Solutions to Suggested Problems
Section 7.1
33. Solving the equation x − y2 = z0 for x, we see that the level curves are
parabolas (with horizontal axes of symmetry) of the form x = y2 + z0 . The
x-section at x = 0 is the parabola z = −y2 ; the y-section at y = −2 is the line
z = x − 4.
(i) level curves
(ii) graph of z=f(x,y)
4
2
2
0
y
–2
1
–4
–6
–8
–2
–1
0
1
2
3
x
4
5
–10
–3
–2
–1
–1
y
–2
0
1
2
3
1
5
4
3
2
1
x
0
–1
–2
35. Writing the equation 1+x2 +y2 = z0 in the form x2 +y2 = z0 −1, we see
that the level curves are circles (centered at the origin), if z0 > 1. If z0 = 1,
then the level curve is the point (0,0). If z0 < 0, then the level curve is the
empty set. The x-section at x = 2 is the parabola z = y2 + 5; the y-section at
y = −3 is the parabola z = x2 + 10.
(i) level curves
(ii) graph of z=f(x,y)
4
30
y
2
25
20
15
–4
–2
2
4
10
x
5
–2
4
2
0
x
–2
–4
–4
2
2
–2
0
y
–4
37. For z0 6= 0, the level curves are of the form xy = z0 . If z0 = 0 the level
curve consists of the two coordinate axes. The x-section at x = 1 is the line
z = y; the y-section at y = −1 is the line z = −x.
(a) level curves
(b) graph of z = f (x, y)
4
15
10
y
2
5
0
–5
–4
–2
0
2
4
–10
x
–15
–4
–2
–2
x
0
2
4
–4
3
–4
–2
2
0
y
4
39. (a) When L = L0 , we have Q = Q(r, L0 ) = πr4 P/(8L0 v) = Kr4 , where
K = πP/(8L0 v) (a positive constant).
(b) When r = r0, we have Q = Q(r0 , L) = πr04 P/(8Lv) = K/L, where K =
πr04 P/(8v) (a positive constant).
(b) graph of Q =
(a) graph of Q = Kr4
K
L
40
2
35
1.5
30
25
1
20
15
0.5
10
5
0
0.2
0.4
0.6
0.8
1
0.2
r
0.4
0.6
L
4
0.8
1
41. (a) The revenue (in dollars) is given by the formula R(x, y) = 2, 000x +
3, 000y, where x is the number of units of product X sold and y is the number
of units of product Y .
(b) If we fix R at 60,000 we get the equation 2, 000x + 3, 000y = 60, 000
or y = − 23 x + 20. When we fix R at 120,000, we get the equation 2, 000x +
3, 000y = 120, 000 or y = − 23 x + 40.
(top line is R=120,000, bottom line is R=60,000)
40
30
y 20
10
0
10
20
30
40
50
60
x
5
42. When the given production level is fixed at 15,000 units, we obtain the
equation 15K 3/4 L1/4 = 15, 000, or K 3/4 L1/4 = 1, 000. Raising each side
12
to the 4th power gives K 3 L = 1012 , and solving for L gives L = 10K 3 . If
Q = 15, 000 and K = 100, then L =
1012
1003
= 106 = 1, 000, 000 units.
5000
4000
3000
L
2000
1000
0
1000
2000
3000
4000
5000
K
59. In this case, (x0 , y0 , z0 ) = (4, −3, 3), a = 2, and b = 5. Therefore, the
equation of the plane is z − 3 = 2(x − 4) + 5(y + 3), or z = 2x + 5y + 10.
63. Let (x0 , y0 , z0 ) = (1, 0, −1). Then z − (−1) = a(x − 1) + b(y − 0) or,
z + 1 = a(x − 1) + by. To find a and b, we substitute
(x, y, z) = (0, 1, 3) to get 3 + 1 = a(0 − 1) + b, or 4 = −a + b
(x, y, z) = (1, −1, −2) to get −2 + 1 = a(1 − 1) + b(−1), or −1 = −b
Thus, a = −3 and b = 1. The equation of the plane is z + 1 = −3(x − 1) + y,
or z = −3x + y + 2.
