Problem 4.5.5
A rectangular area, 800ft2 in size, is to be enclosed along a building, using two walls of pine (wood)
fencing opposite each other, and a third wall of steel fencing, as in the picture.
The cost of the pine fencing is $6 per ‘running foot’. This means the pine fencing (all of a fixed
height, say, 6 ft tall, for example) is purchased by length. If you buy a length of 10 ft of pine fencing,
it costs $60. If you buy 30 feet, it costs you $180.
The cost of the steel fencing is $3 per running foot. Determine the dimensions of the enclosure that
minimize the (total) cost.
Solution:
STEP 1: Draw picture(s) and label variable/unknown quantities:
As I said in class, the picture given in the textbook is very confusing. It makes it seem like the
length of the steel side is exactly the width of the building, which would determine the dimensions.
I already drew a new picture above, but, in general, if you’re not given a picture, you should draw
one. Below, we label the unknowns:
1
w
=
length of wood sides
l
=
length of steel side
STEP 2: Write the quantity to be optimized (max/min’d) in terms of the unknowns:
Here, we are trying to minimize the total cost, C, (in dollars), of the enclosure. So we try to
write C in terms of w and l.
First, we have:
cost of wood used (in dollars)
=
6w + 6w = 12w
cost of steel used (in dollars)
=
3l
Since,
total cost of enclosure = cost of wood used + cost of steel used
=⇒
C = 12w + 3l
STEP 3: Use any constraints to a write a ‘constraint equation’ relating the variables.
The constraint here is that we must make the area enclosed by the fences equal to 800 ft2 . Hence,
w · l = 800
STEP 4: Use the constraint equation(s) to write the quantity you’re trying to optimize
as a function of one variable:
Since the area is required to be 800,
w · l = 800
we can write, say, w in terms of l:
w=
800
l
We can then plug this into our expression for the cost:
800
9, 600
C = 12w + 3l = 12
+ 3l =
+ 3l
l
l
2
Hence, we have expressed the total cost as a function of solely the length of the steel:
C(l) =
9, 600
+ 3l
l
STEP 5: Now optimize, by finding critical points, etc.
Critical Points:
Recall that when denominators are present, it is helpful to write your derivative as a single fraction.
0 0
3l2 − 9, 600
9, 600
−9, 600 3l2
+ 2 =
C0 =
+ 3l = 9, 600 · l−1 + 3l = −9, 600 · l−2 + 3 =
2
l
l
l
l2
C 0 = 0 ⇐⇒ 3l2 − 9, 600 = 0 ⇐⇒ l = ±
p
√
3, 200 = ±40 2
C 0 = undefined ⇐⇒ l2 = 0 ⇐⇒ l = 0
Since l is a length, we have l ≥ 0, but √
since we must have w · l = 800, l = 0 is not allowed. So the
only (pertinent) critical point is l = 40 2.
Endpoints:
This problem is actually a bit peculiar, as there are no ‘endpoints’. Technically, no one forbade
us from letting l = 9, 999, 999 (as long as use w = 800/9, 999, 999).
The only condition on l is that it be (strictly) positive.
In this case, we must rely on an analysis of C 0 , for which we will make a sign chart.
Sign chart for C 0 :
√
Hence, on the interval (0, ∞), C(l) decreases until l = 40 2, then increases (ever) after.
3
√
Consequently, l = 40 2 gives an absolute minimum on (0, ∞).
The corresponding value for w is:
√
√
√
800
40 · 20
20
20
20 · 2
2
√ =√ =√ ·√ =
w= √ =
= 10 2
2
40 2
40 2
2
2
2
4