14.7
7. f (x, y) = (x − y)(1 − xy) = x − y − x2 y + xy 2
⇒ fx = 1−2xy +y 2 , fy = −1−x2 +2xy, fxx = −2y, fxy = −2x+2y, fyy =
2x. Then fx = 0 implies 1−2xy+y 2 = 0 and fy = 0 implies −1−x2 +2xy =
0. Adding the two equations gives 1+y 2 −1−x2 = 0 ⇒ y 2 = x2 ⇒ y = ±x,
but if y = −x then fx = 0 implies 1 + 2x2 + x2 = 0 ⇒ 3x2 = −1 which has
no real solution. If y = x then substitution into fx = 0 gives 1−2x2 +x2 =
0 ⇒ x2 = 1 ⇒ x = ±1, so the critical points are (1, 1) and (−1, −1). Now
D(1, 1) = (−2)(2)−02 = −4 < 0 and D(−1, −1) = (2)(−2)−02 = −4 < 0,
so (1, 1) and (−1, , −1) are saddle points.
9. f (x, y) = y 3 + 3x2 y − 6x2 − 6y 2 + 2 ⇒ fx = 6xy − 12x, fy = 3y 2 + 3x2 −
12y, fxx = 6y − 12, fxy = 6x, fyy = 6y − 12. Then fx = 0 implies
6x(y − 2) = 0, so x = 0 or y = 2. If x = 0 then substitution into fy = 0
gives 3y 2 − 12y = 0 ⇒ 3y(y − 4) = 0 ⇒ y = 0 or y = 4, so we have
critical points (0, 0) and (0, 4). If y = 2, substitution into fy = 0 gives
12 + 3x2 − 24 = 0 ⇒ x2 = 4 ⇒ x = ±2, so we have critical points (±2, 2).
D(0, 0) = (−12)(−12) − 02 = 144 > 0 and fxx (0, 0) = −12 < 0, so
f (0, 0) = 2 is a local maximum. D(0, 4) = (12)(12) − 02 = 144 > 0 and
fxx (0, 4) = 12 > 0, so f (0, 4) = −30 is a local minimum.
D(±2, 2) = (0)(0) − (±12)2 = −144 < 0, so (±2, 2) are saddle points.
16. f (x, y) = ey (y 2 − x2 ) ⇒ fx = −2xey , fy = (2y + y 2 − x2 )ey , fxx =
−2ey , fxy = −2xey , fyy = (2 + 4y + y 2 − x2 )ey . Then fx = 0 implies x = 0
and substituting into fy = 0 gives (2y + y 2 )ey = 0 ⇒ y(2 + y) = 0 ⇒ y = 0
or y = −2, so the critical points are (0, 0) and (0, −2).D(0, 0) = (−2)(2) −
(0)2 = −4 < 0 so (0, 0) is a saddle point. D(0, −2) = (−2e−2 )(−2e−2 ) −
(0)2 = 4e−4 > 0 and fxx (0, −2) = −2e−2 < 0, so f (0, −2) = 4e−2 is a
local maximum.
2
2
20. f (x, y) = x2 ye−x −y ⇒
2
2
2
2
2
2
fx = x2 ye−x −y (−2x) + 2xye−x −y = 2xy(1 − x2 )e−x −y ,
2
2
2
2
2
2
fy = x2 ye−x −y (−2y) + x2 e−x −y = x2 (1 − 2y 2 )e−x −y ,
2
2
fxx = 2y(2x4 − 5x2 + 1)e−x −y ,
2
2
2
2
fxy = 2x(1 − x2 )(1 − 2y 2 )e−x −y , fyy = 2x2 y(2y 2 − 3)e−x −y .
