Ad2014 calvec-industrial-jllf.ps14000302.departamental2

1(ALU) 1IM1 ANGEL DAVID ORTIZ RESENDIZ AD2014-CALVEC-INDUSTRIAL-JLLFAssignment DEPARTAMENTAL2 due 10/19/2014 at 11:47pm CDT

1. (1 pt) Sketch a graph of each surface. Then, match eachsurface to its brief description in words.

? 1. z = x2 +9y2 +2? 2. x2 + y2 + z2 = 25? 3. z = 36x2− y2

? 4. z2 = x2 +9y2

? 5. x2 + y2 = 9

A. Hyperbolic paraboloid (saddle)B. ConeC. Elliptic paraboloidD. SphereE. Circular cylinderF. Elliptical cylinderG. Sinusoidal cylinderH. Skew lineI. Circle

Answer(s) submitted:

CDABI

(score 0.600000023841858)Correct Answers:

CDABE

2. (1 pt) Consider the concentration, C, (in mg/liter) of a drugin the blood as a function of the amount of drug given, x, andthe time since injection, t. For 0≤ x≤ 5 mg and t ≥ 0 hours, wehave

C = f (x, t) = 26te−(5−x)t

f (3,5) =

Give a practical interpretation of your answer: f (3,5) isA. the amount of a 3 mg dose in the blood 5 hours afterinjection.B. the concentration of a 3 mg dose in the blood 5 hoursafter injection.C. the amount of a 5 mg dose in the blood 3 hours afterinjection.

D. the concentration of a 5 mg dose in the blood 3 hoursafter injection.E. the change in concentration of a 3 mg dose in theblood 5 hours after injection.F. the change in concentration of a 5 mg dose in theblood 3 hours after injection.


(incorrect)Correct Answers:

0.00590199B

3. (1 pt) For each surface, decide whether it could be a bowl,a plate, or neither. Consider a plate to be any fairly flat surfaceand a bowl to be anything that could hold water, assuming thepositive z-axis is up.

? 1. x+ y+ z = 3? 2. z = 2? 3. z = x2 + y2

? 4. z = 1− x2− y2

? 5. z =−√

5− x2− y2


PlatePlateBowlNeitherBowl

(correct)Correct Answers:

PLATEPLATEBOWLNEITHERBOWL

4. (1 pt) Sketch a graph of each surface. Then, match eachsurface to its brief description in words.

1

? 1. z = x2 +49y2 +2? 2. z2 = x2 +4y2

? 3. x =−2? 4. z = 16x2− y2

? 5. z = 4? 6. z = 2y2

? 7. x2 + y2 + z2 = 49? 8. x2 + y2 = 16? 9. 3x+3y+6z = 4

A. SphereB. Vertical planeC. PointD. Skew planeE. Elliptical cylinderF. Vertical lineG. ConeH. Elliptic paraboloidI. Circular cylinderJ. Parabolic cylinder

K. Horizontal planeL. Skew line

M. Hyperbolic paraboloid (saddle)


HGFMKJAID


HGBMKJAID

5. (1 pt)

Match each function with its graph. Click on a graphto make it larger.

? 1. f (x,y) =1

x2 + y2

? 2. f (x,y) = x+2y+3? 3. f (x,y) = x3− sin(y)? 4. f (x,y) =−y2

? 5. f (x,y) =−4e−x2−y2

A B C

D EAnswer(s) submitted:

DEACB


DEACB

6. (1 pt)

Match each function with its graph. Click on a graphto make it larger.

? 1. z = 4cos(√

x2 + y2)

? 2. z =− 1x2 + y2

? 3. z = 6xye−x2−y2

? 4. z =4sin(x2 + y2)

x2 + y2

? 5. z = (4x2 +2y2)e1−x2−y2

2

A B C

D EAnswer(s) submitted:

BEDCA


BEDCA

7. (1 pt) A car rental company charges a one-time applicationfee of 20 dollars, 40 dollars per day, and 13 cents per mile forits cars.

(a) Write a formula for the cost, C, of renting a car as a fun-ction of the number of days, d, and the number of miles driven,m.C =

(b) If C = f (d,m), then f (4,730) =Answer(s) submitted:


20+40*d+13*m/100274.9

8. (1 pt) Find a formula for the shortest distance from a point(a,b,c) to the z-axis.distance =

Answer(s) submitted:sqrt(aˆ2+bˆ2)


sqrt(aˆ2+bˆ2)

9. (1 pt) (a) Describe the set of points whose distance fromthe z-axis equals the distance from the xy-plane.

A. A cone opening along the z-axisB. A cylinder opening along the x-axisC. A cylinder opening along the y-axisD. A cone opening along the x-axisE. A cone opening along the y-axisF. A cylinder opening along the z-axis

(b) Find the equation for the set of points whose distance fromthe z-axis equals the distance from the xy-plane.

A. y2 = x2 + z2

B. x2 + y2 = r2

C. y2 + z2 = r2

D. z2 = x2 + y2

E. x2 + z2 = r2

F. x2 = y2 + z2



AD

10. (1 pt)

The domain of the function f (x,y) =√

x+√

y isA. All of the xy-planeB. The first quadrantC. The union of two intervalsD. The area inside a parabolaE. The first and third quadrants

Answer(s) submitted:b


B

11. (1 pt)

The domain of the function f (x,y) = 3x+5yx2+y2−4 is

A. The area inside a circle (not including the circle)B. The union of two intervalsC. The area inside a circle (including the circle)D. The first quadrantE. All the xy-plane except a circle

Answer(s) submitted:e


3

E

12. (1 pt) Find a possible equation for a linear function withthe given contour diagram.

L(x,y) =


-165/2 x-105/2 y


-95 + 5x - 5y

13. (1 pt)

(a) Could contour diagram A represent a linear fun-ction? Why or why not?

A. No, because the lines are not equally spaced.B. No, because the slopes of the lines are positive.C. Yes, because it is in the xy-plane.D. Yes, because it passes through the origin.E. Yes, because the lines are parallel.F. No, because it passes through the origin.

(b) Could contour diagram B represent a linear function?Why or why not?

A. No, because the slopes of the lines are negative.B. No, because the lines are equally spaced.C. Yes, because the lines are parallel and equally spa-ced.D. Yes, because the lines are parallel.E. No, because it passes through the origin.F. Yes, because it passes through the origin.

A B


AA


AC

14. (1 pt) Identify the following surfaces as an elliptical pa-raboloid, hyperbolic paraboloid, a hyperboloid of one sheet, ahyperboloid of two sheets, a cone, a circular cylinder, an ellip-tical cylinder, or a parabolic cylinder, and identify the axis ofsymmetry as the x-axis, the y-axis, or the z-axis.

a. ? ? 2y2−4x2−5z2 +1 = 0

b. ? ? −7x2 +2y2− z2 = 0

c. ? ? 7y2 +7z2 = 1

d. ? ? −5x2−6y2 + z2 = 1

e. ? ? 4x2 +8y2− z = 0



4

hyperboloid of one sheetalong the y-axisconealong the y-axiscircular cylinderalong the x-axishyperboloid of two sheetsalong the z-axiselliptical paraboloidalong the z-axis

15. (1 pt) Decide if the given level surface can be expressedas the graph of a function f (x,y).

? 1. 8x+2y−8z−9 = 0? 2. z− x2−3y2 = 0? 3. x2 + y2 + z2− z−3 = 0? 4. z2 = x2 +7y2


Yes, it canYes, it canNo, it cannotNo, it cannot


YES, IT CANYES, IT CANNO, IT CANNOTNO, IT CANNOT

16. (1 pt) A manufacturer sells aardvark masks at a price of$280 per mask and butterfly masks at a price of $520 per mask.A quantity of a aardvark masks and b butterfly masks is sold ata total cost of $350 to the manufacturer.

(a) Express the manufacturer’s profit, P , as a function of aand b .P(a,b) = dollars.

(b) The curves of constant profit in the ab-plane are

A. circlesB. hyperbolasC. linesD. parabolasE. ellipses


A


280*a+520*b-350C

17. (1 pt) Sketch a contour diagram of each function. Then,decide whether its contours are predominantly lines, parabolas,ellipses, or hyperbolas.