66. z = z0 + ax + by (from equation 2), and a = 3, b = 5. So z = z0 + 3x + 5y,
and ∆z = 3∆x + 5∆y. Setting ∆x = 0.1 and ∆y = −0.2, we find that the
change in z is ∆z = 3(0.1) + 5(−0.2) = −0.7.
72. Let P(x, y) be the profit in selling x units of item X and y units of item
Y . From the information given, we know that P(x, y) = 2x + 5y + c for some
real number c, and P(100, 50) = −500. Therefore, −500 = P(100, 50) =
6
2 · 100 + 5 · 50 + c = 200 + 250 + c = 450 + c, so that c = −950. Thus,
P(x, y) = 2x + 5y − 950, and P(100, 500) = 2(100) + 5(500) − 950 = 200 +
2, 500 − 950 = 1, 750 dollars.
Section 7.2
3.
∂f
∂x
= 3x2 y2 − 4xy, ∂∂yf = 2x3 y − 2x2 + 1.
5.
∂M
∂r
rt
= 1000tert , ∂M
∂t = 1000re .
7.
∂R
∂s
= − 5ts2 , ∂R
∂t =
9.
∂l
∂x
=
11.
2
∂Q
∂K
10t
s .
∂ 2
1
2x
.
(x − 5y + 3) = x2 −5y+3
x2 −5y+3 ∂x
0.4 −0.4 .
= 4K −0.6 L0.6 , ∂Q
∂L = 6K L
13.
∂M
r 360t
r
= 1000(1 +
) [ln(1 +
)] · 360
∂t
360
360
r 360t
r
= 360, 000(1 +
) ln(1 +
).
360
360
∂M
r 360t−1 1
= 1000 · 360t(1 +
)
(
)
∂r
360
360
r 360t−1
= 1000t(1 +
)
.
360
15.
∂f
∂x
= 2x + y, ∂∂xf (2, 1) = 2 · 2 + 1 = 5.
17.
∂f
∂x
= 50x−1/2 y1/2 , ∂∂xf (25, 16) = 50 · 25−1/2 · 161/2 = 50 · 15 · 4 = 40.
7
21. (a)
∂T
∂x
= 10x, ∂T
∂y = −10y.
(b) If x = a, the x-section has the equation T = 5(a2 − y2 ) + 75 = −5y2 +
(75 + 5a2 ). It is a parabola, opening downward, that achieves a maximum
at y = 0.
(c) If y = b, the y-section has the equation T = 5(x2 − b2 ) + 75 = 5x2 +
(75 − 5b2 ). It is a parabola, opening upward, that achieves a minimum at
x = 0.
(d) If there were a local minimum (or maximum), both the x- and y-sections
through the point would have a local minimum (or maximum), which cannot be the case.
23. The equation of the tangent plane has the form
z = f (x0 , y0 ) +
∂f
∂f
(x0 , y0 )(x − x0 ) + (x0 , y0 )(y − y0 ).
∂x
∂y
Since ∂∂xf = 6x and ∂∂yf = 10y, we have ∂∂xf (1, 2) = 6, and ∂∂yf (1, 2) = 20.
Also, f (1, 2) = 23. The equation of the tangent plane is given by z =
23 + 6(x − 1) + 20(y − 2) = −23 + 6x + 20y, and the linear approximation
is f (x, y) ≈ −23 + 6x + 20y.
27. Since
∂f
∂y (4, 1)
∂f
∂x
= 4x−1/2 y2 and
∂f
∂y
= 16x1/2 y, we have
∂f
∂x (4, 1)
= 2 and
= 32. Also, f (4, 1) = 16. The equation of the tangent plane is
given by z = 16 + 2(x − 4) + 32(y − 1) = −24 + 2x + 32y, and the linear
approximation is f (x, y) ≈ −24 + 2x + 32y.
31. z = −2 + 2(x + 3) − 78 (y − 5) = 2x − 87 y + 68
7 . The linear approximation
8
68
is f (x, y) ≈ 2x − 7 y + 7 .
33. The volume of a rectangular box whose height is x inches and whose
base is a square with each side y inches long equals V (x, y) = xy2 .
(a) V (6, 4) = 6 · 42 = 96 cubic inches.