fx = 0 implies x = 0, y = 0, or x = ±1. If x = 0 then fy = 0 for any
y-value, so all points of the form (0, y) are critical points. If y = 0 then
2
fy = 0 ⇒ x2 e−x = 0 ⇒ x = 0, so (0, 0)(already included above)
is a criti
2
1
1
cal point. If x = ±1 then (1 − 2y 2 )e−1−y = 0 ⇒ y = ± √ , so ±1, √
2
2
1
and ±1, − √
are critical points. Now
2
√
1
1
1
D ±1, √
= 8e−3 > 0, fxx ±1, √
= −2 2e−3/2 < 0 and D ±1, − √
=
2 2
2
√
1
1
1
8e−3 > 0, fxx ±1, − √
= 2 2e−3/2 > 0, so f ±1, √
= √ e−3/2
2
2
2
1
1 −3/2
are local maximum points while f ±1, − √
= −√ e
are local
2
2
minimum points. At all critical points (0, y) we have D(0, y), so the
1
Second Derivatives Test gives no information. However, if y > 0 then
2
2
x2 ye−x −y ≥ 0 with equality only when x = 0, so we have local mini2
2
mum valuesf (0, y) = 0, y > 0. Similarly, if y < 0 then x2 ye−x −y ≤ 0
with equality when x = 0 so f (0, y) = 0, y < 0 are local maximum values,
and (0, 0) is a saddle point.
29. Since f is a polynomial it is continuous on D, so an absolute maximum and
minimum exist. Here fx = 2x − 2, fy = 2y, and setting fx = fy = 0 gives
(1, 0) as the only critical point(which is inside D), where f (1, 0) = −1.
Along L1 : x = 0 and f (0, y) = y 2 for −2 ≤ y ≤ 2, a quadratic function
which attains its minimum at y = 0, where f (0, 0) = 0, and its maximum
at y = ±2, where f (0, ±2) = 4.
Along L2 : y = x − 2 for 0 ≤ x ≤ 2, and f (x, x − 2) = 2x2 − 6x + 4 =
2
1
3
3
− , a quadratic which attains its minimum at x = , where
2 x−
2 2
2
3 1
1
f
,−
= − , and its maximum at x = 0, where f (0, −2) = 4.
2 2
2
Along L3 : y = 2 − x for 0 ≤ x ≤ 2, and f (x, 2 − x) = 2x2 − 6x + 4 =
2
1
3
3
− , a quadratic which attains its minimum at x = , where
2 x−
2
2
2
3 1
1
f
,
= − , and its maximum at x = 0, where f (0, 2) = 4. Thus the
2 2
2
adsolute maximum of f on D is f (0, ±2) = 4 and the absolute minimum
is f (1, 0) = −1.
34. fx = y 2 and fy = 2xy, and since fx = 0 ⇔ y = 0, there are no critical
points in the interior of D. Along L1 : y = 0 and f (x, 0) =
p0.
3 − x2 , so let
Along L2 : x = 0 and
f
(0,
y)
=
0.
Along
L
:
y
=
3
p
√
g(x) = f x, 3 − x2 = 3x − x3 for 0 ≤ x ≤ 3. Then g 0 (x) = 3 − 3x2 =
√
0 ⇔ x = 1. The maximum√value is f (1, √
2) = 2 and
√ the minimum occurs
both at x = 0 and x = 3 where f√(0, 3) = f ( 3, 0) = 0. Thus the
absolute maximum of f on D is f (1, 2) = 2, and the absolute minimum
is 0 which occurs at all points along L1 and L2 .
42. The
p distance from the origin to a point (x, y, z) on the surface is d =
x2 + y 2 + z 2 where y 2 = 9 + xz, so we minimize d2 = x2 + 9 + xz + z 2 =
f (x, z). Then fx = 2x + z, fz = x + 2z, and fx = 0, fz = 0 ⇒ x = 0, z = 0,
so the only critical point is (0, 0). D(0, 0) = (2)(2) − 1 = 3 > 0 with
fxx (0, 0) = 2 > 0, so this is a minimum. Thus y 2 = 9 + 0 ⇒ y = ±3 and
the points on the surface cloest to the origin are (0, ±3, 0).
