Math 217: Some Assignment 5 Solutions
14.6 # 16: the unit vector in the direction v = h−1, −2, 2i is
u = v/|v| =
and
1
h−1, −2, 2i,
3
xz
xy
yz
, √
, √
i(3,2,6) · u
Du f (3, 2, 6) = ∇f (3, 2, 6) · u = h √
2 xyz 2 xyz 2 xyz
1
= h1, 3/2, 1/2i · h−1, −2, 2i = −1
3
2 st
st
# 22: Since at (s, t) = (0, 2), ∇f =
√ ht e , (1 + st)e i = h4, 1i, the maximum rate of
change of f at that point is |h4, 1i| = 17, which occurs in the direction of the gradient
< 4, 1 >.
# 56: The point (1, 1, 2) is easily seen to lie on both the ellipsoid and the sphere. A
normal to the ellipsoid at that point,
∇(3x2 + 2y 2 + z 2 )|(1,1,2) = h6x, 4y, 2zi|(1,1,2) = h6, 4, 4i
and a normal to the sphere at that point
∇(x2 + y 2 + z 2 − 8x − 6y − 8z)|(1,1,2) = h2x − 8, 2y − 6, 2z − 8i|(1,1,2) = h−6, −4, −4i
are parallel (since they are multiples of each other), and so the two surfaces have the same
tangent plane there.
14.7 # 10: f (x, y) = xy − x2 y − xy 2 , so the critical point equation is
∇f = hy − 2xy − y 2 , x − x2 − 2xyi = hy(1 − 2x − y), x(1 − x − 2y)i = h0, 0i
whose solutions are (0, 0), (0, 1), (1, 0), (1/3, 1/3). Now
2
D(x, y) := fxx (x, y)fyy (x, y) − fxy
(x, y) = (−2y)(−2x) − (1 − 2x − 2y)2
So D(0, 0) = −1, and this is a saddle. D(0, 1) = D(1, 0) = −4, so these are also saddles.
Finally, D(1/3, 1/3) = 4/9 − 1/9 = 1/3 > 0 and fxx (1/3, 1/3) = −2/3 < 0, so this is a
local maximum.
# 36: The function f (x, y) = x3 − 3x − y 3 + 12y has critical points where
∇f = h3x2 − 3, −3y 2 + 12i = h0, 0i,
1
namely the four points (±1, ±2), of which (±1, 2) are in the region in question. Note
f (1, 2) = 14, f (−1, 2) = 18. Next we deal with the four boundary components. The
top: −2 ≤ x ≤ 3, f (x, 3) = x3 − 3x + 9. 0 = (x3 − 3x + 9)0 = 3x2 − 3 so x = ±1
and f (−1, 3) = 11, f (1, 3) = 7. At the corners, f (−2, 3) = 7, f (2, 3) = 11. The left:
−2 ≤ y ≤ 3, f (−2, y) = −2 − y 3 + 12y. 0 = (−2 − y 3 + 12y)0 = −3y 2 + 12 so y = ±2 and
f (−2, −2) = −18 (one of the corners), f (−2, 2) = 14. The other corner f (−2, 3) = 7 we
already did. The right: 2 ≤ y ≤ 3, f (2, y) = 2 − y 3 + 12y. 0 = (2 − y 3 + 12y)0 = −3y 2 + 12
so y = ±2 and f (2, 2) = 18 (one of the corners). The other corner f (2, 3) = 11 we already
did. The bottom: −2 ≤ x ≤ 2, f (x, x) = 9x is minimized/maximized at the corners as
already computed f (−2, −2) = −18, f (2, 2) = 18. So the absolute minimum is −18, and
the absolute maximum is 18.
# 46: let L, W , and H denote the three dimensions of the box (in cm). Since the
volume is 1000 cm3 , we have H = 1000/(LW ), and so the surface area is
1000 1000
+
),
A(L, W ) = 2(LW + LH + W H) = 2(LW +
W
L
and out job is to minimize A(L, W ) over all L > 0, W > 0. To find critical points:
1000
1000
0 = ∇A(L, W ) = 2hW − 2 , L −
i
L
W2
so W = 1000/L2 and L = 1000/W 2 = L4 /1000 whence H = W = L = (1000)1/3 = 10
yielding surface area A = 600 cm2 . Since this is the only critical point, and A(L, W ) → ∞
if |hL, W i| → ∞ of W → 0 or L → 0, it must be a minimum. So L = W = H = 10 cm
yields minimum area.
#55: given the (xi , yi ), i = 1, 2, . . . , n, we want to minimize the function
n
X
f (m, b) :=
(mxi + b − yi )2 .
i=1
This requires that
0 = fm (m, b) =
n
X
(
2(mxi + b − yi )xi = 2 m
n
X
i=1
x2i + b
i=1
so
m
n
X
x2i
+b
i=1
n
X
xi =
i=1
n
X
n
X
xi −
i=1
n
X
)
xi yi
n=1
xi yi ,
n=1
as well as
0 = fb (m, b) =
n
X
(
2(mxi + b − yi ) = 2 m
i=1
n
X
i=1
so
m
n
X
xi + nb =
i=1
n
X
i=1
Oct. 21, 2013
2
yi .
xi + nb −
n
X
i=1
)
yi
,