8
∂V
2
(b) Since ∂V
∂x = y and ∂y = 2xy, we obtain
The linear approximation formula gives
∂V
∂V
∂x (6, 4) = 16 and ∂y (6, 4) = 48.
∂V
∂V
(6, 4) · (6.02 − 6) +
(6, 4) · (4.01 − 4)
∂x
∂y
= 16 · (0.02) + 48 · (0.01) = 0.8.
V (6.02, 4.01) −V (6, 4) ≈
Section 7.3
1. We must solve the system of equations
∂f
= 2x − 2 = 0
∂x
∂f
= 2y + 4 = 0
∂y
From the first equation we get x = 1, and from the second equation we get
y = −2. Thus, (1, −2) is the only critical point.
3. Setting
equations
∂f
∂x
= −5y − 4x = 0 and
∂f
∂y
= 2y − 5x = 0 we obtain the system of
4x + 5y = 0
5x − 2y = 0,
whose unique solution is x = 0, y = 0. Thus, (0, 0) is the only critical point.
5. Setting ∂∂xf = 3y − 3x2 = 0 and
of equations
∂f
∂y
= 3x − 3y2 = 0 we obtain the system
y = x2
x = y2 ,
which leads to y = y4 , whose solutions are y = 0 and ,y = 1. If y = 0, then
x = 0, and if y = 1 then x = 1. Thus, (0, 0) and (1, 1) are the critical points.
9
9. Set ∂∂xf = 2x = 0 and ∂∂yf = 2yey = 0. The only solution is x = 0 and y = 0.
Thus, (0, 0) is the only critical point.
2
11. Setting ∂g
∂x = 2x − y = 0 and
of equations
∂g
∂y
= −x + 2y + 3 = 0, we obtain the system
2x − y = 0
x − 2y = 3
This system has the unique solution x = −1 and y = −2. Thus, (−1, −2)
2
is the only critical point. The second order partial derivatives are ∂∂xg2 = 2,
∂2 g
∂x∂y
∂ g 2
= −1 and ∂∂yg2 = 2, and D(x, y) = ∂∂xg2 · ∂∂yg2 −( ∂x∂y
) = 2·2−(−1)2 = 3 >
2
2
0. Since D(x, y) > 0 and
∂2 g
∂x2
2
2
= 2 > 0, g has a local minimum at (−1, −2).
15. Setting ∂∂xf = 3x2 − 3y = 0 and
of equations
∂f
∂y
= 3y2 − 3x = 0, we obtain the system
y = x2
x = y2
whose solutions are x = 0, y = 0 and x = 1, y = 1. Thus, (0, 0) and (1, 1)
2
are the critical points. The second order partial derivatives are ∂∂x2f = 6x,
∂2 f
∂x∂y
= −3 and
∂2 f
∂2 f
∂y2
= 6y, and so D(x, y) = 36xy − 9. Thus D(1, 1) = 27 > 0
and ∂x2 (1, 1) = 6 > 0, so f has a local minimum at (1, 1); and D(0, 0) =
−9 < 0, so that f has a saddle point at (0, 0).
29. Set
∂f
∂x
= 4x − 12 = 0 and
∂f
∂y
= 6y − 6 = 0. From the first equation, we
∂2 f
∂2 f
=
4,
2
∂x∂y = 0
∂x
∂2 f
> 0. Therefore,
∂x2
get x = 3, and from the second we get y = 1. In addition,
and ∂∂y2f = 6, and so D(x, y) = 24. Thus D(x, y) > 0 and
f has a local minimum at (3, 1), and since f is a quadratic function the local
minimum is also a global one.
2
10
31. Set
∂f
∂x
= 6 − 6x = 0 and
∂f
∂y
= 4 − 8y = 0. The only solution is x = 1
∂2 f
= −6,
∂x2
2
and ∂∂x2f < 0.
and y = 1/2. Thus, (1, 12 ) is the only critical point. In addition,
∂2 f
∂x∂y
= 0 and
∂2 f
∂y2
= −8, and so D(x, y) = 48. Thus D(x, y) > 0
Therefore, f has a local maximum at (1, 12 ), and since f is a quadratic function the local maximum is also a global one.
