Austin Cholley Zhu MAT 267 ONLINE A Fall 2021
Assignment Section 11.7 due 09/12/2021 at 11:59pm MST
Problem 1. (1 point)
The function fhas continuous second derivatives, and a critical
point at (3,8).
Suppose fxx(3,8) = −10,fxy(3,8) = −8,fyy(3,8) = 10.
Then the point (3,8):
•A. cannot be determined
•B. is a saddle point
•C. is a local minimum
•D. is a local maximum
•E. None of the above
Solution:
SOLUTION
We compute the discriminant
D(3,8) = fxx(3,8)fyy(3,8)−[fxy(3,8)]2= (−10)(10)−(−8)2=
−164.
Since D(3,8)<0, fhas a saddle point at (3,8)by the Second
Derivative Test.
Answer(s) submitted:
•B
(correct)
Correct Answers:
•B
Problem 2. (1 point)
The function fhas continuous second derivatives, and a critical
point at (6,1).
Suppose fxx(6,1) = 13,fxy(6,1) = 3,fyy(6,1) = 12.
Then the point (6,1):
•A. is a saddle point
•B. is a local minimum
•C. cannot be determined
•D. is a local maximum
•E. None of the above
Solution:
SOLUTION
We compute the discriminant
D(6,1) = fxx(6,1)fyy(6,1)−[fxy(6,1)]2= (13)(12)−(3)2=147.
Since D(6,1)>0 and fxx(6,1)>0, fhas a local minimum at
(6,1)by the Second Derivative Test.
Answer(s) submitted:
•B
(correct)
Correct Answers:
•B
1
Problem 3. (1 point)
Each of the following functions has at most one critical point.
Graph a few level curves and a few gradiants and, on this basis
alone, decide whether the critical point is a local maximum (MA),
a local minimum (MI), or a saddle point (S). Enter the appropriate
abbreviation for each question, or N if there is no critical point.
(1) f(x,y) = e−4x2−2y2
Type of critical point:
(2) f(x,y) = e4x2−2y2
Type of critical point:
(3) f(x,y) = 4x2+2y2+4
Type of critical point:
(4) f(x,y) = 4x+2y+4
Type of critical point:
Solution:
SOLUTION
(1) The level curves have the form k=e−4x2−2y2or, for
0<k<1, ln(k) = −4x2−2y2. These equations rep-
resent a family of ellipses centered at the origin. Thus
the origin is either a local maximum or a local mini-
mum. To determine whether it is a maximum or a mini-
mum, we take a look at the gradient. We have ∇f(x,y) =
h−8xe−4x2−2y2,−4ye−4x2−2y2i. By substituting values of
xand ywe can see that the gradient points generally to-
wards the origin. Since the gradient points in the direction
of maximum ascent, we conclude that the origin is local
maximum.
(2) The level curves have the form k=e4x2−2y2or, for 0 <k<
1, ln(k) = 4x2−2y2. These equations represent a family
of hyperbolas, centered at the origin. Thus the origin is a
saddle point.
(3) The level curves have the form k=4x2+2y2+4 or
4x2+2y2=k−4. If k>4, these equations represent a
family of ellipses centered at the origin. Thus the origin
is either a local maximum or a local minimum. To de-
termine whether it is a maximum or a minimum, we take
a look at the gradient. We have ∇f(x,y) = h8x,4yi. By
substituting values of xand ywe can see that the gradient
points away from the origin. Since the gradient points in
the direction of maximum ascent, we conclude that the
origin is local minimum.
(4) The level curves have the form k=4x+2y+4. This is
a family of parallel line. Thus the function does not have
a critical point ( note that the equation z=4x+2y+4
represents a plane.)
Answer(s) submitted:
•ma
•s
•mi
•n
(correct)
Correct Answers:
•MA
•S
•MI
•N
2
Problem 4. (1 point)
Suppose f(x,y) = x2+y2−8x−10y+2
(A) How many critical points does fhave in R2?
(B) If there is a local minimum, what is the value of the discrimi-
nant D at that point? If there is none, type N.
(C) If there is a local maximum, what is the value of the discrimi-
nant D at that point? If there is none, type N.
(D) If there is a saddle point, what is the value of the discriminant
D at that point? If there is none, type N.
(E) What is the maximum value of fon R2? If there is none, type
N.
(F) What is the minimum value of fon R2? If there is none, type
N.
Solution:
SOLUTION
(A) fx(x,y) = 2x−8,fy(x,y) = 2y−10. The partial derivatives
are defined everywhere and they are both zero at the point (4,5).
