202
CHAPTER FOURTEEN /SOLUTIONS
Lastly, when k = 1 the discriminant is zero, so the second derivative test can tell us nothing. Luckily,
we can factor f (x; y) when k = 1. When k = 1,
f (x; y) = x2 , 2xy + y2 = (x , y)2 .
This is always greater than or equal to zero. So f (0; 0) = 0 is a minimum and the surface is a trough-shaped
parabolic cylinder with its base along the line x = y.
When k = ,1,
f (x; y) = ,x2 , 2xy , y2 = ,(x + y)2 .
This is always less than or equal to zero. So f (0; 0)
with its top ridge along the line x = ,y.
y
k = ,2
,16
,12
,8
,4
=
0 is a maximum. The surface is a parabolic cylinder,
y
k = ,1
,30
,20
,1
x
,5
,1
12
8
,5
4
x
,10
,8
,20
y
,4
1
,1
,1
1
,4
,8
x
4
8
,12
,30
k=1
,16
,12
16
,10
,1
y
k=0
,16
k=2
12
16
y
16
12
30
20
8
10
4
5
1
1
x
x
1
5
10
20
30
Figure 14.6
Solutions for Section 14.2
1. Mississippi lies entirely within a region designated as “80s” so we expect both the high and low daily
temperatures within the state to be in the 80s. The South-Western most corner of the state is close to a region
designated as “90s”, so we would expect the temperature here to be in the high 80s, say 87-88. The northern
most portion of the state is relatively central to the “80s” region. We might expect the temperature there to
be between 83-87.
14.2 SOLUTIONS
203
Alabama also lies completely within a region designated as “80s” so both the high and low daily
temperatures within the state are in the 80s. The south-eastern tip of the state is close to a “90s” region so we
would expect the temperature here to be 88-89 degrees. The northern most part of the state is relatively
central to the “80s” region so the temperature there is 83-87 degrees.
Pennsylvania is also in the “80s” region, but it is touched by the boundary line between the “80s” and
a “70s” region. Thus we expect the low daily temperature to occur there and be about 80 degrees. The state
is also touched by a boundary line which contains a “90s” region so the high will occur there and be 89-90
degrees.
New York is split by a boundary between an “80s” and a “70s” region, so the northern portion of the
state is likely to be about 74-76 while the southern portion is likely to be in the low 80s, maybe 81-84 or so.
California contains many different zones. The northern coastal areas will probably have the state daily
low at 65-68, although without another contour on that side, it is difficult to judge how quickly the temperature
is dropping off to the west. The tip of Southern California is in a 100s region, so there we expect the state
daily high to be 100-101.
Arizona will have a low around 85-87 in the northwest corner and a high in the 100s, perhaps 102-107
in its southern regions.
Massachusetts will probably have a high around 81-84 and a low at 70.
2. Since f (x; y) 0 for all x; y and sincef (0; 0) = 0, the function has a global maximum (it is 0) and no global
minimum.
3. Suppose x is fixed. Then for large values of y the sign of f is determined by the highest power of y, namely
y3 . Thus,
f (x; y) ! 1 as y ! 1
f (x; y) ! ,1 as y ! ,1:
So f does not have a global maximum or minimum.
4. The function f has no global maximum or global minimum; g has a global minimum (it is 0) but no global
maximum; h has no global maximum or minimum.
5. To maximize z = x2 + y2 , it suffices to maximize x2 and y2 . We can maximize both of these at the same
time by taking the point (1; 1), where z = 2. It occurs on the boundary of the square. (Note: We also have
maxima at the points (,1; ,1); (,1; 1) and (1; ,1) which are on the boundary of the square.)
To minimize z = x2 + y2 , we choose the point (0; 0), where z = 0. It does not occur on the boundary
of the square.
6. To maximize this function, it suffices to maximize x2 and minimize y2 . We can do this by choosing the point
(1; 0), or (,1; 0) where z = 1. These occur on the boundary of the square.
To minimize z = x2 , y2 , it suffices to maximize y2 and minimize x2 . We can do this by taking the
point (0; 1), or (0; ,1) where z = ,1. These occur on the boundary of the square.
7. To maximize z = ,x2 , y2 it suffices to minimize x2 and y2 . Thus, the maximum is at (0; 0), where z = 0.
It doesn’t occur on the boundary of the square.