35. Setting ∂P
∂x = 8 − 0.02x − 0.01y = 0 and
we obtain the system of equations
∂P
∂y
= 11 − 0.01x − 0.04y = 0,
2x + y = 800
x + 4y = 1100
The system has the solution x = 300 and y = 200. In addition,
∂2 P
∂2 P
∂y2
∂2 P
∂x2
=
−0.02 < 0, ∂x∂y = −0.01 and
= −0.04, and so D(x, y) = 0.0007 > 0.
Thus, the profit is maximized if x = 300 and y = 200, with the maximum
profit being
P(300, 200) = 2400+2200−900−0.01(3002 +300·200+2·2002 ) = 1400.
39. (a) The weekly revenue of the toy store is
1
R(x, y) = x(50 − 3x + 2y) + y(6 + x − y)
2
5
= 50x + 6y + xy − 3x2 − y2 .
2
5
(b) Setting ∂R
∂x = 50 + 2 y − 6x = 0 and
system of equations
∂R
∂y
= 6 + 52 x − 2y = 0, we obtain the
12x − 5y = 100
5x − 4y = −12
whose solution is x = 20 and y = 28. In addition,
∂2 R
∂y2
= −2. Since D(x, y) = −6(−2) − ( 52 )2 =
11
23
4
∂2 R
∂x2
= −6,
> 0 and
∂2 R
∂x2
∂2 R
∂x∂y
=
5
2
and
= −6 < 0, the
revenue function has a local maximum at x = 20 and y = 28. Because the
function is quadratic, it also has a global maximum at that point.
(c) The weekly profit of the toy store is
P(x, y) = R(x, y) − 9(50 − 3x + 2y) − 14(6 + 0.5x − y)
5
= 70x + 2y + xy − 3x2 − y2 − 534.
2
5
Setting ∂P
∂x = 70 + 2 y − 6x = 0 and
system of equations
∂P
∂y
= 2 + 52 x − 2y = 0, we obtain the
12x − 5y = 140
5x − 4y = −4
748
whose solution is x = 580
23 ≈ 25.22 and y = 23 ≈ 32.52. Applying again
the second derivative test, we see that those are the prices the store should
charge for the standard and deluxe games, respectively, to maximize its
profit.
41.
stove is placed at the point (x, y), the total distance is r(x, y) =
p (a) If thep
∂r
∂r
2
2
x + y + (x − 3)2 + (y − 5)2 . Setting ∂x
= 0 and ∂y
= 0 leads to the
system of equations
x
p
x2 + y2
y
p
x2 + y2
= p
= p
3−x
(x − 3)2 + (y − 5)2
5−y
(x − 3)2 + (y − 5)2
Assuming y 6= 0 and y 6= 5 (which must be true for the second equation to
be valid), we can divide the first equation by the second to get
x 3−x
=
y 5−y
which reduces to x(5 − y) = y(3 − x) and further simplifies to 5x = 3y, or
y = 35 x. Conversely, if y = 53 x and 0 < x < 3, then
∂r
x
x−3
= p
+p
∂x
(x − 3)2 + (y − 5)2
x2 + y2
3x
3(x − 3)
3
3
= √
+p
= √ −√ =0
2
2
34
34
34(x − 3)
34x
12
and similarly
∂r
∂y
= 0.
(b) We have seen that any pair (x, y) satisfying y = 35 x and 0 < x < 3 is a
critical point. These are the points of the line segment between (0, 0) and
(3, 5). A geometric argument shows that choosing any such point minimizes
the total distance. In fact, this exercise is an analytic verification of the wellknown geometric principle that the shortest distance between two points is
along the straight line between them.
47. Setting
∂E
∂a
∂E
∂b
= 2(a + b − 2) + 2(2a + b − 2) · 2 + 2(3a + b − 4) · 3 = 28a + 12b − 36 = 0
= 2(a + b − 2) + 2(2a + b − 2) + 2(3a + b − 4) = 12a + 6b − 16 = 0
we obtain the system of equations
7a + 3b = 9
6a + 3b = 8.
This system has the solution a = 1 and b = 23 . In addition,
∂2 E
= 28,
∂a2
∂2 E
= 12,
∂a∂b
∂2 E
= 6.