? 1. z = x2−5y2

? 2. z = x2 +2y2

? 3. z = y−2x2

? 4. z =−5x2


HyperbolasEllipsesEllipsesParabolas


HYPERBOLASELLIPSESPARABOLASLINES

18. (1 pt)

Match each function with its contour plot. Click ona graph to make it larger. Darker areas represent lowerelevations and lighter areas represent higher elevations.

? 1. f (x,y) = x+ y? 2. f (x,y) = sin(x)? 3. f (x,y) = y2

? 4. f (x,y) = (x− y)2

5

A B

C D


ADBC


ADBC

19. (1 pt)

Match each function with its contour plot. Click ona graph to make it larger. Darker areas represent lowerelevations and lighter areas represent higher elevations.

? 1. f (x,y) = 3−√

x2 + y2

? 2. f (x,y) = cos(

x2 + y2

4

)? 3. f (x,y) = 3− x2− y2

? 4. f (x,y) =1

1+ x2 + y2

A B

C D


BCDA


BCDA

20. (1 pt)

Match each function with its contour plot. Click ona graph to make it larger. Darker areas represent lowerelevations and higher elevations represent higher eleva-tions.

? 1. f (x,y) = x− y2

? 2. f (x,y) = y+ x2

? 3. f (x,y) = y− x2

? 4. f (x,y) = x+ y2

6

A B

C D



DCBA

21. (1 pt)Match the functions with the verbal description of the level

surfaces by placing the letter of the verbal description to the leftof the letter of the function.

1. w = x2 + y2 + z2

2. w =√

(x2 + y2 + z2)3. w = x2 +2y2 +3z2

4. w = x2− y2− z2

5. w = x+2y+3z6. w =

√(x+2y+3z)

A. a collection of equally spaced parallel planesB. a collection of concentric ellipsoidsC. a collection of equally spaced concentric spheresD. two cones and two collections of hyperboloidsE. a collection of unequally spaced concentric spheresF. a collection of unequally spaced parallel planes


ec

bdfa


ECBDAF

22. (1 pt)Match the surfaces with the verbal description of the level

curves by placing the letter of the verbal description to the leftof the letter of the surface.

1. z = xy2. z = 1

x−13. z = 2x2 +3y2

4. z =√

(x2 + y2)5. z = x2 + y2

6. z = 2x+3yA. a collection of concentric ellipsesB. two straight lines and a collection of hyperbolasC. a collection of unequally spaced concentric circlesD. a collection of unequally spaced parallel linesE. a collection of equally spaced concentric circlesF. a collection of equally spaced parallel lines


bdaecf


BDAECF

23. (1 pt)Match the functions with the verbal description of the level

surfaces by placing the letter of the verbal description to the leftof the number of the function.

1. w = x2 + y2 + z2

2. w =√

(x2 + y2 + z2)3. w = x2− y2− z2

4. w =√

(x+2y+3z)5. w = x2 +2y2 +3z2

6. w =√

(x2 +2y2 +3z2)7

7. w = x+2y+3zA. a collection of equally spaced concentric spheresB. two cones and two collections of hyperboloidsC. a collection of unequally spaced concentric spheresD. a collection of unequally spaced parallel planesE. a collection of equally spaced parallel planesF. a collection of concentric ellipsoids

Answer(s) submitted:cabeffd


CABDFFE

24. (1 pt) Consider the level surface given by

x2− y2 + z2 = 2.

Match the slices with their correct plots below.1. Slice for x = 12. Slice for x =

√2

3. Slice for y = 04. Slice for x = 0

A B C

D E FAnswer(s) submitted:

cbfd


C

BFD

25. (1 pt) The graphs below show level sets for six differentfunctions, where the red areas represent the the lowest heightsand the purple areas are the highest heights.

A B C

D E F

Match the following functions with the proper graphs above.

1. f (x,y) = 4− x2−3y2

2. f (x,y) = y2−3x3. f (x,y) = 3y− x2

4. f (x,y) = 4−3x2− y2


abde


ECDA

26. (1 pt) Consider the level surface given by

2x2 +5y2 + z2 = 4.

Match the slices with their correct plots below.

1. Slice for x = 02. Slice for z = 1,53. Slice for x = 14. Slice for z = 0

8

A B C

D E F


efcb


EFCB

27. (1 pt) The graphs below show level sets for six differentfunctions, where the red areas represent the the lowest heightsand the purple areas are the highest heights.

A B C

D E F

Match the following functions with the proper graphs above.

1. f (x,y) =√

x2 + y2

2. f (x,y) = e−(x2+y2)/2

3. f (x,y) = 1−√

x2 + y2

4. f (x,y) = 1− x2− y2



EFDB

28. (1 pt) Match the surfaces (a) - (f) below with the contourdiagrams (1) - (6) below those.

(a) (b) (c)

(d) (e) (f)

(a) Surface (a) matches contour ?(b) Surface (b) matches contour ?(c) Surface (c) matches contour ?(d) Surface (d) matches contour ?(e) Surface (e) matches contour ?(f) Surface (f) matches contour ?

(1) (2) (3)

(4) (5) (6)SOLUTION(a) 2

(b) 3(c) 6(d) 1(e) 5(f) 4

Answer(s) submitted:236154


9

236154

29. (1 pt) Find an equation for the contour of f (x,y) =2x2y+13x+10 that goes through the point (5,−1).

Equation:SOLUTIONSince f (5,−1) = 2 · 52 · −1 + 13 · 5 + 10 = 25, an equation

for the contour is

2x2y+13x+10 = 25.



2*y*xˆ2+13*x+10 = 25

30. (1 pt) Match each of the tables shown below with thecontour diagrams below them.

(a)

y\x −1 0 1−1 2 1 2

0 1 0 11 2 1 2

(b)

y\x −1 0 1−1 0 1 0

0 1 2 11 0 1 0

(c)

y\x −1 0 1−1 2 0 2

0 2 0 21 2 0 2

(d)

y\x −1 0 1−1 2 2 2

0 0 0 01 2 2 2

Table (a) : graph ?Table (b) : graph ?Table (c) : graph ?Table (d) : graph ?

1. 2. 3.

4. 5. 6.SOLUTIONThe correct matches are:(a): Table (a) matches graph 6.

(b): Table (b) matches graph 4.(c): Table (c) matches graph 3.(d): Table (d) matches graph 2.



6432

31. (1 pt) Values of f (x,y) = 12 (x+ y−4)(x+ y−3)+ y are

in the table below.(a) Find a pattern in the table. Make a conjecture and use it to

complete the table without computation. Check by using the for-mula for f . (b) Using the formula, check that the pattern holdsfor all x≥ 1 and y≥ 1.

x\y 1 2 3 4 5 61 2 2 3 5 8 122 1 2 4 7 113 1 3 6 104 2 5 95 4 86 7

SOLUTION(a) We notice that the values increment by one as we read

along diagonals of the table with y decreasing as x increases.The pattern restarts at y = 1 on each diagonal with a value oneless than the value at the end of the previous diagonal. Thus,we see that with the value f (1,1) =2, the next diagonal has thevalues f (2,1) =1 and f (1,2) =2; the diagonal after that conti-nues with f (3,1) =1, f (2,2) =2 and f (1,3) =3. This patterncontinues through the rest of that table, giving the table below:

x\y 1 2 3 4 5 61 2 2 3 5 8 122 1 2 4 7 11 163 1 3 6 10 15 214 2 5 9 14 20 275 4 8 13 19 26 346 7 12 18 25 33 42

(b) It appears that the value of f increases by 1 whenever x isdecreased by 1 and y is increased by 1. To check this, compute

f (x−1,y+1)= (1/2)((x−1)+(y+1)−4)((x−1)+(y+1)−4)+(y+1)

= (1/2)(x+ y−4)(x+ y−3)+ y+1 = f (x,y)+1.

So this is correct. Then, it appears that the value of f changesby -1 when moving from the point (1,y) to the point (y + 1,1).We can similarly check this:

f (y+1,1)= (1/2)((y+1)+1−4)((y+1)+1−3)+1 =(1/2)((1+y−4)+1)((1+y−3)+1)+1

=(1/2)((1+y−4)+1)(1+y−3))+(1/2)((1+y−4)+(1+y−3)+1)+1= (1/2)((1+ y−4)(1+ y−3))+ y+1− (3+4)/2+1/2+1

= f (1,y)−1.10



161521142027131926341218253342

32. (1 pt) The temperature T (in ◦C) at any point in the region−10≤ x≤ 10, −10≤ y≤ 10 is given by the function

T (x,y) = 100− x2− y2.

