



In the
surface type problem, we usually need to find the minimum surface
area



needed to
enclose a certain volume of some stuff. 


The key
to this problem is the shape of the container. 


Often,
the container is a cylinder (like a can). 


The
equations for this shape container are a little more complex 


than
those for a box. 





Example: 





What is
the least surface area can (cylinder) 


that will enclose 15 cubic
inches of stuff. 


The can
DOES have a lid. 





To solve
this, 


we first need the equation for the surface area and volume of
a cylinder. 











So we
have two unknown variables, the radius (r) and the height (h). 


We need
to solve for one in terms of the other, 


so that
we can do calculus on a Surface Area equation 


with just one
independent variable. 


We know
the total volume (15 in^{3}), so we can use the volume
equation to do that... 





Volume
(V) = pr^{2}h



15 = pr^{2}h 








Substituting
this into the surface area equation for h, we get ... 





A = 2pr( 
15 
)
+ 2pr^{2} 

pr^{2} 



Simplifying
this: 








Taking
the derivative of this, 


Remember,
r is the variable and p
is just a number 





A
= 30r^{1} + 2pr^{2}



A'
= 30r^{2} + 4pr






Find
where the slope is zero ... 


0
= 30r^{2} + 4pr



30r^{2}
+ 4pr



Multiply
both sides by r ^{2} 


30
+ 4pr^{3}



2.387 = r^{3} 



= r 



1.336
»
r






When we
solved for h, we found that 





So 








h
» 2.675






So the
surface area of the can is: 





A 
=
2prh 
+
2pr^{2} 
A 
= 2p(1.336)(2.675) 
+ 2p(1.336)^{2} 
A 
= 22.455 
+
11.215 
A 
=
33.670 square inches 







To check
to see if this is a minimum or maximum area, find the second
derivative 


find the
second derivative of the Area equation and check for concavity. 





POP QUIZ: 


We are
looking for a MINIMUM area, 


do we
want the second derivative value to be positive or negative? 





A'
= 30r^{2} + 4pr



A''
= 60r^{3} + 4p



Which we
can write as ... 








Looking
at this, you can probably see that the only place that A'' can be
negative 


is where
r is negative. 


And a
negative radius has no meaning in this problem. 





The
radius value at the A' = 0 point was 1.336, 


so plug that into the
second derivative equation ... 





A''
= 
60 
+ 4p 

(1.336)^{
3} 






A''
= 44.910 + 12.566



A''
= 57.476 





57.476 is
a positive number so the area is a minimum. 


And that
area was 33.670 square inches. THAT IS OUR ANSWER. 


It is WAY
common in these problems to work everything out, 


but not specify
what your answer is. 


Some
teachers will cut you slack on that one, 


but most will hit you for
at least a few points. 





Let's
look at those dimensions again. 


The
radius we found was 1.336, that makes the diameter 2.672 


The
height we found was 2.675 





Except
for rounding error, that is the same value. 


Interesting
eh? 





The
smallest surface area of any shape for a given volume is a sphere. 


The
smallest surface area of a box type shape for a given volume is a
cube. 


A can
with the same diameter and height 


is the cylinder equivalent of a
sphere or cube. 





A
variation on this problem 


is where the material used for the
different parts of the container 


cost
different amounts of money per square area unit. 


The
problem is to spend the least amount of money on the container. 


This is a
nearly identical problem to the one we just did. 





If we
changed the problem we did 


and said
that the material for the top and bottom of the cylinder 


cost $3 per
square inch and the
material for the sides cost $2 per square inch. 





Now to
find the least amount of money we would spend building the
container, 


the area
equation: 


A
= 2prh
+ 2pr^{2}






Would
become the cost equation: 





$
= ($2) x 2prh
+ ($3) x 2pr^{2}






And we
would work the problem the exact same way. 





1)
Substitute for h in the volume equation 


2) Find
the derivative of the $ equation, that is, $' 


3) Set $'
equal to zero and solve for r 


4) Use
the volume equation to find h 


5) Find
the second derivative of $, that is, $'' and check for concavity. 





Hmmm,
sounds like a good problem to use in homework, eh? 





copyright 2005 Bruce Kirkpatrick 