To minimize z = ,x2 , y2 , it suffices to maximize x2 and y2 . Do this by taking the point (1; 1),
(,1; ,1), (,1; 1), or (1; ,1) where z = ,2. These occur on the boundary of the square.
8. (a) This tells us that an increase in the price of either product causes a decrease in the quantity demanded
of both products. An example of products with this relationship is tennis rackets and tennis balls. An
increase in the price of either product is likely to lead to a decrease in the quantity demanded of both
products as they are used together. In economics, it is rare for the quantity demanded of a product to
increase if its price increases, so for q1 , the coefficient of p1 is negative as expected. The coefficient of
p2 in the expression could be either negative or positive. In this case, it is negative showing that the two
products are complementary in use. If it was positive, however, it would indicate that the two products
are competitive in use, for example Coke and Pepsi.
204
CHAPTER FOURTEEN /SOLUTIONS
(b) The revenue for A would be q1 p1 = 150p1 , 2p21 , p1 p2 , and the revenue for
200p2 , p1 p2 , 3p22 . The total sales revenue of both products, R, would be
B would be q2 p2
=
R(p1; p2) = 150p1 + 200p2 , 2p1p2 , 2p21 , 3p22 :
Note that R is a function of p1 and p2 . To find the critical points of R, set rR = 0, i.e.,
@R = @R = 0:
@p1 @p2
This gives
and
@R
@p1
=
150 , 2p2 , 4p1
=
0
@R
@p2
=
200 , 2p1 , 6p2
=
0
Solving simultaneously, we have p1 = 25 and by substituting in, we get p2
(25; 25) is a critical point for R. Further,
@ 2 R = ,4; @ 2 R = ,6; @ 2 R
@p1 @p2
@p21
@p22
=
=
25. Therefore the point
,2;
so the discriminant at this critical point is
D = (,4)(,6) , (,2)2 = 20:
Since D > 0 and @@pR2 < 0, this critical point is a local maximum. Since R is quadratic in p1 and
1
this is a global maximum. Therefore the maximum possible revenue is
2
p2,
R = 150(25) + 200(25) , 2(25)(25) , 2(25)2 , 3(25)2
2
2
= (6)(25) + 8(25)
=
4375:
This can be obtained when p1
9. The total revenue is
and as q = q1 + q2 , this gives
=
, 7(25)2
p2 = 25. Note that at these prices, q1 = 75 units, and q2 = 100 units.
R = pq = (60 , 0:04q)q = 60q , 0:04q2;
R = 60q1 + 60q2 , 0:04q12 , 0:08q1q2 , 0:04q22:
Therefore, the profit is
P (q1; q2) = R , C1 , C2
2
2
= ,13:7 + 60q1 + 60q2 , 0:07q1 , 0:08q2 , 0:08q1q2 :
At a local maximum point, we have grad P = ~0 :
@P = 60 , 0:14q , 0:08q = 0;
1
2
@q1
@P
@q = 60 , 0:16q2 , 0:08q1 = 0:
2
205
14.2 SOLUTIONS
Solving these equations, we find that
q1 = 300
q2 = 225:
and
To see whether or not we have found a local maximum, we compute the second-order partial derivatives:
@2P
@q12
Therefore,
=
2
,0:14; @@qP2
2
2
2
2
P
D = @ P2 @ P2 , @q@ @q
@q1 @q2
1
2
P = ,0:08:
,0:16; @q@ @q
1
2
2
=
,0:14)(,0:16) , (,0:08)2 = 0:016;
=(
and so we have found a local maximum point. The graph of P (q1 ; q2) has the shape of an upside down
paraboloid since P is quadratic in q1 and q2 , hence (300; 225) is a global maximum point.
10.
R , C = p1q1 + p2 q2 , 2q12 , 2q22 , 10.
p1
@P
@q1 = p1 , 4q1 = 0 gives q1 = 4
@P = p , 4q = 0 gives q = p2
2
2
@q2 2
4
(a) The revenue R = p1 q1 + p2 q2 . Profit = P
2
Since @@qP2
@ 2 P2
@q2
1
2
(,4)(,4) > 0 and @ P2
@q1
,4,
,4 and @q@ @qP
=
p =4; p2=4) we have that the discriminant, D =
< 0, thus P has a local maximum value at (q1; q2) = (p1 =4; p2=4). Since P is
p p
p p
p
p
quadratic in q1 and q2 , this is a global maximum. So P = 4 + 4 , 2 16 , 2 16 , 10 = 8 + 8 , 10 is
=
=
2
1
2
=
0, at
( 1
2
1
2
2
2
1
2
2
2
1
2
2
the maximum profit.
