A cylindrical can is to be made to hold2000cm³of oil. Find the dimensions that will minimize the cost of the metal to manufacture the can. SOLUTIONDraw the first diagram, whereris the radius andhthe height (both in centimeters). In order to minimize the cost of the metal, we minimize the total surface area of the cylinder (top, bottom, and sides). From the second figure, we see that the sides are made from a rectangular sheet with dimensions

2r

andh. So the surface area is

A= 2r²+

.

To eliminatehwe use the fact that the volume is given as2000cm³. Thus

r²h=2000

which gives

h=

.

Substitution of this into the expression forAgives

A	=	2r2+ 2r
	=	2r2+ .

Therefore the function we want to minimize is

A(r) =

, r> 0.

To find the critical numbers, we differentiate:

A'(r)	=	4r
	=	4 r2 .

Then

A'(r) = 0

when

r³= ,

so the only critical number is

r=

.

Since the domain ofAis

(0,),

we can't use the argument ofthis exampleconcerning endpoints. But we can observe that

A'(r) < 0

for

r<

and

A'(r) > 0

for

r>

,

soAis decreasing for allrto the left of the critical number and increasing for allrto the right. Thus

r=

must give give the absolute minimum. [Alternatively, we could argue that

A(r)

as

r0⁺

and

A(r)

as

r,

so there must be a minimum value of

A(r),

which must occur at the critical number. See the graph.] The value ofhcorresponding to

r=

3

1000/

is

h	=	2000 r2 = 2000
	=	= 2r.

Thus, to minimize the cost of the can, the radius should be

3

1000/

cm and the height should be equal to twice the radius, namely, the diameter.