(a) On a sheet of paper, sketch isothermal curves (curves ofconstant temperature) for T = 100◦ C, T = 75◦ C, T = 50◦ C,T = 25◦ C, and T = 0◦ C.

(b) Suppose a heat-seeking bug is put down at each of thefollowing points on the xy-plane. Thinking of the positive y di-rection as north and the positive x direction as east, pick thedirection in which it should move to increase its temperaturefastest.(i) (0,5) the bug should go ?(ii) (−5,0) the bug should go ?(iii) (−5,−5) the bug should go ?

(Think about how your answers to these questions are relatedto the level curves through that point!)

SOLUTION(a) To find the level curves, we let T be a constant.

T = 100− x2− y2, so x2 + y2 = 100−T,

which is an equation for a circle of radius√

100−T centered atthe origin.

At T = 100◦, we have a circle of radius 0 (a point).At T = 75◦, we have a circle of radius 5.At T = 50◦, we have a circle of radius 5

√2.

At T = 25◦, we have a circle of radius 5√

3.At T = 0◦, we have a circle of radius 10.

No matter where we put the bug, it should go straight towardthe origin—the hottest point on the xy-plane. Its direction ofmotion is perpendicular to the tangent lines of the level curves.Thus,at (0,5) the bug should go S,at (−5,0) the bug should go E,and at (−5,−5) the bug should go NE.



SENE

33. (1 pt) On a piece of paper, sketch each of the followingsurfaces:

(i) z = 1− (x2 + y2)(ii) z = 2x2

Use your graphs to fill in the following descriptions of cross-sections of the surfaces.

(a) For (i) (z = 1− (x2 + y2)):Cross sections with x fixed give ?Cross sections with y fixed give ?Cross sections with z fixed give ?

(b) For (ii) (z = 2x2):Cross sections with x fixed give ?Cross sections with y fixed give ?Cross sections with z fixed give ?

SOLUTIONThe graphs of these equations are:

(i)

Thus,Cross sections with x fixed give a downward opening parabolain the yz-planeCross sections with y fixed give a downward opening parabolain the xz-planeCross sections with z fixed give an empty set, or a circle centeredon the origin in the xy-plane

11

(ii)

Thus,Cross sections with x fixed give a horizontal line in the yz-planeCross sections with y fixed give an upward opening parabola inthe xz-planeCross sections with z fixed give two vertical lines in the xy-plane



a downward opening parabola in the yz-planea downward opening parabola in the xz-planean empty set, or a circle centered on the origin in the xy-planea horizontal line in the yz-planean upward opening parabola in the xz-planetwo vertical lines in the xy-plane

34. (1 pt) Without a computer or calculator, match the equa-tions (a)–(f) with the graphs (1)-(6) below.

(a) z = xye−(x2+y2) : ?(b) z = sin(y) : ?(c) z =− 1

x2+y2 : ?

(d) z = cos2(x)cos2(y) : ?(e) z = cos(x) : ?(f) z = 1√

(x−2)2+y2+1: ?

1. 2. 3.

4. 5. 6.

SOLUTION(a) is 1

(b) is 5(c) is 4(d) is 2

(e) is 3(f) is 6


154236


154236

35. (1 pt)

Find the limit

lı́m(x,y)→(−14,2)

xycos(x+7y)

(Enter ’DNE’ if the limit does not exist)Answer(s) submitted:

-28


-28

36. (1 pt)

The largest set on which the funtion f (x,y) =√

x+ y−√x− y is continuous is

A. {(x,y)|− x≤ y≤ x}B. {(x,y)|− x < y≤ x}C. {(x,y)|− x < y < x}D. {(x,y)|x≥ y}E. the whole xy-plane


a


A

37. (1 pt)

Find the limit (enter ’DNE’ if the limit does not exist)

lı́m(x,y)→(0,0)

(6x+ y)2

36x2 + y2

1) Along the x-axis:2) Along the y-axis:3) Along the line y = x :

12

4) The limit is:



111.32432432432432dne

38. (1 pt) Find the limits, if they exist, or type DNE for anywhich do not exist.

lı́m(x,y)→(0,0)

1x2

5x2 +4y2

1) Along the x-axis:2) Along the y-axis:3) Along the line y = mx :4) The limit is:



0.201/(5 + 4*m**2)dne

39. (1 pt)

The largest set on which the function f (x,y) = 1/(17− x2−y2) is continuous is

A. The interior of the circle x2 + y2 = 17, plus the circleB. All of the xy-planeC. The exterior of the circle x2 + y2 = 17D. The interior of the circle x2 + y2 = 17E. All of the xy-plane except the circle x2 + y2 = 17


e


E

40. (1 pt)

Find the limit (enter ’DNE’ if the limit does not exist)Hint: rationalize the denominator.

lı́m(x,y)→(0,0)

(−9x2−2y2)√(−9x2−2y2 +1)−1


2


2

41. (1 pt)Find the limit, if it exists, or type N if it does not exist.

lı́m(x,y,z)→(0,0,0)

xy+5yz+3xzx2 +25y2 +9z2 =


N


N


lı́m(x,y,z)→(2,4,5)5zex2+y2

2x2+4y2+5z2 =Answer(s) submitted:

(25eˆ20)/197


61569187.234745

43. (1 pt)Find the limit, if it exists, or type DNE if it does not exist.

A. lı́m(x,y)→(−5,−4)

e√

(2x2+1y2) =

B. lı́m(x,y)→(0,0)

4x2

4x2 +5y2 =


3374.621351DNE


3374.62135072886DNE

13

44. (1 pt)Find the limit, if it exists, or type DNE if it does not exist.

A. lı́m(x,y)→(0,0)

(x+14y)2

x2 +196y2 =

B. lı́m(x,y)→(0,0)

9x3 +6y3

x2 + y2 =

(Hint for B: use polar coordinates, that is x = r cos(θ),y =r sin(θ) )


DNE0


DNE0


lı́m(x,y,z)→(0,0,0)

3xy+3yz+5xz9x2 +9y2 +25z2 =


N


N

46. (1 pt) In this problem we show that the function

f (x,y) =4x− y2

x+ y2

does not have a limit as (x,y)→ (0,0).(a) Suppose that we consider (x,y)→ (0,0) along the curve

y = 2x1/2. Find the limit in this case:lı́m

(x,2x1/2)→(0,0)

4x−y2

x+y2 =

(b) Now consider (x,y)→ (0,0) along the curve y = 3x1/2.Find the limit in this case:

lı́m(x,3x1/2)→(0,0)

4x−y2

x+y2 =

(c) Note that the results from (a) and (b) indicate that f hasno limit as (x,y)→ (0,0) (be sure you can explain why!) .To show this more generally, consider (x,y)→ (0,0) along thecurve y = mx1/2, for arbitrary m. Find the limit in this case:

lı́m(x,mx1/2)→(0,0)

4x−y2

x+y2 =

(Be sure that you can explain how this result also indicates thatf has no limit as (x,y)→ (0,0).

SOLUTIONWe want to show that f does not have a limit as (x,y) approa-

ches (0,0).(a) If (x,y) tends to (0,0) along the curve y = 2x1/2, then

f (x,y) = f (x,2x1/2) =4x−

(2x0,5

)2

x+(2x0,5)2 = 0.

Thereforelı́mx→0

f (x,2x1/2) = 0.

(b) Similarly, if (x,y) tends to (0,0) along the curve y =3x1/2, then

f (x,y) = f (x,3x1/2) =4x−

(3x0,5

)2

x+(3x0,5)2 =−12

,

and so

lı́mx→0

f (x,3x1/2) =−12

.

Because we have two different values for these limits, weknow that the function doesn’t have a limit at (0,0). No matterhow close we are to the origin, there will be points on f (x,y)that are close to 0 and points that are close to −1

2 , whereas forthe limit to exist the function must get close to a single value.

