HW14-HW15
Qinfeng Li
October 4
Problem 1. (HW 16, # 2) Find the local maximum and minimum values and
saddle point(s) of the function. f (x, y) = xy + x1 + y1 .
Solution. fx = y − x12 , fy = x − y12 , fxx = x23 , fxy = 1, fyy = y23 . Then fx = 0
implies y = x12 and fy = 0 implies x = y12 . Substituting the first equation into
the second gives x = (1/x1 2 )2 = x4 , thus x(x3 − 1) = 0, and hence x = 0 or
x = 1. f is not defined when x = 0, and when x = 1 we have y = 1, so the only
critical point is (1, 1). D(1, 1) = (2)(2) − 12 = 3 > 0 and fxx (1, 1) = 2 > 0, so
f (1, 1) = 3 is a local minimum.
Problem 2. (HW 16, # 4) Find the local maximum and minimum values and
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2
saddle point(s) of the function. f (x, y) = (x2 + y 2 )ey −x .
Solution.
fx = 2xey
2
−x2
2
−x2
(−2x) = 2xey
2
2
2
2
−x2
(1 − x2 − y 2 ),
2
2
−x2
2
2
2
2
(−2y) + 2x(2y)ey −x (1 − x2 − y 2 ) = −4xyey −x (x2 + y 2 ),
2
2
2
2
2
2
2
2
= 2yey −x (2y)+(1+x2 +y 2 ) 2y(2yey −x ) + 2ey −x = 2ey −x (1 + x2 + y 2 )(1 + 2y 2 ) + 2y 2 ).
fxy = 2xey
fyy
2
+ (x2 + y 2 )ey
+ (x2 + y 2 )ey −x (2y) = 2yey −x (1 + x2 + y 2 ),
2
2
2
2
2
2
2
2
= 2xey −x (−2x)+(1−x2 −y 2 ) 2x(−2xey −x ) + 2ey −x = 2ey −x (1 − x2 − y 2 )(1 − 2x2 ) − 2x2 ),
fy = 2yey
fxx
−x2
2
fy = 0 implies y = 0, and substituting into fx = 0 gives 2xe−x (1 − x2 ) = 0,
hence x = 0 or x = ±1. Thus the critical points are (0, 0) and (±1, 0). Now
D(0, 0) = (2)(2) − 0 > 0 and fxx (0, 0) = 2 > 0, so f (0, 0) = 0 is a local
minimum. D(±1, 0) = (−4e−1 )(4−1 ) − 0 < 0, so (±1, 0) are saddle points.
Problem 3. (HW 17, # 1) Find the absoltue maximum and minimum values
of f on the set D.
f (x, y) = 4x + 6y − x2 − y 2 , D = {(x, y) : 0 ≤ x ≤ 4, 0 ≤ y ≤ 5}.
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Solution. fx (x, y) = 4 − 2x and fy (x, y) = 6 − 2y, so the only critical point is
(2, 3) (which is in D) where f (2, 3) = 13. Let L1 , L2 , L3 , L4 be the lower, right,
upper, left boundary of D.
Along L1 : y = 0, so f (x, y) = f (x, 0) = 4x − x2 , 0 ≤ x ≤ 4. Using
calculus 1 knowledge, we know this function has maximum value when x = 2
where f (2, 0) = 4 and a minimum value both when x = 0 and x = 4, where
f (0, 0) = f (4, 0) = 0.
Along L2 : x = 4, so f (x, y) = f (4, y) = 6y − y 2 , 0 ≤ y ≤ 5. Using calculus 1 knowledge, we know this function has maximum value when y = 3 where
f (4, 3) = 9 and a minimum value both when y = 0 where f (4, 0) = 0.
Along L3 : y = 5, so f (x, y) = f (x, 5) = −x2 + 4x + 5, 0 ≤ x ≤ 4. Using
calculus 1 knowledge, we know this function has maximum value when x = 2
where f (2, 5) = 9 and a minimum value both when x = 0 and x = 4, where
f (0, 5) = f (4, 5) = 5.
Along L4 : x = 0, so f (x, y) = f (0, y) = 6y − y 2 , 0 ≤ y ≤ 5. Using calculus 1 knowledge, we know this function has maximum value when y = 3 where
f (0, 3) = 9 and a minimum value both when y = 0 where f (0, 0) = 0.
Thus the absolute maximum is f (2, 3) = 13 and the absolute minimum is
attained at both (0, 0) and (4, 0), where f (0, 0) = f (4, 0) = 0.
Problem 4. (HW 17, # 4) Find the points on the surface y 2 = 9 + xz that are
cloest to the origin.
Solution.
The distance from the origin to a point (x, y, z) on the surface is
p
d = x2 + y 2 + z 2 where y 2 = 9 + xz, so we minimize d2 = x2 + 9 + xz + z 2 =
f (x, y). Then fx = 2x + z, fz = x + 2z. Hence solving fx = 0, fz = 0 imply
x = z = 0, so the only critical point is (0, 0). D(0, 0) = (2)(2) − 1 = 3 > 0 with
fxx = 2 > 0, so this is a minimum. Thus y 2 = 9 + 0, and hence y = ±3 and
the points on the surface closest to the origin are (0, ±3, 0).
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