(b) The rate of change of the maximum profit as p1 increases is
@ (max P ) = 2p1
@p1
8
=
p1 :
4
11. We calculate the partial derivatives and set them to zero.
@ (range) = ,10t , 6h + 400 = 0
@t
@ (range) = ,6t , 6h + 300 = 0:
@h
10t + 6h = 400
6t + 6h = 300
solving we obtain
so
4t = 100
t = 25
Solving for h, we obtain 6h = 150, yielding h = 25. Since the range is quadratic in h and t, the second
derivative test tells us this is a local and global maximum. So the optimal conditions are h = 25% humidity
and t = 25 C.
206
CHAPTER FOURTEEN /SOLUTIONS
12. We want to maximize the theater’s profit, P , as a function of the two variables (prices) pc and pa . As always,
P = R , C , where R is the revenue, R = qcpc + qa pa , and C is the cost, which is of the form C = k(qc + qa )
for some constant k. Thus,
P (pc; pa ) = qc pc + qa pa , k(qc + qa)
,4
,1
,2
,3
= rpc , krpc + spa , kspa
To find the critical points, solve
@P = ,3rp,4 + 4krp,5 = 0
c
c
@pc
@P = ,sp,2 + 2ksp,3 = 0:
a
a
@pa
We get pc = 4k=3 and pa = 2k.
This critical point is a global maximum by the following useful, general argument. Suppose that
F (x; y) = f (x) + g(y), where f has a global maximum at x = b and g has a global maximum at y = d.
Then for all x; y:
F (x; y) = f (x) + g(y) f (b) + g(d) = F (b; d);
so F has global maximum at x = b; y = d.
The profit function in this problem has the form
P (pc ; pa) = f (pc ) + g(pa );
and the usual single-variable calculus argument using f 0 and g0 shows that pc = 4k=3 and pa = 2k are global
maxima for f and g, respectively. Thus the maximum profit occurs when pc = 4k=3 and pa = 2k. Thus,
pc 4k=3 2
pa = 2k = 3 :
13. The function f (x; y) in Example 3 is given by
80
f (x; y) = xy
+ 10x + 10xy + 20y:
This has critical points when fx
=
fy = 0.
fx (x; y) = ,x280y + 10 + 10y:
fy (x; y) = ,xy802 + 10x + 20:
Substituting x = 2, y
=
1 gives
fx (2; 1) = 2,2 80
1 + 10 + 10:1 = 0
fy (2; 1) = 2,8012 + 10:2 + 20 = 0:
So our point, (2; 1), is a critical point.
14.2 SOLUTIONS
207
To determine if this critical point is a minimum we use the second derivative test.
fxx = x160y ;
160
fyy = xy
;
fxy = x80y + 10;
3
3
2 2
So D
=
20 80 , 302
=
fxx (2; 1) = 20;
fyy (2; 1) = 80;
fxy (2; 1) = 30:
700 > 0 and fxx (2; 1) > 0, therefore the point (2; 1) is a local minimum.
14.
h
w
l
Figure 14.7
The box is shown in Figure 14.7. Cost of four sides = (2hl + 2wh)(1)c/. Cost of two bottoms = (2wl)(2)c/.
Thus the total cost C (in cents) of the box is
C = 2(hl + wh) + 4wl:
But volume wlh = 512, so l
=
512=(wh), thus
2048
C = 1024
w + 2wh + h :
To minimize C , find the critical points of C by solving
Ch = 2w , 2048
h2 = 0;
Cw = 2h , 1024
w 2 = 0:
We get
2wh2 = 2048
2hw2 = 1024:
Since w, h 6= 0, we can divide the first equation by the second giving
2wh2
2hw2
=
2048
;
1024
so
h = 2;
w
thus
h = 2w:
208
CHAPTER FOURTEEN /SOLUTIONS
Substituting this in Ch = 0, we obtain h3 = 2048, so h = 12:7 cm. Thus w = h=2 = 6:35 cm, and
l = 512=(wh) = 6:35 cm. Now we check that these dimensions minimize the cost C. We find that
4096 2048
2
2
=(
D = Chh Cww , Chw
h3 )( w3 ) , 2 ;
and at h = 12:7, w = 6:35, Chh > 0 and D = 16 , 4 > 0, thus C has a local minimum at h = 12:7
and w = 6:35. Since C increases without bound as w; h ! 0 or 1, this local minimum must be a global
minimum.