(c) We can see this more generally by considering the limitas (x,y) tends to (0,0) along the curve y = mx1/2. Then

f (x,y) = f (x,mx1/2) =4−m2

1+m2 ,

so that

lı́mx→0

f (x,mx1/2) =4−m2

1+m2 ,

and we see for every different value of m we will get a differentlimiting value for the function. Thus the function does not havea limit at (0,0).


000


(4 - 2ˆ2)/(1 + 2ˆ2)(4 - 3ˆ2)/(1 + 3ˆ2)(4-mˆ2)/(1+mˆ2)

47. (1 pt) For the function f (x,y) below, determine whetherthere is a value for c making the function continuous everywhe-re. If so, find it.

f (x,y) =

{c+ y x≤ 2,

2− y x > 2

c =(If there is no value of c that works, enter none, and be sure thatyou can explain why there is no such value.)

SOLUTIONThe function f is continuous at all points (x,y) with x 6= 2.

So let’s analyze the continuity of f at the point (2,a). We have

lı́m(x,y)→(2,a)x<2

f (x,y) = lı́my→a

(c+ y) = c+a

andlı́m

(x,y)→(2,a)x>2

f (x,y) = lı́my→a

(2− y) = 2−a.

14

So we need to see if we can find one value for c such thatc + a = 5− a for all a. This would require that c = 5− 2a, butthen c would depend on a, which is exactly what we don’t want.Therefore, we cannot make the function continuous everywhere.

Graphically we can see this by thinking about what the fun-ction looks like: for x ≤ 2, it is a plane with positive y-slope,passing through the line y = 0, z = c. For x > 2, it is a planewith negative y-slope, pssing through the line y = 0, z = y0 = 2.This is shown in the figure below.


none


none

48. (1 pt) Show that the function

f (x,y) =x2y

x4 + y2 .

does not have a limit at (0,0) by examining the following limits.(a) Find the limit of f as (x,y)→ (0,0) along the line y = x.

lı́m(x,y)→(0,0)y=x

f (x,y) =

(b) Find the limit of f as (x,y)→ (0,0) along the line y = x2.lı́m

(x,y)→(0,0)y=x2

f (x,y) =

(Be sure that you are able to explain why the results in (a)and (b) indicate that f does not have a limit at (0,0)!

SOLUTION(a) Let us suppose that (x,y) approaches (0,0) along the line

y = x. Then

f (x,y) = f (x,x) =x3

x4 + x2 =x

x2 +1.

Thereforelı́m

(x,y)→(0,0)y=x

f (x,y) = lı́mx→0

xx2 +1

= 0.

(b) On the other hand, if (x,y) approaches 48 48 501 5020,0)the curve y = x2 we have

f (x,y) = f (x,x2) =x4

2x4 = 0,5

and solı́m

(x,y)→(0,0)y=x2

f (x,y) = lı́mx→0

f (x,x2) = 0,5.

Thus no matter how close they are to the origin, there will bepoints (x,y) such that f (x,y) is close to 0 and points (x,y) suchthat f (x,y) is close to 0,5. So the limit

lı́m(x,y)→(0,0)

f (x,y)

does not exist.Answer(s) submitted:

01/2


01/2

49. (1 pt) Consider the function f whose graph is shownbelow.

This function is given by

f (x,y) =

{6xy

x2+y2 , (x,y) 6= (0,0)

0, (x,y) = (0,0)

(a) Find a formula for the single variable function f (0,y).f (0,y) =What is f (0,0) for this function?f (0,0) =Find its limit as y→ 0:lı́my→0

f (0,y) =

(b) Based on your work in (a), is the single variable functionf (0,y) continuous? ?

(c) Next, similarly consider f (x,0).f (x,0) =

f (0,0) =lı́mx→0

f (x,0) =

(d) Based on this work in (a), is the single variable functionf (x,0) continuous? ?

(e) Finally, consider f along rays emanating from the origin.Note that these are given by y = mx, for some (constant) valueof m.

Find and simplify f on the ray y = x:f (x,x) =

(Notice that this means that y = x is a contour of f . Be sureyou can explain why this is.)

15

Find and simplify f on any ray y = mx.f (x,mx) =

(Again, notice that this means that any ray y = mx is a con-tour of f ; be sure you can explain why.)

(f) Is f (x,y) continuous at (0,0)? ?SOLUTION(a)-(d) We have f (x,0) = 0 for all x and f (0,y) = 0 for all

y, and obviously lı́mx→0

f (x,0) = 0 = lı́my→0

f (0,y), so these are both

continuous (constant) functions of one variable.(e) Along y = x, we have

f (x,y) = f (x,x) = 3.

Because this is a constant value, this is a contour of f . Similarly,along any ray y = mx, we have

f (x,y) = f (x,mx) =6m

m2 +1,

which is also a constant, and we therefore have that all such raysare contours of the function f .

(f) Finally, notice that because we know f (x,y) = 3 alongthe line y = x, and f (x,0) = 0, we can see that no matter howclose we are to the origin, we can find points (x,y) where thevalue f (x,y) is 3 and points (x,y) where the value f (x,y) is 0.So the limit lı́m

(x,y)→(0,0)f (x,y) does not exist. Thus, f is not con-

tinuous at (0,0), even though the one-variable functions f (x,0)and f (0,y) are continuous at (0,0).



000yes000yes6/26*m/(mˆ2 + 1)no

50. (1 pt)Find the limit, if it exists, or type N if it does not exist.lı́m(x,y)→(−3,−2) e

√(4x2+2y2) =


759.94


eˆ(sqrt(4*(-3)ˆ2 + 2*(-2)ˆ2))

51. (1 pt) Suppose f (x,y) =√

tan(x)+ y and u is the unitvector in the direction of 〈0,1〉. Then,

(a) ∇ f (x,y) =

(b) ∇ f (−0,1,6) =

(c) fu (−0,1,6) = Du f (−0,1,6) =Answer(s) submitted:

<(secˆ2(x))/(2 sqrt(tan(x)+y)), 1/(2sqrt(tan(x)+y))><0.2079249084, 0.2058525809>0.2058525809


<1/[2*sqrt(tan(x)+y)]*[sec(x)]ˆ2,1/[2*sqrt(tan(x)+y)]><0.207925,0.205853>0.205853

52. (1 pt) Suppose f (x,y) =4

x2 + y2 and u is the unit vector

in the direction of 〈−3,2〉. Then,

(a) ∇ f (x,y) =

(b) ∇ f (1,2) =

(c) fu (1,2) = Du f (1,2) =Answer(s) submitted:

<-(8x)/(xˆ2+yˆ2)ˆ2, -(8y)/(xˆ2+yˆ2)ˆ2><-8/25, -16/25>-0.0887520314


<-(8*x/[(xˆ2+yˆ2)ˆ2]),-(8*y/[(xˆ2+yˆ2)ˆ2])><-0.32,-0.64>-0.088752

53. (1 pt) Suppose f (x,y) = sin(

2yx

)and u is the unit vector

in the direction of 〈−3,−3〉. Then,

(a) ∇ f (x,y) =

(b) ∇ f (3,π) =

(c) fu (3,π) = Du f (3,π) =Answer(s) submitted:

<(-2ycos(2y/x))/xˆ2, (2cos(2y/x)/x)><pi/9, -1/3>-0.0111245695

16


<-2*y/(xˆ2)*cos(2*y/x),2*x/(xˆ2)*cos(2*y/x)><0.349066,-0.333333>-0.0111246

54. (1 pt) View the curve (y− x)2 + 2 = xy− 3 as a contourof f (x,y).

(a) Use ∇ f (2,3) to find a vector normal to the curve at (2,3).

(b) Use your answer to part (a) to find an implicit equationfor the tangent line to the curve at (2,3).



<-5,0>x = 2

55. (1 pt) The temperature at any point in the plane is given

by T (x,y) =190

x2 + y2 +1.

(a) What shape are the level curves of T ?A. parabolasB. ellipsesC. hyperbolasD. circlesE. linesF. none of the above

(b) At what point on the plane is it hottest?

What is the maximum temperature?

(c) Find the direction of the greatest increase in temperatureat the point (2,−2).What is the magnitude of that greatest increase?

(d) Find the direction of the greatest decrease in temperatureat the point (2,−2).What is the magnitude of that greatest decrease?