45. Center the sphere at the origin so that its equation is x2 + y 2 + z 2 = r2 ,
and orient the inscribed rectangular box so that its edges are parallel to
the coordinate axes. Any vertex of the box satisfies x2 + y 2 + z 2 = r2 ,
so take (x, y, z) to be the vertex in thepfirst octant. Then the box has
length 2x, width 2y, and
2z = 2 r2 − x2 − y 2 with volume given
height
p
p
2
by V (x, y) = (2x)(2y) 2 r − x2 − y 2 = 8xy r2 − x2 − y 2 for 0 <
x < r, 0 < y < r. Then
2
p
1
8y(r2 − 2x2 − y 2 )
Vx = (8xy)· (r2 −x2 −y 2 )−1/2 (−2x)+ r2 − x2 − y 2 ·8y = p
2
r2 − x2 − y 2
2
2
2
8x(r − x − 2y )
and Vy = p
.
r 2 − x2 − y 2
Setting Vx = 0 gives y = 0 or 2x2 + y 2 = r2 , but y > 0 so only the latter
solution applies. Similarly, Vy = 0 with x > 0 implies x2 + 2y 2 = r2 .
2
Substituting, we have 2x2 + y 2 = x2 + 2y 2p⇒ x2 = y√
⇒ x = y.
2
2
2
2
2
2
r /3 = r/ 3 = y.Thus
Then x + 2y = r ⇒ 3x√ = r √ ⇒ x =
the only critical point is (r/ 3, r/ 3). There must be a maximum volume and here it must occur at a critical point, so the maximum
volume
√
r
r
occurs when x = y = r/ 3 and the maximum volume is V √ , √
=
3
3
s
2 2
r
r
r
r
8
2
√
r − √
8 √
− √
= √ r3 .
3
3
3
3
3 3
53. Let x, y, z be the dimensions
p of the rectangular box. Then the volume of
the boxpis xyz and L = x2 + y 2 + z 2 ⇒ L2 = x2 + y 2 + z 2
⇒ z = L2 − x2 − y 2 .
p
Substituting, we have volume V (x, y) = xy L2 − x2 − y 2 (x, y > 0).
p
p
1
Vx = xy· (L2 −x2 −y 2 )−1/2 (−2x)+y L2 − x2 − y 2 = y L2 − x2 − y 2 −
2
x2 y
,
L2 − x2 − y 2
p
xy 2
Vy = x L2 − x2 − y 2 − p
. Vx = 0 implies y(L2 − x2 − y 2 ) =
L2 − x2 − y 2
x2 y ⇒ y(L2 − 2x2 − y 2 ) = 0 ⇒
2x2 + y 2 = L2 (since y > 0), and Vy = 0 implies x(L2 − x2 − y 2 ) = xy 2 ⇒
x(L2 − x2 − 2y 2 ) = 0 ⇒
x2 +2y 2 = L2 (since x > 0). Substituting y 2 = L2 −2x2 into x2 +2y 2 = L2
gives x2 + 2L2 − 4x2 = L2 ⇒
q
√
√
2
2
3x = L ⇒ x = L/ 3 (since x > 0) and then y = L2 − 2(L/ 3)2 =
√
L/ 3.
√
√
So the only critical point is (L/ 3, L/ 3) which, from the geometrical
nature of the problem, must give an absolute
maximum. Thus the maxi√
√
√ 2q
√
√
mum volume is V (L/ 3, L/ 3) = (L/ 3) L2 − (L/ 3)2 − (L/ 3)2 =
√
L3 /(3 3) cubic units.
55. Note that the variables are m and b, and f (m, b) =
Then fm =
n
X
n
X
[yi − (mxi + b)]2 .
i=1
−2xi [yi − (mxi + b)] = 0
i=1
implies
n
n
n
n
X
X
X
X
(xi yi − mx2i − bxi ) = 0 or
xi yi = m
x2i + b
xi and
i=1
fb =
n
X
i=1
−2[yi − (mxi + b)] = 0 implies
i=1
3
i=1
i=1
n
X
yi = m
i=1
n
X
xi +
n
X
i=1
n
X
b=m
i=1
!
xi +nb. Thus we have the two desired
i=1
equations.
Now fmm =
n
X
2x2i , fbb =
i=1
n
X
2 = 2n and fmb =
i=1
n
X
2xi . And fmm (m, b) >
i=1
0 always and
D(m, b) = 4n
n
X
i=1
!
x2i
−4
n
X
!2
xi
i=1

= 4 n
n
X
i=1
!
x2i
−
n
X
!2 
xi
>
i=1
0 always so the solutions of these two equations do indeed minimize
n
X
i=1
4
d2i .