Optimization II
LRT
03/22/2017
Additional or alternative guidelines.
1. What are we trying to do?
2. If max/min, what quantity are we trying to max.min?
3. Find a formula for this quantity, possibly in terms of
several variables which occur in the problem.
4. Using additional information in the problem eliminate
all but one of the variables in your formula from part 3.
5. Determine a relevant interval for the problem.
6. See 4.4.
#5
The picture is a little misleading since it seems to indicate
that one side of the enclosure is the width of the store. If
that’s the case we would need to know the width of the
store and we don’t. And then it would be an arithmetic
problem, not a calculus problem.
Better picture:
Here w is the width of the
rectangle and h is the height.
The area is A = w · h = 800.
The cost is
C = 6w +3h+6w = 12w +3h.
S
t
o
r
e
Enclosure
w
h
#5 continued
Here w is the width of the
rectangle and h is the height.
The area is A = w · h = 800.
The cost is C = 12w + 3h.
S
t
o
r
e
Enclosure
w
The goal is to minimize the
h cost.
The formula for the cost has 2
variables in it, w and h.
We must find a formula for the cost which only involves a
single variable. It does not much matter whether we choose
w or h. Let’s use the constraint equation to solve for w as a
function of h.
#5 continued
Here w is the width of the
rectangle and h is the height.
The area is A = w · h = 800.
The cost is C = 12w + 3h.
S
t
o
r
e
Enclosure
Solve w · h = 800 for w as
800
.
h a function of h so w =
h
Then the cost is
w
C = 12
800
9600
+ 3h =
+ 3h
h
h
#5 continued
We want to minimize
9600
C=
+ 3h.
h
S
t
o
r
e
Enclosure
h
There are additional constraints. Clearly w > 0 and
h > 0.
w
More subtle is that if h is greater than the store width, we
will leave a gap in the enclosure. Since the problem does
not tell us what to do in this case, we have
h 6 width of store but since we don’t know this we’ll
denote it by S. Then the interval for h is (0, S].
#5 continued
C=
9600
+ 3h
h
dC
9600
=− 2 +3
dh
h
9600
We are looking for critical points so solve − 2 + 3 = 0
h
√
9600
2
= 3 3200 = h h = ±40 2 and of course
h2 √
h = 40 2.
d2 C
9600
= 2 · 3 > 0 so the curve is concave down on
2
dh
h
√
(0, ∞). Hence h = 40
√ 2 is a local minimum and
√ C is
decreasing on (0, 40 2) and increasing on (40 2, ∞).
For the interval (0, S] check lim+ C(h) = ∞ and C(S) is
h→0
the cost at the other end point.
Since C is concave
value of C
√ up everywhere the minimum
√
occurs at h = 40 2. Hence √
C(S) > C(40 2). √
If the store is wider than 40 2 ft then h = 40 2 is√
the
solution but if the width of the store is less than 40 2 then
h = S.
The question asked for the dimensions.
√
√
800
If h = 40 2, w = √ = 10 2.
40 2
If h = S, w =
800
.
S
The picture referred to below is the picture for problem #5
in the book.
Without the picture, I might have been tempted to put the
steel fence in the back.
In this case, the cost function becomes
C = 6(h + w) + 3w = 6h + 9w
C = 6h + 9w
Solve 800 = wh for h as a function of w this time. Then
800
4800
800
+ 9w =
+ 9w.
h=
so C(w) = 6
w
w
w
4800
dC
= − 2 + 9 so solve
dw
w
−
4800
+ 9 = 0,
w2
40
w = ±√
3
4800
= 9,
w2
40
w=√ .
3
d 2C
4800
= 2 · 3 > 0 on (0, ∞).
2
dw
w
w2 =
4800
1600
=
,
9
3
800
800
so
, the constraint h 6 S implies w >
h
h
800
the relevant interval for this problem is I =
,∞ .
S
Since w =
40
Since C is concave up everywhere on (0, ∞), w = √ is the
3
40
absolute minimum on this interval. On √ , ∞ , C is
3
increasing.
40
800
40
If √ >
, w = √ is the minimum location.
S
3
3
40
800
800
If √ 6
,w=
is the minimum location.
S
S
3
800
40
Notice that the crucial value for S is √ =
or
S
3
√
S = 20 3.
The dimensions are
√
40
w = √ , h = 20 3
3
√
if S > 20 3.
800
,h=S
S
√
if S 6 20 3.
w=
Notice that the store does not have to be as wide with this
configuration
in √
order to achieve the absolute minimum
√
since 20 3 < 40 2.
You were not asked, but the costs for these two
configurations can be determined as well.
In the first case.
√
√
√
√
h = 40 2, w = 10 2, C = 240 2
if S > 40 2.
h = S, w =
800
6S 2 + 2400
,C=
S
S
In the second case.
√
√
40
h = √ , w = 20 3, C = 260 3
3
h = S, w =
800
6S 2 + 7200
,C=
S
S
√
if S 6 40 2.
√
if S > 20 3.
√
if S 6 20 3.
In the first case.
√
√
√
h = 40 2, w = 10 2, C = 240 2
√
if S > 40 2.
800
6S 2 + 2400
,C=
S
S
√
if S 6 40 2.
h = S, w =
In the second case.
√
√
40
h = √ , w = 20 3, C = 260 3
3
h = S, w =
800
6S 2 + 7200
,C=
S
S
√
if S > 20 3.
Looking at the data, √
we
see that if S > 40 2
then it is cheaper to
build the first configura√
tion since 240 < 260 3.
√
if S 6 20 3.
√
If S 6 20 3 it is cheaper to to build the first configuration
6S 2 + 2400
6S 2 + 7200
<
.
S √
S √
What about 20 3 < S < 40 2? We can build the first
since
configuration at a cost of
√
cost of 260 3.
6S 2 + 2400
and the second for a
S
Hence we are asking when is
√
√
20 3 < S < 40 2.
√
6S 2 + 2400
6 260 3 for
S
Ignore the interval for a moment and find out where
√
6S 2 + 2400
6 260 3.
S
√
6S 2 + 2400
= 260 3,
√S
6S 2 − 260 3S + 2400 = 0.
Solve
√
6S 2 + 2400 = 260 3S,
The roots are r1 =≈ 5.7735026919 and r2 ≈ 69.2820323028.
Take S = 10, which lies between the two roots and check
√
6(10)2 + 2400
3000
=
= 30 < 260 3.
10
10
Hence
√
6S 2 + 2400
< 260 3 between r1 and r2 .
S
√
6S 2 + 2400
< 260 3 between the two roots and since
S
√
5.7735026919
≈ r1 < 20 3 and
√
40 2 < r2 ≈ 69.2820323028, the first configuration is
cheaper no matter what the width of the store is.
Since
We talked some about problem #29 and how you needed
some interpretation in order to make it doable.