D(0,0)190<-9.38272,9.38272>13.2692<9.38272,-9.38272>-13.2692

56. (1 pt)

The contour plot of z = f (x,y) is shown below. De-termine if the given quantity in each part is positive, ne-gative, or zero.

? 1. The value of ∇ f ·~j at the point P = (−4,1).? 2. The value of ∇ f ·~i at the point P = (−4,1).

? 3.∂ f∂y

at the point Q = (2,−2).

? 4.∂ f∂x

at the point Q = (2,−2).

(Click on graph to enlarge)


NegativePositivePositivePositive


POSITIVENEGATIVENEGATIVEPOSITIVE

17

57. (1 pt)

Use the contour diagram for f (x,y) shown below toestimate the directional derivative of f in the direction~vat the point P.

(a) At the point P = (2,2) in the direction ~v =~i, thedirectional derivative is approximately

(b) At the point P = (2,3) in the direction ~v = −~j,the directional derivative is approximately

(c) At the point P = (1,4) in the direction ~v = (~i +~j)/√

2, the directional derivative is approximately

(d) At the point P = (3,0) in the direction ~v = −~i, thedirectional derivative is approximately



46


2-26/sqrt(2)0

58. (1 pt) Find the directional derivative of f (x,y,z) =4xy + z2 at the point (2,−4,−2) in the direction of a vectormaking an angle of π/2 with ∇ f (2,−4,−2).

fu (2,−4,−2) = Du f (2,−4,−2) =Answer(s) submitted:


0

59. (1 pt) Check that the point (−5,5,4) lies on the surfacecos(x+ y) = exz+20.

(a) View this surface as a level surface for a functionf (x,y,z). Find a vector normal to the surface at the point(−5,5,4).

(b) Find an implicit equation for the tangent plane to the surfaceat (−5,5,4).



<-4,0,5>5*z-4*x = 40

60. (1 pt) Are the following statements true or false?

? 1. If~u is perpendicular to ∇ f (a,b), then f~u (a,b) = 〈0,0〉.? 2. If ~u is a unit vector, then f~u(a,b) is a vector.? 3. f~u (a,b) = ||∇ f (a,b)||.? 4. ∇ f (a,b) is a vector in 3-dimensional space.? 5. Suppose fx(a,b) and fy(a,b) both exist. Then there is

always a direction in which the rate of change of f at(a,b) is zero.

? 6. If f (x,y) has fx(a,b) = 0 and fy(a,b) = 0 at the point(a,b), then f is constant everywhere.

? 7. The gradient vector ∇ f (a,b) is tangent to the contourof f at (a,b).

? 8. f~u (a,b) is parallel to ~u.Answer(s) submitted:

FalseTrueFalseFalseFalseFalseFalseTrue


FALSEFALSEFALSEFALSETRUE

18

FALSEFALSEFALSE

61. (1 pt) For each function f (x,y,z), determine whether itsgradient points radially outward from the origin, radially inwardtoward the origin, radially outward from the z-axis, or radiallyinward toward the z-axis.

? 1. f (x,y,z) =1

x2 + y2

? 2. f (x,y,z) = x2 + y2

? 3. f (x,y,z) =1

x2 + y2 + z2

? 4. f (x,y,z) = x2 + y2 + z2



Points radially inward toward the z-axisPoints radially outward from the z-axisPoints radially inward toward the originPoints radially outward from the origin

62. (1 pt) If z = sin(

xy

)and x = ln(u) and y = v4, find the

following partial derivatives using the chain rule. Enter youranswers as functions of u and v.

∂z∂u

=

∂z∂v

=Answer(s) submitted:

(cos((log(u))/vˆ4))/(u vˆ4)-(4 log(u) cos((log(u))/vˆ4))/vˆ5


1/u*vˆ4/[(vˆ4)ˆ2]*cos([ln(u)]/(vˆ4))-4*vˆ3*ln(u)/[(vˆ4)ˆ2]*cos([ln(u)]/(vˆ4))

63. (1 pt) If z = (x+ y)ey and x = u2 + v2 and y = u2− v2,find the following partial derivatives using the chain rule. Enteryour answers as functions of u and v.

∂z∂u

=

∂z∂v


4 u (uˆ2+1) eˆ(uˆ2-vˆ2)-4 uˆ2 v eˆ(uˆ2-vˆ2)


2*2*u*eˆ(uˆ2-vˆ2)+2*uˆ2*eˆ(uˆ2-vˆ2)*2*u*ln(e)-2*uˆ2*eˆ(uˆ2-vˆ2)*2*v*ln(e)

64. (1 pt) If z = tan−1(

xy

)and x = u2 + v2 and y = u2− v2,

find the following partial derivatives using the chain rule. Enteryour answers as functions of u and v.

∂z∂u

=

∂z∂v


-(2 u vˆ2)/(uˆ4+vˆ4)(2 uˆ2 v)/(uˆ4+vˆ4)


1/(1+[(uˆ2+vˆ2)/(uˆ2-vˆ2)]ˆ2)*[2*u*(uˆ2-vˆ2)-(uˆ2+vˆ2)*2*u]/[(uˆ2-vˆ2)ˆ2]1/(1+[(uˆ2+vˆ2)/(uˆ2-vˆ2)]ˆ2)*[2*v*(uˆ2-vˆ2)+(uˆ2+vˆ2)*2*v]/[(uˆ2-vˆ2)ˆ2]

65. (1 pt) Suppose z = x2 siny, x =−3s2 +4t2, y = 10st.A. Use the chain rule to find ∂z

∂s and ∂z∂t as functions of x, y, s

and t.

∂z∂s =

∂z∂t =B. Find the numerical values of ∂z

∂s and ∂z∂t when (s, t) =

(−5,−5).

∂z∂s (−5,−5) =

∂z∂t (−5,−5) =


10t(4tˆ2-3sˆ2)ˆ2 cos(10st)+s(36sˆ2 - 192)sin(10st)


( 4 * -3 * s * x * sin(y) ) + ( 2 * 5 * t * x**2 * cos(y) )( 4 * 4 * t * x * sin(y) ) + ( 2 * 5 * s * x**2 * cos(y) )-8986.67656947704-5589.82850108072

19

66. (1 pt) Use the chain rule to find dzdt , where

z = x2y+ xy2, x = 1− t6, y =−4+ t2

First the pieces:

∂z∂x =∂z∂y =dxdt =dydt =

End result (in terms of just t):dzdt =

Answer(s) submitted:2xy+yˆ2xˆ2+2xy-6tˆ52t2t(7tˆ12-24tˆ10-5tˆ8+24tˆ6-24tˆ4+2tˆ2-7)


2 x y + yˆ2xˆ2 + 2 x y-1 6 tˆ(6 - 1)1 2 tˆ(2 - 1)(2*(1+-1*t**6)*(-4+1*t**2) + (-4+1*t**2)**2)*-1*6*tˆ(6 - 1) +(2*(1+-1*t**6)*(-4+1*t**2) + (1+-1*t**6)**2)*1*2*tˆ(2 - 1)

67. (1 pt)Use the chain rule to find dw

dt , where

w = x5y1 + y1z3,x = et ,y = et sin t,z = et cos t

First the pieces:∂w∂x =∂w∂y =∂w∂z =dxdt =dydt =dzdt =Now all together:

dwdt = ∂w

∂xdxdt + ∂w

∂ydydt + ∂w

∂zdzdt is too horrible to write down (co-

rrectly).Answer(s) submitted:

5yxˆ4xˆ5+zˆ33yzˆ2eˆteˆt (sin(t)+cos(t))eˆt (cos(t)-sin(t))


5*x**(5-1)*y**11*x**5 * y**(1-1) + 1*y**(1-1)*z**33*y**1 * z**(3-1)exp(t)exp(t)*(cos(t) + sin(t))exp(t)*(cos(t) - sin(t))

68. (1 pt) The radius of a right circular cone is increasing ata rate of 5 inches per second and its height is decreasing at arate of 5 inches per second. At what rate is the volume of thecone changing when the radius is 20 inches and the height is 50inches?

cubic inches per secondAnswer(s) submitted:

8000/3 pi


8377.58040957278

69. (1 pt) Suppose z = x2 siny, x = 3s2 +3t2, y = 4st.A. Use the chain rule to find ∂z

