 cm^2 area.

Those are two equations with two unknowns of r and h. Solve the equations as a system.

LET ME TRY ANALYZING A SOLUTION AGAIN BUT DIFFERENTLY.
r is radius, h is height.

The volume is given and for this cylinder shaped can, volume is cubic centimeters.

What about the curved circular side? If you cut it from top to bottom and flatten it into a rectangle, then you see the top to bottom height, h, and now the circumference, , is the "length". This area was given in the problem description and is cubic centimeters. The r and the h are the same as for the volume.

The system of equations to solve for r and h is this:

The equations really should be simplified before working with them to solve the system. Both equations have a factor of on their left and right sides. No need! For both equations, divide both sides by pi. Also, in the area equation, you should divide both sides by 2. You do those things and you obtain:

SIMPLIFIED SYSTEM

The route to a finished solution should now be very simple.
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