Therefore, the dimensions of the box that minimize the cost are w = 6:35 cm, l = 6:35 cm and h = 12:7
cm.
15.
h
w
l
Figure 14.8
Let w, h and l be width, height and length of the suitcase in cm. Then its volume V = lwh, and
w + h + l 135. To maximize the volume V , choose w + h + l = 135, and thus l = 135 , w , h,
V = wh(135 , w , h)
2
2
= 135wh , w h , wh
Differentiating gives
Vw = 135h , 2wh , h2;
Vh = 135w , w2 , 2wh:
Find the critical points by solving Vw = 0 and Vh = 0:
Vw = 0 gives 135h , h2 = 2wh;
Vh = 0 gives 135w , w2 = 2wh:
As hw 6= 0, we cancel h (and w respectively) in the above equations and get
135 , h = 2w
135 , w = 2h
Subtracting gives
w , h = 2(w , h)
hence w = h. Therefore, substituting into the equation vw = 0
135h , h2 = 2h2
and therefore
3h2 = 135h:
14.2 SOLUTIONS
Since h 6= 0, we have
So w = h = 45 cm. Thus, l
209
h = 135
= 45:
3
=
135 , w , h = 45 cm. To check that this critical point is a maximum, we find
Vww = ,2h; Vhh = ,2w;
Vwh = 135 , 2w , 2h;
so
2
2
= 4hw , (135 , 2w , 2h) :
D = Vww Vhh , Vwh
At w = h = 45, we have Vww = ,2(45) < 0 and D = 4(45)2 , (135 , 90 , 90)2 = 6075 > 0, hence V
maximum at w = h = l = 45.
is
Therefore, the suitcase with maximum volume is a cube with dimensions width = height = length = 45
cm.
16. Let P (K; L) be the profit obtained using K units of capital and L units of labor. The cost of production is
given by
C (K; L) = kK + `L;
and the revenue function is given by
R(K; L) = pQ = pAK a Lb :
Hence, the profit is
P = R , C = pAK a Lb , (kK + `L):
In order to find local maxima of P , we calculate the partial derivatives and see where they are zero. We have:
@P = apAK a,1 Lb , k;
@K
@P = bpAK a Lb,1 , `:
@L
The critical points of the function P (K; L) are solutions (K; L) of the simultaneous equations:
k = pAK a,1 Lb ;
a
`
a b,1
b = pAK L :
Multiplying the first equation by K and the second by L, we get
kK
a
and so
=
`L ;
b
`a L:
K = kb
Substituting for K in the equation k=a = pAK a,1 Lb , we get:
k = pA `a a,1 La,1 Lb :
a
kb
210
CHAPTER FOURTEEN /SOLUTIONS
We must therefore have
Hence, if a + b 6= 1,
a ` a,1
L1,a,b = pA ka
:
b
" a a,1 #1= 1,a,b
`
L = pA a
;
(
k
)
(
)
b
" a (a,1)#1=(1,a,b)
`a
`
`a
:
K = kb L = kb pA ak
b
and
To see if this is really a local maximum, we apply the second derivative test. We have:
@ 2 P = a(a , 1)pAK a,2 Lb ;
@K 2
@ 2 P = b(b , 1)pAK a Lb,2;
@L2
@ 2 P = abpAK a,1 Lb,1:
@K@L
Hence,
@2P @2P , @2P 2
D = @K
2 @L2
@K@L
2 2 2a,2 2b,2
= ab(a , 1)(b , 1)p A K
L , a2b2 p2A2 K 2a,2L2b,2
2 2 2a,2 2b,2
= ab((a , 1)(b , 1) , ab)p A K
L
2 2 2a,2 2b,2
= ab(1 , a , b)p A K
L :
Now a, b, p, A, K , and L are positive numbers. So, the sign of this last expression is determined by the sign
of 1 , a , b.
(a) We assumed that a + b < 1, so D > 0, and as 0 < a < 1, then @ 2 P=@K 2 < 0 and so we have a
unique local maximum. To verify that the local maximum is a global maximum, we focus on the cost.