Thus this is the only critical point.
(B) D(4,5) = fxx(4,5)fyy(4,5)−[fxy(4,5)]2=2(2)−02=4.
Since D(4,5)>0 and fxx(4,5) = 2>0, the point (4,5)is a lo-
cal minimum.
(C) There are no local maxima.
(D) There are no saddle points.
(E) Since R2is unbounded and there are no local maximum, there
is no maximum value of fin R2.
(F) When (x,y)→(±∞,±∞), the function f(x,y)grows un-
bounded, thus the minimum value of fis attained at the local
minimum (4,5)and it is given by
f(4,5) = (4)2+ (5)2−(8)(4)−(10)(5) + 2=−39
Answer(s) submitted:
•1
•4
•n
•n
•n
•2-4ˆ(2)-5ˆ(2)
(correct)
Correct Answers:
•1
•4
•N
•N
•N
•2 - 4**2 - 5**2
3
Problem 5. (1 point)
Consider the function
f(x,y) = y√x−y2−5x+19y.
Find and classify the critical point of the function in the interior
of its domain.
fx=
fy=
fxx =
fxy =
fyy =
The critical point is
( , )
Classification:
•
• local minimum
• local maximum
• saddle point
• cannot be determined
Solution:
SOLUTION
fx(x,y) = y
2√x−5
fy(x,y) = √x−2y+19
fxx(x,y) = −y
4x3/2
fxy(x,y) = 1
2√x
fyy(x,y) = −2
The interior of the domain of the function fis the set of (x,y)
points with x>0.
Both fxand fyare continuous and defined for x>0. Thus, to find
the critical points, we solve fx=0,fy=0, that is,
(y
2√x−5=0
√x−2y+19 =0
⇔y−10√x=0
√x−2y+19 =0
From the first equation, we have y=10√xand, substituting into
the second, yields −19√x+19 =0.
Thus √x=1⇒x=1 and y=10.
The only critical point in the interior of the domain is (1,10).
Use the second derivative test to classify the critical point:
D(x,y) = fxx ·fyy −(fxy)2=y
2x3/2−1
4x, and
D(1,10) = 5−1
4>0.
Since fxx(1,10) = −5
2<0, the critical point is a local maximum.
Answer(s) submitted:
•(y/(5sqrt(x)))-1
•sqrt(x)-5y+19
•-(y/(20xˆ(((19)/5))))
•(1/(2sqrt(x)))
•-2
•1
•5
•local maximum
(score 0.5)
Correct Answers:
•y/(2*sqrt(x)) - 5
•sqrt(x) - 2*y + 19
•-y/(4*x**(3/2))
•1/(2*sqrt(x))
•-2
•1
•10
•local maximum
4
Problem 6. (1 point)
Suppose f(x,y) = xy(1−8x−9y).
f(x,y)has 4 critical points. List them in increasing lexographic
order. By that we mean that (x, y) comes before (z, w) if x<zor
if x=zand y<w. Also, determine whether the critical point a
local maximum, a local minimim, or a saddle point.
First point ( , ).
Classification:
•
• local minimum
• local maximum
• saddle point
• cannot be determined
(local minimum, local maximum, saddle point, cannot be deter-
mined).
Second point ( , ).
Classification:
•
• local minimum
• local maximum
• saddle point
• cannot be determined
(local minimum, local maximum, saddle point, cannot be deter-
mined).
Third point ( , ) .
Classification:
•
• local minimum
• local maximum
• saddle point
• cannot be determined
(local minimum, local maximum, saddle point, cannot be deter-
mined).
Fourth point ( , ).
Classification:
•
• local minimum
• local maximum
• saddle point
• cannot be determined
(local minimum, local maximum, saddle point, cannot be deter-
mined).
Solution:
SOLUTION
fx(x,y) = y(1−16x−9y),fy(x,y) = x(1−8x−18y).
Then fx=0 implies y=0 or y=1
9−16
9x.
If y=0, then substitution into fy=0 gives x(1−8x) = 0⇒x=0
or x=1
8, so we have critical points (0,0)and 1
8,0.
If y=1
9−16
9x, then substitution into fy=0 gives x(24x−1) =
0⇒x=0 or x=1
24 .
If x=0, then y=1
9, and if x=1
24 , then y=1
27 , so 0,1
9and
1
24 ,1
27 . are critical points.
Thus the four critical points are (0,0),0,1
9,1
24 ,1
27 ,1
8,0,
We use the second derivative test to classify the points: fxx(x,y) =
−16y,fyy(x,y) = −18x,fxy(x,y) = 1−16x−18y. Thus D(x,y) =
(−16y)(−18x)−(1−16x−18y)2.