∂s and ∂z∂t as functions of x, y, s

and t.∂z∂s =∂z∂t =

B. Find the numerical values of ∂z∂s and ∂z

∂t when (s, t) = (1,3).∂z∂s (1,3) =∂z∂t (1,3) =

Answer(s) submitted:(2(3sˆ2+3tˆ2)sin(4st))(6s) + (3sˆ2+3tˆ2)ˆ2cos(4st)(4t)0


( 4 * 3 * s * x * sin(y) ) + ( 2 * 2 * t * x**2 * cos(y) )( 4 * 3 * t * x * sin(y) ) + ( 2 * 2 * s * x**2 * cos(y) )8920.456503830762458.3754999965

70. (1 pt)Use the chain rule to find dz

dt , where

z = x2y+ xy2,x =−4− t3,y =−5− t3

First the pieces:∂z∂x =∂z∂y =dxdt =dydt =dzdt =

Answer(s) submitted:2xy+yˆ2xˆ2+2xy-3tˆ2

20

-3tˆ2(2(-4-tˆ3)(-5-tˆ3)+(-5-tˆ3)ˆ2)(-3tˆ2) + ((-4-tˆ3)ˆ2+2(-4-tˆ3)(-5-tˆ3))(-3tˆ2)


2*x*y + y**2x**2 + 2*x*y-1*3*t**(3 - 1)-1*3*t**(3 - 1)(2*(-4+-1*t**3)*(-5+-1*t**3) + (-5+-1*t**3)**2)*-1*3*t**(3 - 1) +(2*(-4+-1*t**3)*(-5+-1*t**3) + (-4+-1*t**3)**2)*-1*3*t**(3 - 1)

71. (1 pt)Use the chain rule to find ∂z

∂s and ∂z∂t , where

z = exy tany,x = 2s+5t,y =5s3t

First the pieces:∂z∂x = ∂z

∂y =∂x∂s = ∂x

∂t =∂y∂s = ∂y

∂t =And putting it all together:∂z∂s = ∂z

∂x∂x∂s + ∂z

∂y∂y∂s and ∂z

∂t = ∂z∂x

∂x∂t + ∂z

∂y∂y∂t


ytan(y)eˆ(xy)(sec(y)ˆ2)eˆ(xy) +xtan(y)eˆ(xy)255/(3t)(-5s)/(3tˆ2)


y*tan(y)*exp(x*y)exp(x*y)*sec(y)**2 + tan(y)*x*exp(x*y)255/(3*t)-5*s/(3*t**2)

72. (1 pt) If

z = xey, x = u2− v2, y = u2 + v2,

find ∂z/∂u and ∂z/∂v. The variables are restricted to domains onwhich the functions are defined.

∂z/∂u =∂z/∂v =SOLUTIONSince z is a function of two variables x and y which are fun-

ctions of two variables the two chain rule identities which applyare:

∂z∂u

=∂z∂x

∂x∂u

+∂z∂y

∂y∂u

= (ey)(2u)+(xey ln(e)) · (2u)

= eu2+v2 ·2u+(u2− v2)eu2+v2

ln(e) ·2u.

and∂z∂v

=∂z∂x

∂x∂v

+∂z∂y

∂y∂v

= (ey)(−(2v))+(xey ln(e)) · (2v)

=(u2− v2)eu2+v2

ln(e) ·2v− eu2+v2 ·2v.Answer(s) submitted:


eˆ(uˆ2+vˆ2)*2*u+(uˆ2-vˆ2)*eˆ(uˆ2+vˆ2)*ln(e)*2*u(uˆ2-vˆ2)*eˆ(uˆ2+vˆ2)*ln(e)*2*v-eˆ(uˆ2+vˆ2)*2*v

73. (1 pt) Given z = f (x,y), x = x(u,v), y = y(u,v), withx(5,6) = 3 and y(5,6) = 4, calculate zv(5,6) in terms of someof the values given in the table below.

fx(5,6) = b fy(5,6) = a xu(5,6) = s yu(5,6) = cfx(3,4) = d fy(3,4) = 2 xv(5,6) = 3 yv(5,6) = r

zv(5,6) =SOLUTIONUsing the chain rule,

zu(u,v) = fx(x,y) · xv(u,v)+ fy(x,y) · yv(u,v).

Since x(5,6) = 3 and y(5,6) = 4, substituting gives

zu(5,6) = fx(3,4) · xv(5,6)+ fy(3,4) · yv(5,6) = d ·3+2r.



(d)*(3) + (2)*(r)

74. (1 pt) If

z = (x+ y)ex, x = 4t, y = 5− t2,

find dz/dt using the chain rule. Assume the variables are res-tricted to domains on which the functions are defined.

dzdt =SOLUTIONSubstituting into the chain rule gives

dzdt

=∂z∂x

dxdt

+∂z∂y

dydt

= ((1+ x+ y)ex)(4)+(ex)(−2t)

= 4(1+4t +5− t2)e4t −2te4t .


4eˆx(x + y + 1) - 2teˆx


4*(1+4*t+5-tˆ2)*eˆ(4*t)-2*t*eˆ(4*t)

21

75. (1 pt)Use the chain rule to find ∂z

∂s and ∂z∂t , where

z = x2 + xy+ y2,x = 3s+4t,y = 5s+6t

First the pieces:∂z∂x = ∂z

∂y =∂x∂s = ∂x

∂t =∂y∂s = ∂y

∂t =And putting it all together:∂z∂s =∂z∂t =


2x+yx+2y345611(3s+4t)+13(5s+6t)14(3s+4t)+16(5s+6t)


2*x + y2*y + x3456(2*(3*s + 4*t) + 5*s + 6*t)*3 +(2*(5*s + 6*t) + 3*s + 4*t)*5(2*(3*s + 4*t) + 5*s + 6*t)*4 +(2*(5*s + 6*t) + 3*s + 4*t)*6

76. (1 pt) An unevenly heated metal plate has temperatureT (x,y) in degrees Celsius at a point (x,y). If T (2,1) = 127,Tx (2,1) = 12, and Ty (2,1) = −8, estimate the temperature atthe point (2,04,0,96).

T (2,04,0,96) ≈ . Please include unitsin your answer.



127.8 degC

77. (1 pt) A student was asked to find the equation of the tan-gent plane to the surface z = x2− y3 at the point (x,y) = (5,2).The student’s answer was z = 17+2x(x−5)−

(3y2)(y−2).

(a) At a glance, how do you know this is wrong. What mis-takes did the student make? Select all that apply.

A. The (x - 5) and (y - 2) should be x and y.B. The 17 should not be in the answer.C. The partial derivatives were not evaluated a the point.D. The answer is not a linear function.E. All of the above

(b) Find the correct equation for the tangent plane.z =



CD17+10*(x-5)-12*(y-2)

78. (1 pt) Find the implicit equation of the tangent plane toz = ln

(x2 +1

)+ y2 at the point (0,7,49).

Answer(s) submitted:<14y-49z+2302>


14*y-z = 49

79. (1 pt)

Find equations of the tangent plane and normal line to thesurface x = 4y2 +3z2−159 at the point (-3, 6, -2).Tangent Plane: (make the coefficient of x equal to 1).

= 0.Normal line: 〈−3, , 〉+t〈1, , 〉.

Answer(s) submitted:x+4yˆ2+3zˆ2-1596-2

-12


x - -3 - 2*4*6*(y - 6) - 2*3*-2*(z--2)6-2-4812

80. (1 pt) Find the equation of the tangent plane to the surfacez = cos(3x)cos(4y) at the point (3π/2,3π/2,0).

z =Answer(s) submitted:


-3 (x-3*pi/2)

22

81. (1 pt)

Find an equation of the tangent plane to the surface z =−1x2 +3y2 +3x−3y+2 at the point (2, 3, 22).z =



22 + -1*(x - 2) + 15 * (y - 3)

82. (1 pt)

Find the linearization of the function f (x,y) =√54−1x2−1y2 at the point (-2, -5).

L(x,y) =Use the linear approximation to estimate the value off (−2,1,−4,9) =


2/5x + y +54/55.05766


5 + 0.4*(x - -2) + 1 * (y - -5)5.06

83. (1 pt)Consider the surface xyz = 60.

A. Find the unit normal vector to the surface at the point(3,5,4) with positive first coordinate.( , , )

B. Find the equation of the tangent plane to the surface at thegiven point. Express your answer in the form ax+by+cz+d =0, normalized so that a = 20.