Let C = kK + `L. Since K 0 and L 0, K C=k and L C=`. Therefore the profit satisfies:
P = pAK a Lb , (kK + `L)
C a C b
pA k
` ,C
a+b , C
= mC
where m = pA(1=k)a (1=`)b . Since a + b < 1, the profit is negative for large costs C , say C C0
(C0 = m1,a,b will do). Therefore, in the KL-plane for K 0 and L 0, the profit is less than or
equal to zero everywhere on or above the line kK + `L = Co . Thus the global maximum must occur
inside the triangle bounded by this line and the K and L axes. Since P 0 on the K and L axes as
well, the global maximum must be in the interior of the triangle at the unique local maximum we found.
In the case a + b < 1, we have decreasing returns to scale. That is, if the amount of capital and
labor used is multiplied by a constant > 0, we get less than times the production.
14.2 SOLUTIONS
211
(b) Now suppose a + b 1. If we multiply K and L by for some > 0, then
Q(K; L) = A(K )a (L)b = a+b Q(K; L):
We also see that
So if a + b = 1, we have
C (K; L) = C (K; L):
P (K; L) = P (K; L):
Thus, if = 2, so we are doubling the inputs K and L, then the profit P
is doubled and hence there
can be no maximum profit.
If a + b > 1, we have increasing returns to scale and there can again be no maximum profit:
doubling the inputs will more than double the profit. In this case, the profit increases without bound as
K; L go toward infinity.
17. Let the line be in the form y = b + mx. When x equals ,1, 0 and 1, then y equals b , m, b, and b + m,
respectively. The sum of the squares of the vertical distances, which is what we want to minimize, is
f (m; b) = (2 , (b , m))2 + (,1 , b)2 + (1 , (b + m))2 :
To find the critical points, we compute the partial derivatives with respect to m and b,
fm = 2(2 , b + m) + 0 + 2(1 , b , m)(,1)
= 4 , 2b + 2m , 2 + 2b + 2m
= 2 + 4m;
fb = 2(2 , b + m)(,1) + 2(,1 , b)(,1) + 2(1 , b , m)(,1)
= ,4 + 2b , 2m + 2 + 2b , 2 + 2b + 2m
= ,4 + 6b:
Setting both partial derivatives equal to zero, we get a system of equations:
2 + 4m = 0;
,4 + 6 b = 0 :
The solution is m
y = 23 , 12 x.
=
,1=2 and b = 2=3. One can check that it is a minimum. Hence, the regression line is
18.
y = mx + b
x
Figure 14.9
(a) Points which are directly above or below each other share the same x coordinate, therefore, the point
on the least squares line which is directly above or below the point in question will have x coordinate
xi and from the formula for the least squares line, it will have y coordinate b + mxi .
212
CHAPTER FOURTEEN /SOLUTIONS
p
(b) The general distance formula in two dimensions is d = (x2 , x1 )2 + (y2 , y1 )2 , so d2 = (x2 ,
x1)2 + (y2 , y1 )2. Since the x coordinates are identical for the two points in question, the first term in
the square root is zero. This yields d2 = (yi , (b + mxi ))2 .
(c) In both cases we use the chain rule and our knowledge of summations to show the relationship.
@f
@b
n @
n
X
@ (X
2
2
(yi , (b + mxi )) ) =
(yi , (b + mxi ))
@b i=1
@b
i=1
n
X
@
=
2(yi , (b + mxi )) (yi , (b + mxi ))
@b
i=1
=
=
=
n
X
i=1
,2
2(yi , (b + mxi )) (,1)
n
X
yi , (b + mxi ))
(
i=1
n
X
n @
@f = @ ( (y , (b + mx ))2 ) = X
2
(yi , (b + mxi ))
i
i
@m @m i=1
@m
i=1
n
X
@ (y , (b + mx ))
=
2(yi , (b + mxi )) i
i
@m
i=1
=
=
(d) We can separate
@f
@b
n
X
i=1
,2
2(yi , (b + mxi )) (,xi )
n
X
i=1
yi , (b + mxi )) xi
(
into three sums as shown:
@f
@b
Similarly we can separate
=
,2
n
X
i=1
yi , b
n
X
i=1
1,m
n !
