



So we spent
all that time with maxima and minima



because most of the problems
you will see deal with
finding



the maximum or minimum amount of something with a given
amount of input. 





That is
actually a good definition of what engineering is all about. 


Things
like: 


How big a
bridge can you build with a certain amount of materials. 


How many
people can you push through the Haunted Mansion ride in a day. 


What's
the cheapest computer that can fly a plane safely. 


How many
cows can you keep alive on a certain farm 


(Note: this is not the exact same problem as #2 above). 





Anyway, the
drill is we will be given some problem, generally as a paragraph. 


We
convert that to a function 


(like it was in Algebra, this is usually
the hardest part). 


And use
the maxima and minima techniques from the last few chapters 


to find
the answer. 





Area
problems are some of the least complex types so we'll start with
those. 


There are
two types of area problems. Fence types and surface types. 


Generally
fence types deal with finding 2 dimensional areas 


and
surface types deal with finding 3 dimensional volumes. 





In the
fence type, we usually need to enclose the maximum area possible 


with a
certain amount of fence, or we
might need to enclose a certain area 


with the minimum amount of
fence. 





This will
give us these two equations: 





2W
+ 2L = Total Length of Fence



Area
Fenced = W x L






Sometimes
we are given the total amount of fence 


and want to know how much
area can be fenced. 





We are
going to do the maxima and minima stuff with our equations
above, 


but they
have too many variables. 


We can
cut that down by solving the first equation for L ... 





2W
+ 2L = 
Total
Length of Fence 
2L
= 
Total
Length of Fence  2W 



L
= 
Total
Length of Fence 

W 

2 






Remember,
in this style of problem the total length of fence is given, 


it's a
number so that
messy fraction in the middle will just be some number. 





So now we
just substitute this value in for L in the second equation and get: 





Area
Fenced = 
W x L 






Area
Fenced = 
W x
( 
Total
Length of Fence 

W) 

2 






Multiply
this out and rearrange a bit, and we get: 





Area
Fenced = 
W^{2}
+ ( 
Total
Length of Fence 
)W 

2 






This is
actually just an ordinary quadratic function in disguise. 


An
ordinary quadratic generally looks like this: 





F(X) = AX^{2}
+ BX + C 





Area
Fenced is playing the part of F(X) 


W is
playing the part of X. 


A is 1 


B is the
goofy Total Length of Fence/2 thing 


C is
zero, so it doesn't even show up 





So take
the derivative of this equation, find where the slope is zero 


and that
or an end point will be the answer candidates. 





There are
two common variations of this problem. 


In one
variation, one of the sides of the area to be fenced is a river 


or
something else that
doesn't need to be fenced 








Since we
get one of the W's for free, that makes the "Length of
Fence" equation: 





W
+ 2L = Total Length of Fence






Other
than that, it's the same problem. 





In the
other variation, you have to fence in two areas 





Since we
need an extra L, that makes the "Length of Fence"
equation: 





2W
+ 3L = Total Length of Fence 





Other
than that, it's the same problem. 





Neither
of these variations change the area equation. 


It will
still be: 


Area
Fenced = W x L






Yeesh,
enough already, let's do one! 


OK, we
have 1200 feet of fence 


and we need to enclose as big an area as we
can into two sections... 











Here we
go. Since we have two sections and 1200 feet of fence, 


one of our
equations will be: 





2W
+ 3L = 1200






The area
equation is still: 


Area
Fenced = W x L






We want
to find the maximum area, 


so we need to solve the first equation for
one of the variables 


(either
one will do), 


then we
need to substitute that into the area equation, 


find the
first derivative of the area equation, 


and do
our maxima stuff. 


We are
only looking for the maximum area, so we don't need the minima this
time. 





OK, lets
solve the first equation for W 





2W
+ 3L = 
1200 


2W
= 
1200
 
3L 





W
= 
600
 
3 
L 

2 






So
substituting for W in the area equation and multiplying out ... 





A
= 
L
x W 







A
= 
L
x (600
 
3 
L) 

2 





A
= 
3 
L^{2}
+ 
600L 

2 











A local
minima or maxima will happen anywhere the slope is zero, 


so we
need the derivative of the area equation, we'll call it A': 





A
= 
3 
L^{2}
+ 600L 

2 


A'
= 
3L + 600 






Now find
the L value where A' = 0 ... 





A'
= 
0 = 3L +
600 
3L
= 
600 
L
= 
200 






So where
the slope is zero, L = 200. 


Now find
W at that point. 





We know
that 2W + 3L = 1200 so: 





2W
+ 3L = 1200



2W
+ 3(200) = 1200



2W
+ 600 = 1200



2W
= 600 


W
= 300 





So the
first suspect point is W = 300, L = 200 





What are
the end points? 


Well the
lengths can't be negative, 


so the endpoints are where one or the
other (L or W) is zero. 


Find
these points ... 





2W
+ 3L = 1200 
2W
+ 3L = 1200 
2W
+ 3(0) = 1200 
2(0)
+ 3L = 1200 
2W
= 1200 
3L
= 1200 
W
= 600 
L
= 400 






So the
candidates for the maximum area are: 





W
= 600, L = 0 
an endpoint 
W
= 0, L = 400 
the other
endpoint 
W
= 300, L = 200 
the A' = 0
point 






Solving
the area equation for these points: 





A = W x L 
A = W x L 
A = W x L 
A = 600 x 0 
A = 0 x 400 
A = 300 x
200 
A = 0 
A = 0 
A = 60,000 
(not very
big) 
(not any
better) 
(A Winner!) 






copyright 2005 Bruce Kirkpatrick 