= 0.Answer(s) submitted:

(3sqrt(2))/10sqrt(2)/2(2sqrt(2))/5


0.7212184459746190.4327310675847710.5409138344809645*4*(x - 3) + 3*4*(y - 5) + 3*5*(z - 4)

84. (1 pt)

Consider a function f : R2 → R2 for which f (−1,−1) =

(1,2) and D f (−1,−1) =(

3 −11 0

).

The linear approximation of f at the point (−1,−1) is (written

as a row vector)L(x,y) = ( , ).



1 + 3 * (x - -1) + -1 * (y - -1)2 + 1 * (x - -1) + 0 * (y - -1)

85. (1 pt)

Consider the function f : R2 → R2 given by f (x,y) =(e2x, sin(9xy)).

The derivative (Jacobian) matrix D f (x) =(

a bc d

)where

a = , b = ,c = , d = .



2 * exp(2 * x)09 * y * cos(9 * x * y)9 * x * cos(9 * x * y)

86. (1 pt)

Find an equation of the tangent plane to the surface z =5y2−5x2 at the point (2, 2, 0).z =



0 + -20 * (x - 2) + 20 * (y - 2)

87. (1 pt) Find the equation of the tangent plane to the surfacedetermined by

x2y4 + z−20 = 0at x = 4, y = 3.

z =SOLUTIONThe surface is given by

z = f (x,y) = 20− x2y4,

and f (4,3) =−1276. We have

fx(x,y) =−(2xy4) , so fx(4,3) =−648,

and

fy(x,y) =−(x2 ·4y3) , so fy(4,3) =−1728.

23

Thus, the tangent plane is

z = f (4,3)+ fx(4,3)(x−4)+ fy(4,3)(y−3),

or

z =−1276−648(x−4)−1728(y−3)=−(1276+648(x−4)+1728(y−3)) .



-[1276+648*(x-4)+1728*(y-3)]

88. (1 pt) (a) Check the local linearity of f (x,y) = e−y cos(x)near x = 1,5, y = 1 by filling in the following table of values off for x = 1,4, 1,5, 1,6 and y = 0,9, 1, 1,1. Express values of fwith 4 digits after the decimal point.

x = 1.4 1.5 1.6y = 0,9y = 1

y = 1,1

(b) Next, fill in the table for the values x = 1,49, 1,5, 1,51and y = 0,99, 1, 1,01, again showing 4 digits after the decimalpoint.

x = 1.49 1.5 1.51y = 0,99

y = 1y = 1,01

Notice if the two tables look nearly linear, and whether thesecond looks more linear than the first (in particular, think abouthow you would decide if they were linear, or if the one were mo-re closely linear than the other).

(c) Give the local linearization of f (x,y) = e−y cos(x) at(1,5,1):Using the second of your tables:f (x,y)≈Using the fact that fx(x,y) = −e−y sin(x) and fy(x,y) =−e−y ln(e)cos(x):f (x,y)≈

SOLUTION(a),(b) The two tables of values are shown below

x = 1.4 1.5 1.6y = 0,9 0.0691 0.0288 -0.0119y = 1 0.0625 0.026 -0.0107

y = 1,1 0.0566 0.0235 -0.0097

x = 1.49 1.5 1.51y = 0,99 0.03 0.0263 0.0226

y = 1 0.0297 0.026 0.0224y = 1,01 0.0294 0.0258 0.0221

Both tables are nearly linear. To check this, observe that theincrements in each row (column) are equal, or nearly so. Thesecond table is more linear due to finer data.

(c) From either table we see that f (1,5,1) ≈ 0,026. Also,using the second table,

fx(1,5,1)≈ f (1,51,1)− f (1,49,1),02

=0,0224−0,0297

0,02=−0,365,

fy(1,5,1)≈ f (1,5,1,01)− f (1,5,0,99),02

=0,0258−0,0263

0,02=−0,025.

Using these estimates for the partial derivatives, we get that thelocal linearization of f around (1,5,1) is

f (x,y)≈ f (1,5,1)+ fx(1,5,1)(x−1,5)+ fy(1,5,1)(y−1)= 0,026+(−0,365)(x−1,5)+(−0,025)(y−1).

Now we use fx =−e−y sin(x) and fy =−e−y ln(e)cos(x), gi-ving

fx(1,5,1) =−0,366958, fy(1,5,1) =−0,0260228,

we have the local linearization of f around (1,5,1)

f (x,y)≈ 0,0260228+(−0,366958)(x−1,5)+(−0,0260228)(y−1) .

Note that these agree to a reasonable number of decimal places.Answer(s) submitted:


0.06910.0288-0.01190.06250.026-0.01070.05660.0235-0.00970.030.02630.0226

24

0.02970.0260.02240.02940.02580.02210.026 + (0.0224-0.0297)*(x-1.5)/0.02 + (0.0258-0.0263)*(y-1)/0.020.0260228 + -0.366958*(x - 1.5) + -0.0260228*(y - 1)

89. (1 pt) Find the equation of the tangent plane to

z = ex + y+ y2 +4

at the point (0,4,25).z =SOLUTIONWe have

z = ex + y+ y2 +4.

The partial derivatives are

∂z∂x

∣∣∣∣(x,y)=(0,4)

= ex ln(e)∣∣∣∣(x,y)=(0,4)

= 1,

and

∂z∂y

∣∣∣∣(x,y)=(0,4)

= 1+2y∣∣∣∣(x,y)=(0,4)

= 9.

So the equation of the tangent plane is

z = 25+1(x−0)+9(y−4) = 25+ x+9(y−4) .



25+x+9*(y-4)

90. (1 pt) Find the equation of the tangent plane to the surfacez = e−2x ln(2y) at the point (−1,3, 13.24 ).

z =Note: Your answer should be an expression of x and y; e.g.

“5x + 2y - 3” Please use E instead of e=2.71828 for exponentialnotation, eg. .0003 = 3E-04.



-26.478822467832607 * x + 2.463018699643550 * y + -20.628467332846952

91. (1 pt) If z = f (x)+ yg(x), what can we say about zyy?

A. zyy = zxxB. zyy = 0C. zyy = g(x)D. zyy = yE. We cannot say anything


B


B

92. (1 pt) Calculate all four second-order partial derivativesof f (x,y) = 5x2y+4xy3.

fxx (x,y) =

fxy (x,y) =

fyx (x,y) =

fyy (x,y) =Answer(s) submitted:

10y10x+12yˆ210x+12yˆ224xy


10*y5*2*x+4*3*yˆ25*2*x+4*3*yˆ24*x*3*2*y

93. (1 pt)

Use the level curves of the function z = f (x,y) todetermine if each partial derivative at the point P is po-sitive, negative, or zero.

? 1. fy (P)? 2. fx (P)? 3. fxx (P)? 4. fyy (P)

25



PositiveNegativePositivePositive


POSITIVENEGATIVEPOSITIVEPOSITIVE

94. (1 pt)

Use the level curves of the function z = f (x,y) todetermine if each partial derivative at the point P is po-sitive, negative, or zero.

? 1. fx (P)? 2. fyx (P)? 3. fy (P)? 4. fxx (P)? 5. fyy (P)




ZeroZeroPositiveZeroPositive

95. (1 pt)

The figure shows the level curves of a function f (x,y)around a maximum or a minimum M. One of the pointsP or Q has coordinates (x1,y1) and the other has coor-dinates (x2,y2). Suppose b > 0 and c > 0. Consider thetwo linear approximations to f given by

f (x,y) ≈ a+b(x− x1)+ c(y− y1),f (x,y) ≈ k +m(x− x2)+n(y− y2).

(a) What is the relationship between the values of a andk?

A. a > kB. a = kC. a =−kD. a < k

(b) What are the coordinates of point P?A. (x2,y2)B. (x1,y1)

(c) Is M a maximum or a minimum? ?

(d) Is the sign of m positive or negative? ?

(e) Is the sign of n positive or negative? ?