X
i=1
@f after multiplying through by x :
i
@m
xi
!
n
n
n
X
X
@f = ,2 X
2
yi xi , b xi , m xi
@m
i=1
i=1
i=1
Setting
@f
@b
and
@f
@m equal to zero we have:
bn + m
b
n
X
i=1
n
X
xi =
i=1
n
X
xi + m
i=1
n
X
yi
i=1
n
X
x2i =
i=1
xi yi
14.2 SOLUTIONS
(e)
213
Pn
To solvePthis pair of linear equations, we multiply the first equation by i 1 x2i , multiply the second
n
one by i 1 xi, and subtract; we get
n
n
n
n X
n
n
X
X
X
X
X
2
2
2
=
=
bn
So,
b=
i=1
xi , b(
n
n
X
X
2
i=1
xi
Similarly,
m= n
i=1
n
X
i=1
i=1
yi ,
xi)
=
i=1
n X
n
X
i=1
xi
i=1
i=1
xi ,
i=1
xi yi
i=1
xi;
1
! 0 X
n
n !2
X
xiyi = @n x2i ,
xi A
i=1
i=1
0 n
1
n X
n !
n !2
X
X
X
xiyi , xi yi = @n x2i ,
xi A
i=1
i=1
(f) Applying the formula to the given data, we have
perfect agreement with the example.
19.
yi
i=1
i=1
b = , 13 , m = 1 which gives y
=
,(1=3) + x, in
(a) Let t be the number of years since 1960 and let P (t) be the population in millions in the year 1960 + t.
We assume that P = Ceat, and therefore
ln P
=
at + ln C:
So, we plot ln P against t and find the line of best fit. Our data points are (0; ln 180), (10; ln 206), and
(20; ln 226). Applying the method of least squares to find the best-fitting line, we find that
, ln 180 0:0114;
a = ln 226 20
ln C
Then, C
= 5:20 =
e
181:3 and so
=
ln 206
3
5 ln 180
, ln 206
+
5:20
6
6
P (t) = 181:3e0:0114t:
In 1990, we have t = 30 and the predicted population in millions is
P (30) = 18:3e0:01141(30) = 255:2:
(b) The difference between the actual and the predicted population is about 6 million or 2 12 %. Given that
only three data points were used to calculate a and c, this discrepancy is not surprising. Thus, the 1990
census, data does not mean that the assumption of exponential growth is unjustified.
(c) In 2010, we have t = 50 and P (50) = 320:6.
20.
(a) Let the line take the form of y = mx + b, where x equals the number of years since 1920, which is the
original year, and y equals the postage corresponding to the year. When the year is 1920, we have x = 0,
so the postage equals b. When the year is 1932, the postage equals b + 12m, because the difference in
years is 12. And, when the year is 1995, the postage equals b + 75m. The sum of the squares of the
vertical distances, which is what we wish to minimize, is
f (m; b) = (0:02 , b)2 + (0:03 , (b + 12m))2 + (0:04 , (b + 38m))2
2
2
2
+(0:05 , (b + 43m)) + (0:06 , (b + 48m)) + (0:08 , (b + 51m))
2
2
2
+(0:1 , (b + 54m)) + (0:13 , (b + 55m)) + (0:15 , (b + 58m))
2
2
2
+(0:2 , (b + 61m)) + (0:22 , (b + 65m)) + (0:25 , (b + 68m))
2
2
+(0:29 , (b + 71m)) + (0:32 , (b + 75m))
214
CHAPTER FOURTEEN /SOLUTIONS
To find the critical points, calculate the partial derivatives fm and fb .
fm = 0 + 2(0:03 , (b + 12m))(,12) + 2(0:04 , (b + 38m))(,38)
+2(0:05 , (b + 43m))(,43) + 2(0:06 , (b + 48m))(,48) + 2(0:08 , (b + 51m))(,51)
+2(0:1 , (b + 54m))(,54) + 2(0:13 , (b + 55m))(,55) + 2(0:15 , (b + 58m))(,58)
+2(0:2 , (b + 61m))(,61) + 2(0:22 , (b + 65m))(,65) + 2(0:25 , (b + 68m))(,68)
+2(0:29 , (b + 71m))(,71) + 2(0:32 , (b + 75m))(,75)
= 1398b + 81766m , 240:66
fb = 2(0:02 , b)(,1) + 2(0:03 , (b + 12m))(,1) + 2(0:04 , (b + 38m))(,1)
+2(0:05 , (b + 43m))(,1) + 2(0:06 , (b + 48m))(,1) + 2(0:08 , (b + 51m))(,1)
+2(0:1 , (b + 54m))(,1) + 2(0:13 , (b + 55m))(,1) + 2(0:15 , (b + 58m))(,1)
+2(0:2 , (b + 61m))(,1) + 2(0:22 , (b + 65m))(,1) + 2(0:25 , (b + 68m))(,1)
+2(0:29 , (b + 71m))(,1) + 2(0:32 , (b + 75m)(,1)
= 28b + 1398m , 3:88
Setting both partial derivatives equal to zero, we get a system of two equations:
1398b + 81766m = 240:66
28b + 1398m = 3:88
with solutions m 0:0066 and b ,0:2135. Thus the equation of the line is y = 0:0066x , 0:2135
To predict the cost of a postage stamp in the year 2010, we substitute x = 2010 , 1920 into the
equation we just created and obtain:
y = 0:0066(2010 , 1920) , 0:2135
y = 0:3805
Therefore, the cost of a postage stamp in the year 2010 would be $0:38.