D(0,0) = 0−12=−1<0.
D0,1
9=0−(−1)2=−1<0
Hence (0,0)and 0,1
9are saddle points.
D1
24 ,1
27 =4
9−−1
32=1
3>0
fxx 1
24 ,1
27 =−16
27 <0.
Thus 1
24 ,1
27 is a local maximum.
D1
8,0=0−(−1)2=−1<0
Thus 1
8,0is a saddle point.
Answer(s) submitted:
•0
•0
•saddle point
•0
•(1/9)
•saddle point
•(1/(24))
•(1/(27))
•local maximum
•(1/8)
•0
•saddle point
(correct)
Correct Answers:
•0
•0
•saddle point
•0
•0.111111
•saddle point
•0.0416667
•0.037037
•local maximum
•0.125
•0
•saddle point
5
Problem 7. (1 point)
Consider the function f(x,y) = xy −4y−16x+64 on the region
on or above y=x2and on or below y=20.
Find the absolute minimum value:
Find the points at which the absolute minimum value is attained.
List your answer sas points in the form (a,b).
.
Find the absolute maximum value:
Find the points at which the absolute maximum value is attained.
List your answers as points in the form (a,b)
.
Solution:
SOLUTION
f=xy −4y−16x+64 ⇒fx(x,y) = y−16,fy(x,y) = x−4, so
the only critical point is (4,16)(which is in the domain) where
f(4,16) = 0.
Along the parabola y=x2,−√20 ≤x≤√20:
g(x) = f(x,x2) = xx2−4x2−16x+64. Plotting this cubic poly-
nomial in the interval −√20,√20and solving g0(x) = 0, we can
see that it attains its minimum value at x=−√20 and its maxi-
mum value at x=−4
3.
The minimum value is f(−√20,20) = −4√20 −16 ≈−33.8885.
The maximum value is f−4
3,16
9=2048
27
Along the line y=20,−√20 ≤x≤√20:
h(x) = f(x,20)=+4x−16 This is an increasing line that at-
tains its minimum value at x=−√20 and its maximum value at
x=√20.
The minimum value is the same one we evaluated along the
parabola.
The maximum value is h(√20) = f(√20,20) = +4√20 −16 ≈
1.88854
Thus the absolute minimum value of the function is −4√20 −
16 ≈ −33.8885 attained at (−√20,20).
The absolute maximum value of the function is 2048
27 attained at
(−4
3,16
9).
Answer(s) submitted:
•-64
•(-sqrt(20),20)
•
•
(score 0.25)
Correct Answers:
•-33.8885
•(-4.47214,20)
•75.8519
•(-1.33333,1.77778)
Problem 8. (1 point)
Consider the function f(x,y) = 2x3+y4on the region {(x,y)|x2+
y2≤64}.
Find the absolute minimum value:
Find the point(s) at which the absolute minimum is attained.
List your answer as comma separated list, e.g. (1, 1), (2,3)
).
Find the absolute maximum value:
Find the point(s) at which the absolute maximum is attained.
List your answer as comma separated list, e.g. (1,1), (2,3)
).
Solution:
SOLUTION
f(x,y) = 2x3+y4⇒fx(x,y) = 6x2,fy(x,y) = 4y3. Thus the only
critical point is (0,0), which is in the interior of the domain, and
f(0,0) = 0
The boundary of the region is x2+y2=64 ⇒y=
±√64 −x2,−8≤x≤8. Substituting in the function fyields
g(x) = f(x,±√64 −x2) = 2x3+ (64 −x2)2,−8≤x≤8.
Plotting the function g(x)in the given interval, we can see that it
attains its minimum at x=−8 and its maximum at x=0.
We have f(−8,0) = −1024 and f(0,−8) = f(0,8) = 4096
Thus the absolute minimum value is −1024 attained at (−8,0)
and the absolute maximum value is 4096 attained at (0,−8)and
(0,8).
Answer(s) submitted:
•-1024
•(-8,0)
•4098
•(0,-8),(0,8)
(correct)
Correct Answers:
•-1024
•(-8,0)
•4096
•(0,-8), (0,8)
6
Problem 9. (1 point)
Find the coordinates of the point (x,y,z)on the plane z=
4x+4y+2 which is closest to the origin.
x=
y=
z=
Solution:
SOLUTION
Let dbe the distance from (0,0,0)to any point (x,y,z)on the
plane z=4x+4y+2, so d=px2+y2+z2where z=4x+4y+2.