∂b2
Therefore, D(a, b) = 28 · 6 − 122 = 24 > 0 and ∂∂aE2 = 28 > 0. Since E is a
quadratic function, it has a global minimum at (1, 23 ).
2
Section 7.5
1. Given f (x, y) = x2 + y2 with constraint g(x, y) = 2x + 6y − 2000 = 0, we
∂f
∂g
set ∂∂xf = λ ∂g
∂x and ∂y = λ ∂y , leading to the system of equations
2x = 2λ
2y = 6λ
2x + 6y = 2000
From the first two equations we obtain x = λ and y = 3λ, so that y = 3x.
Substituting that into the third equation gives 2x + 18x = 2000 or x = 100,
13
and it follows that y = 300. Therefore, if there is a local minimum or maximum, subject to the constraint, it must occur at (100, 300).
From geometric considerations, we can conclude that f (x, y) does, in fact,
have a minimum at (100, 300), subject to the constraint. For the level sets
of f are circles centered at the origin, and the one tangent to the constraint
line g(x, y) = 0 at (100, 300) has the smalles radius of all those that intersect
the line.
5. Given f (x, y) = (x − 1)2 + (y − 2)2 − 4 and g(x, y) = 3x + 5y − 47, we
∂F
∂F
define F(x, y, λ) = f (x, y) − λg(x, y) and set ∂F
∂x = 0, ∂y = 0, and ∂λ = 0.
These are equivalent to
system of equations
∂f
∂x
∂f
∂g
= λ ∂g
∂x , ∂y = λ ∂y , and g(x, y) = 0, and lead to the
2(x − 1) = 3λ
2(y − 2) = 5λ
3x + 5y = 47.
From the first two equations, we obtain 10(x − 1) = 6(y − 2), which reduces
to 5x−3y = −1. Combining that with the third equation leads to the solution
x = 4, y = 7.
We next apply the 2nd derivative test. The partial derivatives are:
∂F
∂2 F
∂2 F
∂2 F
∂F
= 2(x−1)−3λ,
= 2(y−2)−5λ,
=
2,
=
0,
= 2.
∂x
∂y
∂x2
∂x∂y
∂y2
Since D(x, y) = 2 · 2 − 02 = 4 > 0 and ∂∂xF2 = 2 > 0, f (x, y) has a local minimum at (4, 7), subject to the constraint g(x, y) = 0.
2
We can also reach the same conclusion from geometric considerations similar to that of exercise 1 involving the level sets, which are circles centered
at (1, 2). In fact, we reach the stronger conclusion that there is a global minimum at (4, 7).
7. Given f (x, y) = xy with the constraint g(x, y) = x2 + y2 − 1 = 0, we set
14
∂f
∂x
= λ ∂g
∂x and
∂f
∂y
= λ ∂g
∂y , leading to the system of equations
y = 2xλ
x = 2yλ
2
x + y2 = 1.
By multiplying the first equation by y and the second by x, we√obtain y2 =
2xyλ = x2 . That reduces the third equation√to 2x2√= 1, or x =√± 2/2.√Thus,
there
√ are four
√ possible√extreme√points: ( 2/2, 2/2), (− 2/2, − 2/2),
(− 2/2, 2/2) and ( 2/2, − 2/2).
The constraint curve is the circle of radius 1 about the origin, and the level
curves of f are of the form xy = c. If c > 0, each curve has two branches,
one in the first quadrant and the other in the third. Among those curves that
intersect the constraint√circle,
is 1/2, occuring at the
√the largest value
√ of c √
two tangency points ( 2/2, 2/2), and (− 2/2, − 2/2).
Thus, f has a global maximum, subject to the constraint, at each of those
points. Similarly,√ f has√a global minimum,
subject
to the constraint, at each
√
√
of the points (− 2/2, 2/2) and ( 2/2, − 2/2), where c = −1/2.
9. Given f (x, y) = x2 + y2 with the constraint g(x, y) = xy − 1 = 0, we set
∂f
∂f
∂g
∂g
∂x = λ ∂x and ∂y = λ ∂y , leading to the system of equations
2x = λy
2y = λx
xy = 1.