(b) Looking at the data in Figure 14.10 you can see that it does not appear linear over the whole graph, but
does look linear after about 1972.
cost
0:35
0 :3
0:25
0 :2
0:15
0 :1
0:05
0
year
1940
1960
1980
Figure 14.10
2000
14.2 SOLUTIONS
(c)
215
ln cost
,1:5
,2
,2:5
,3
,3:5
year
1940
1960
1980
2000
Figure 14.11
Looking at the data in Figure 14.11 you can see that it appears linear after 1960. If it were linear
over the entire range, say ln y = mx + b, then y = emx+b so the price of a stamp is exponential, and is
increasing rapidly as time goes on.
To find the line that best fits this data, we use the same method as before. This time
f (m; b) = (ln 0:02 , b)2 + (ln 0:03 , (b + 12m))2 + (ln 0:04 , (b + 38m))2
2
2
2
+(ln 0:05 , (b + 43m)) + (ln 0:06 , (b + 48m)) + (ln 0:08 , (b + 51m))
2
2
2
+(ln 0:10 , (b + 54m)) + (ln 0:13 , (b + 55m)) + (ln 0:15 , (b + 58m))
2
2
2
+(ln 0:20 , (b + 61m)) + (ln 0:22 , (b + 65m)) + (ln 0:25 , (b + 68m))
2
2
+(ln 0:29 , (b + 71m)) + (ln 0:32 , (b + 75m))
Setting the partial derivatives equal to zero, we get:
fm = 0 + 2(ln 0:03 , (b + 12m))(,12) + 2(ln 0:04 , (b + 38m))(,38)
+2(ln 0:05 , (b + 43m))(,43) + 2(ln 0:06 , (b + 48m))(,48)
+2(ln 0:08 , (b + 51m))(,51) + 2(ln 0:10 , (b + 54m))(,54)
+2(ln 0:13 , (b + 55m))(,55) + 2(ln 0:15 , (b + 58m))(,58)
+2(ln 0:20 , (b + 61m))(,61) + 2(ln 0:22 , (b + 65m))(,65)
+2(ln 0:25 , (b + 68m))(,68) + 2(ln 0:29 , (b + 71m))(,71)
+2(ln 0:32 , (b + 75m))(,1)
= 1398b + 8166m + 2572:72 = 0;
fb = 2(ln 0:02 , b)(,1) + 2(ln 0:03 , (b + 12m))(,1) + 2(ln 0:04 , (b + 38m))(,1)
+2(ln 0:05 , (b + 43m))(,1) + 2(ln 0:06 , (b + 48m))(,1)
+2(ln 0:08 , (b + 51m))(,1) + 2(ln 0:10 , (b + 54m))(,1)
+2(ln 0:13 , (b + 55m))(,1) + 2(ln 0:15 , (b + 58m))(,1)
+2(ln 0:20 , (b + 61m))(,1) + 2(ln 0:22 , (b + 65m))(,1)
+2(ln 0:25 , (b + 68m))(,1) + 2(ln 0:29 , (b + 71m))(,1)
+2(ln 0:32 , (b + 75m))(,1)
= 28b + 1398 + 64:20 = 0:
These equations have solutions m 0:0529 and b ,4:9329. Thus, the equation of this line is
ln y = 0:0529x , 4:9329;
216
CHAPTER FOURTEEN /SOLUTIONS
where y is the price of a stamp.
To predict the cost of the stamp in 2010, we substitute x = 2010 , 1920 into the equation and get
ln y
ln y
21.
(a)
=
=
0:0529(2010 , 1920) , 4:9329
,0:1719
Therefore, y = e,:1719 0:8421, and so the cost would be $0:84.