We minimize d2=f(x,y) = x2+y2+ (4x+4y+2)2.
Then fx(x,y) = 2x+4·2(4x+4y+2) = 34x+32y+16 and
fy(x,y) = 2y+4·2(4x+4y+2) = 32x+34y+16
Solving 34x+32y+16 =0 and 32x+34y+16 =0 simultaneously
gives x=−8
33 ,y=−8
33 , so the only critical point is −8
33 ,−8
33 .
An absolute minimum exists (since there is a minimum distance
from the point to the plane) and it must occur at a critical point.
The corresponding zvalue is z=4−8
33 +4−8
33 +2=2
33 .
So the point on the plane which is closest to the origin is
−8
33 ,−8
33 ,2
33
Answer(s) submitted:
•((11)/(sqrt(10)))
•
•
(incorrect)
Correct Answers:
•-0.242424
•-0.242424
•0.0606061
Problem 10. (1 point)
Find the point(s) on the surface z2=xy +1 which are closest to
the point (10,11,0).
List points as a comma-separated list, (e.g., (1,1,-1), (2, 0, -1),
(2,0, 3)).
Solution:
SOLUTION
49 Let dbe the distance from (10,11,0)to any point (x,y,z)
on the surface z2=xy +1, so d=p(x−10)2+ (y−11)2+z2
where z2=xy +1.
We minimize d2=f(x,y) = (x−10)2+ (y−11)2+ (xy +1).
Then fx(x,y) = 2x+y−20 and
fy(x,y) = x+2y−11
Solving 2x+y−20 =0 and x+2y−11 =0 simultaneously gives
x=6,y=8, so the only critical point is (6,8).
An absolute minimum exists (since there is a minimum distance
from the point to the surface) and it must occur at a critical point.
The corresponding zvalues are z=±p1+6(8) = ±7.
So the points on the surface which are closest to the point
(10,11,0)are
(6,8,7),(6,8,−7)
Answer(s) submitted:
•(10,0,4),(10,0,-4)
(incorrect)
Correct Answers:
•(6,8,-7), (6,8,7)
7
Problem 11. (1 point)
Find the volume of the largest rectangular box in the first octant
with three faces in the coordinate planes, and one vertex in the
plane x+6y+10z=60.
Largest volume is
Solution:
SOLUTION
Let x,y,z,be positive numbers. Then x=60 −6y−10zand
we want to maximize V=xyz = (60 −6y−10z)yz =f(y,z).
Then fy(y,z) = (−12y−10z+60)zand fz(y,z) = (−20z−6y+
60)y.
Since y>0 and z>0, to find the critical points, we can solve
−12y−10z+60 =0,−20z−6y+60 =0.
Solving the two equations simultaneously, yields the solution
y=10
3,z=2.
Then x=60 −610
3−10(2) = 20 and the largest possible vol-
ume is given by
V=2010
3(2) = 400
3
Answer(s) submitted:
•
(incorrect)
Correct Answers:
•133.333
Problem 12. (1 point)
The contours of a function fare shown in the figure below.
For each of the points shown, indicate whether you think it is a
local maximum, local minimum, saddle point, or none of these.
(a) Point P is
• ?
• a local maximum
• a local minimum
• a saddle point
• none of these
(b) Point Q is
• ?
• a local maximum
• a local minimum
• a saddle point
• none of these
(c) Point R is
• ?
• a local maximum
• a local minimum
• a saddle point
• none of these
(d) Point S is
• ?
• a local maximum
• a local minimum
• a saddle point
• none of these
Solution:
SOLUTION
(a) Point P is a local maximum
(b) Point Q is a local maximum
8
(c) Point R is none of these
(d) Point S is a saddle point
Answer(s) submitted:
•a local minimum
•a local maximum
•a saddle point
•a saddle point
(score 0.75)
Correct Answers:
•a local maximum
•a local maximum
•none of these
•a saddle point
Problem 13. (1 point)
A contour diagram for a function f(x,y)is shown below.
Estimate the position and approximate value of the global maxi-
mum and global minimum on the region shown.
Global maximum at of
Global minimum at of
Solution:
SOLUTION
The maximum value, which is slightly above -17, say -16.5, oc-
curs approximately at (0,0). The minimum value, which is about
-27.5, occurs approximately at (2.5,5).
Answer(s) submitted:
•(0,0)
•-16
•(2.5,5)
•-27
(correct)
Correct Answers:
•(0,0)
•-16.5
•(2.5,5)
•-27.5
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