Multiplying the first equation by x and the second by y, we obtain 2x2 =
λxy = 2y2 , so that y = ±x. However, y 6= −x, otherwise the third equation
would become −x2 = 1. Therefore, y = x, and by substituting that into the
third equation we obtain x2 = 1, or x = ±1.
Thus, the possible extreme points are : (1, 1), and (−1, −1). By considering the relation of the level curves of f to the constraint curve as we have
done in the previous exercises, we conclude that, subject to the constraint, f
has a global minimum at each of the points (1, 1) and (−1, −1), with value
f (±1, ±1) = 2.
15
13. With f (x, y) = x2 + y2 and g(x, y) = 3x + y − 1, we set
∂f
∂g
∂y = λ ∂y , leading to the system of equations
∂f
∂x
= λ ∂g
∂x and
2x = 3λ
2y = λ
3x + y = 1.
From the first two equations we get y = 31 x, and substituting that into the
3 1
3
. Thus, the only extreme point is ( 10
, 10 ).
third equation gives x = 10
An examination of the level curves, similar to that in exercise 1, shows that
f has a global minimum, subject to the constraint g(x, y) = 0, at the point
3 1
( 10
, 10 ), which is therefore the point on the line closest to the origin.
As an alternative, we can apply the second derivative test to F(x, y, λ) =
3 1 1
, 10 , 5 ).
(x2 + y2 ) − λ(3x + y − 1) at the point (x, y, λ) = ( 10
23. We need to maximize P(x, y) subject to the constraint Q(x, y) = x + y −
∂Q
∂Q
∂P
100 = 0. Setting ∂P
∂x = λ ∂x and ∂y = λ ∂y , we obtain the system of equations
10 − 0.2x = λ
20 − 0.2y = λ
x + y = 100,
whose unique solution is x = 25, y = 75, and λ = 5. Applying the second
derivative test with F(x, y, λ) = 10x + 20y − 0.1(x2 + y2 ) − λ(x + y − 100),
we find that
∂F
∂F
∂2 F
∂2 F
∂2 F
= 10−0.2x−λ,
= 20−0.2y−λ,
=
−0.2,
=
0,
= −0.2.
∂x
∂y
∂x2
∂x∂y
∂y2
Then D(x, y) = (−0.2) · (−0.2) − 02 = 0.04 > 0 and ∂∂xF2 = −0.2 < 0, from
which we conclude that P(x, y) has a maximum at (25, 75). When the company manufactures 25 units of X and 75 units of Y , it achieves its maximum
profit P(25, 75) = 1, 125.
2
25. We need to maximize Q(K, L) under the constraint 70K +65L = 4, 000, 000.
∂g
∂Q
Letting g(K, L) = 70K + 65L − 4, 000, 000 and setting ∂Q
∂K = λ ∂K and ∂L =
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∂g
, we obtain the system of equations
λ ∂L
14K −0.65 L0.65 = 70λ
26K 0.35 L−0.35 = 65λ
70K + 65L = 4, 000, 000.
whose solution is K = 20, 000, and L = 40, 000. Thus, the combination of
20,000 units of capital and 40,000 units of labor gives the maximum output.
(Note: You must still verify that this is indeed a maximum).
30. (a) We want to optimize S(x, y) = x3/4 y1/4 under the constraint x+y = 4.
∂T
∂S
∂T
Write T (x, y) = x + y − 4 and set ∂S
∂x = λ ∂x and ∂y = λ ∂y . We obtain the
system of equations
3 −1/4 1/4
x
y
= λ
4
1 3/4 −3/4
x y
= λ
4
x + y = 4.
whose solution is x = 3, y = 1.
(b) There is a maximum at (x, y) = (3, 1), subject to the assumptions that
x ≥ 0, y ≥ 0 and x + y = 4. To see why, observe that S(x, y) ≥ 0 all along
the segment of the line x + y = 4 connecting the points (0, 4) and (4, 0), and
S(0, 4) = S(4, 0) = 0. Therefore, there must be a maximum at some point
of the segment, and the only candidate is (3, 1).
(c) This crepuscular mammal should spend 3 of the dawn hours in foraging
and 1 in defending. In other words, 3/4 of the dawn time should be spent
foraging and 1/4 in defending.
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