We have the following contour diagram for f :
625
25
1
5
125
Figure 14.12
(b) We first compute grad f :
fx = 4(x + 1)3 , (x2y22xy+ 1)2
2
fy = 4(y , 1)3 , (x2y22yx+ 1)2 :
2
and
These equations are difficult to solve simultaneously and this is why we need to use a gradient search
method. We choose (x0 ; y0 ) = (,1; 1) as our starting point and compute
grad f (,1; 1) = 0:5~i
, 0:5~j :
Since we wish to minimize f , we move from (x0 ; y0 ) in the opposite direction of the gradient to a point
x t ; y1(t)) = (x0 ; y0) , t grad f (x0 ; y0) = (,1; 1) , t(0:5; ,0:5);
such that f (x1 ; y1 ) < f (x0 ; y0 ). We should choose t so that the function f (x1 (t); y1 (t)) is minimized.
Since the function f (x; y) is complicated and only an approximate answer is required, we will try
t = 0:5 for each iteration of the gradient search method. Therefore,
(x1 ; y1 ) = (,1:25; 1:25);
grad f (,1:25; 1:25) 0:27~i , 0:27~j :
Repeating this step, again with t = 0:5, gives
(x2 ; y2 ) = (,1:25; 1:25) , (0:5)(0:27; ,0:27) (,1:38; 1:38);
grad f (,1:38; 1:38) 0:02~i , 0:02~j :
( 1( )
Performing one more iteration, we get
x ; y3) = (,1:38; 1:38) , (0:5)(0:02; ,0:02) (,1:39; 1:39);
grad f (,1:39; 1:39) 0:002~i , 0:002~j ;
so we are already very close to a critical point. We find that f (,1:39; 1:39) 0:2575 and verify that
( 3
this is a global minimum using the contour diagram from part (a).
14.3 SOLUTIONS
217
22. To execute the gradient search we first evaluate the two first partial derivatives of C :
@C
@n
@C
@d
=
=
0:15 , 3
4d 1:8
n,2
,5:87
4d 0:8
4d 0:8
+ 1:44
+ 4:32
n,1
,11:688 45d
5
5
5
If we choose as our starting point for the gradient search a pipe diameter of 1 meter with three pumping
stations, after numerous iterations, we find the optimum values to be about 1:61 meter diameter pipe with six
pumping stations. This produces a minimum cost of 4.14 million dollars.
23. We have
fx = 2x(y + 1)3 = 0 only when x = 0 or y = ,1
fy = 3x2(y + 1)2 + 2y = 0 never when y = ,1 and only for y = 0 when x = 0
We conclude that fx = 0 and fy = 0 only when x = 0; y = 0, so f has only one critical point, namely (0; 0).
The second derivative test at (0; 0) gives
D = fxx fyy , (fxy )2 = 2(y + 1)3(6x2(y + 1) + 2) , (6x(y + 1)2)2
= 2(1)(2) , 0 > 0 when x = 0; y = 0
Since fxx > 0 at (0; 0), this means f has a local minimum at (0; 0).
[Alternatively, if we expand (y + 1)3 , then we can view f (x; y) as x2 + y2 + (terms of degree 3 or greater
in x and y), which means that f behaves likes x2 + y2 near (0; 0).]
Although (0; 0) is a local minimum, it cannot be a global minimum since for fixed x, say x = 1, the
function f (x; y) is a cubic polynomial in y and cubics take on arbitrarily large positive and negative values.
In the single-variable case, suppose a function f defined on the real line is differentiable and its derivative
is continuous. Then if f has only one critical point, say x = 0, then if that critical point is a local minimum,
it must also be a global minimum. This is because f 0 cannot change sign without f 0 = 0 so we must have
f 0 < 0 for x < 0 and f 0 > 0 for x > 0. Thus f is decreasing for all x < 0 and increasing for all x > 0,
which makes x = 0 the global minimum for f .
Solutions for Section 14.3
1. Our objective function is f (x; y) = x + y and our equation of constraint is g(x; y) = x2 + y2 = 1. To
optimize f (x; y) with Lagrange multipliers, we solve rf (x; y) = rg(x; y) subject to g(x; y) = 1. The
gradients of f and g are
rf (x; y) = ~i + ~j ;
rg(x; y) = 2x~i + 2y~j :
So the equation rf
=
rg becomes
~i + ~j
Solving for gives
=
(2x~i + 2y~j )
= 21x = 21y ;