A closed can, in a shape of a circular, is to contain 500cm^3 of liquid when full. The cylinder, radius r cm and height h cm, is. made from thin sheet metal. The total external surface area of the cylinder is A cm^2. Show that A=2πr^2 + 1000/r.

To express the height as a function of the volume and the radius, we are going to use the volume formula for a cylinder: [tex]V= \pi r^2h[/tex]
where
[tex]V[/tex] is the volume 
[tex]r[/tex] is the radius 
[tex]h[/tex] is the height 

We know for our problem that the cylindrical can is to contain 500cm^3 when full, so the volume of our cylinder is 500cm^3. In other words: [tex]V=500cm^3[/tex]. We also know that the radius is r cm and height is h cm, so [tex]r=rcm[/tex] and [tex]h=hcm[/tex]. Lets replace the values in our formula:
[tex]V= \pi r^2h[/tex]
[tex]500cm^3= \pi (rcm^2)(hcm)[/tex]
[tex]500cm^3=h \pi r^2cm^3[/tex]
[tex]h= \frac{500cm^3}{ \pi r^2cm^3} [/tex]
[tex]h= \frac{500}{ \pi r^2} [/tex]

Next, we are going to use the formula for the area of a cylinder: [tex]A=2 \pi rh+2 \pi r^2[/tex]
where
[tex]A[/tex] is the area 
[tex]r[/tex] is the radius 
[tex]h[/tex] is the height

We know from our previous calculation that [tex]h= \frac{500}{ \pi r^2} [/tex], so lets replace that value in our area formula:
[tex]A=2 \pi rh+2 \pi r^2[/tex]
[tex]A=2 \pi r(\frac{500}{ \pi r^2})+2 \pi r^2[/tex]
[tex]A= \frac{1000}{r} +2 \pi r^2[/tex]
By the commutative property of addition, we can conclude that:
[tex]A=2 \pi r^2+\frac{1000}{r}[/tex]