BAMinimum

26

NegativePositive

96. (1 pt)

Find the partial derivatives of the function

f (x,y) =7x−7y2x+7y

fx(x,y) =fy(x,y) =


y/((4xy)/9 + (4xˆ2)/63 +(7yˆ2)/9)-(63 x)/(2 x+7 y)ˆ2


((2*x + 7*y)*7 - (7*x - 7*y)*2)/(2*x + 7*y)**2((2*x + 7*y)*(- 7) - (7*x - 7*y)*7)/(2*x + 7*y)**2

97. (1 pt)Find the first partial derivatives of f (x,y,z) = z arctan( y

x ) atthe point (1, 1, -1).

A. ∂ f∂x (1,1,−1) =

B. ∂ f∂y (1,1,−1) =

C. ∂ f∂z (1,1,−1) =


1/2-1/2pi/4


0.5-0.50.785398163397448

98. (1 pt)


f (x,y) =Z x

ycos(8t2−3t +2)dt

fx(x,y) =fy(x,y) =



cos(8*x**2 + -3*x + 2)-cos(8*y**2 + -3*y + 2)

99. (1 pt)Find all the first and second order partial derivatives of

f (x,y) =−4sin(2x+ y)+4cos(x− y).A. ∂ f

∂x = fx =B. ∂ f

∂y = fy =

C. ∂2 f∂x2 = fxx =

D. ∂2 f∂y2 = fyy =

E. ∂2 f∂x∂y = fyx =

F. ∂2 f∂y∂x = fxy =


-8cos(2x+y)-4sin(x-y)-4cos(2x+y) + 4sin(x-y)16sin(2x+y)-4cos(x-y)4sin(2x+y) - 4cos(x-y)8sin(2x+y) + 4cos(x-y)8sin(2x+y) +4cos(x-y)


2*-4*cos(2*x+y) - 4*sin(x-y)-4*cos(2*x+y) + 4*sin(x-y)-4*(-4 * sin(2*x + y)) - 4*cos(x-y)- -4*sin(2*x+y) - 4*cos(x-y)-2*(-4*sin(2*x+y)) + 4*cos(x-y)-2*(-4*sin(2*x+y)) + 4*cos(x-y)

100. (1 pt)


w =√

2r2 +7s2 +5t2

∂w∂r =∂w∂s =∂w∂t =


(2r)/(sqrt(2rˆ2 +7sˆ2 + 5tˆ2))(7s)/(sqrt(2rˆ2 +7sˆ2 + 5tˆ2))(5t)/(sqrt(2rˆ2 +7sˆ2 + 5tˆ2))


2*r/sqrt(2*r**2 + 7*s**2 + 5*t**2)7*s/sqrt(2*r**2 + 7*s**2 + 5*t**2)5*t/sqrt(2*r**2 + 7*s**2 + 5*t**2)

101. (1 pt) Consider the function

f (x,y) = e2x−x2−10y−y2.

Find and classify all critical points of the function. If there aremore blanks than critical points, leave the remaining entriesblank.

27

fx =fy =fxx =fxy =fyy =

The critical point with the smallest x-coordinate is( , ) Classification: (local minimum,local maximum, saddle point, cannot be determined)

The critical point with the next smallest x-coordinate is( , ) Classification: (local minimum,local maximum, saddle point, cannot be determined)



(2 - 2*x)*exp(2*x - x**2 + -10*y - y**2)(-10 - 2*y)*exp(2*x - x**2 + -10*y - y**2)((2 - 2*x)**2 -2)*exp(2*x - x**2 + -10*y - y**2)(2 - 2*x)*(-10 - 2*y)*exp(2*x - x**2 + -10*y - y**2)((-10 - 2*y)**2 -2)*exp(2*x - x**2 + -10*y - y**2)1-5local maximum

102. (1 pt)

Consider the function

f (x,y) = (2x− x2)(8y− y2).


fx =fy =fxx =fxy =

fyy =

There are several critical points to be listed. List them lexico-grahically, that is in ascending order by x-coordinates, and forequal x-coordinates in ascending order by y-coordinates (e.g.,(1,1), (2, -1), (2, 3) is a correct order)








(2 - 2*x)*(8*y - y**2)(2*x - x**2)*(8 - 2*y)-2*(8*y - y**2)(2 - 2*x)*(8 - 2*y)-2*(2*x - x**2)00saddle point08saddle point1

28

4local maximum20saddle point28saddle point

103. (1 pt) Consider the function

f (x,y) = y√

x− y2−2x+7y.


fx =fy =fxx =fxy =fyy =





y/(2*sqrt(x)) - 2sqrt(x) - 2*y + 7-y/(4*x**(3/2))1/(2*sqrt(x))-214local maximum

104. (1 pt)

The function f has continuous second derivatives, and a cri-tical point at (-5, -2).

Suppose fxx(−5,−2)=−2, fxy(−5,−2)=−4, fyy(−5,−2)=5.Then the point (-5, -2):

A. is a saddle pointB. cannot be determinedC. is a local maximumD. is a local minimumE. None of the above



A

105. (1 pt) Find the absolute maximum and absolute mini-mum of the function f (x,y) = xy−5y−25x+125 on the regionon or above y = x2 and on or below y = 28.

As usual, ignore unneeded answer blanks, and list points le-xicographically.

Absolute minimum value:attained at ( , ) and ( , ).

Absolute maximum value:attained at ( , ) and ( , ).



-30.8745078663875-5.2915026221291828

148.148148148148-1.666666666666672.77777777777778

29

106. (1 pt) The function

f (x,y) = xy(1−10x−2y)

has 4 critical points. List them and select the type of criticalpoint.

Points should be entered as ordered pairs and listed in increasinglexicographic order. By that we mean that (x,y) comes before(z,w) if x < z or if x = z and y < w.

First point of type ?Second point of type ?Third point of type ?Fourth point of type ?


0Saddle0Saddle1/6Maximum1/30Saddle


(0,0)Saddle(0, 1/2)Saddle(1/30, 1/6)Maximum(1/10, 0)Saddle

107. (1 pt) For each of the following functions, find the maxi-mum and mimimum values of the function on the circular disk:x2 +y2 ≤ 1. Do this by looking at the level curves and gradients.

(A) f (x,y) = x+ y+3:maximum value =minimum value =

(B) f (x,y) = 3x2 +4y2:maximum value =minimum value =

(C) f (x,y) = 3x2−4y2:maximum value =minimum value =



3+sqrt(2)3-sqrt(2)403-4

108. (1 pt) Find the absolute maximum and minimum of thefunction f (x,y) = y

√x− y2 − x + 3y on the domain 0 ≤ x ≤

9, 0≤ y≤ 9.Absolute minimum value: attained atAbsolute maximum value: attained atIn each case, your second answer should be one or more

points (x,y). If there is more than one point at which the ma-ximum or minimum value is attained, enter them all separatedby commas (e.g. (1,2),(1,3) ).



-54(0,9)3(1,2)

109. (1 pt) Consider the function z = x11y + 22x10− 2048y.Find and classify all critical points of the function. If there aremore blanks than critical points, leave the remaining entriesblank.

The critical point with the smallest x-coordinate is( , ) Classification: ? (local minimum, local maxi-mum, saddle point, cannot be determined)

The critical point with the next smallest x-coordinate is( , ) Classification: ? (local minimum, local maxi-mum, saddle point, cannot be determined)

The critical point with the next smallest x-coordinate is( , ) Classification: ? (local minimum, local maxi-mum, saddle point, cannot be determined)


saddle point


2

30

-10SADDLE POINT

110. (1 pt) Find the absolute maximum and absolute mini-mum of the function f (x,y) = 2x3 +y4 on the region {(x,y)|x2 +y2 ≤ 49}

Absolute minimum value: attained atAbsolute maximum value: attained atIn each case, your second answer should be one or more

points (x,y). If there is more than one point at which the ma-ximum or minimum value is attained, enter them all separatedby commas (e.g. (1,2),(1,3) ).



-686(-7,0)2401(0,-7), (0,7)

111. (1 pt)The level curves of a function f (x,y) consist of a collection

of hyperbolas and two lines. If the lines intersect at a point P,what are the possibilities for P? Type the letters of all possibili-ties, with no punctuation, in alphabetical order.

A. P is a local maximum, that is, f (P)≥ f (Q) for all Q nearP.

B. P is a local minimum, that is, f (P)≤ f (Q) for all Q nearP.

C. P is neither a local maximum nor a local minimum.


c,b,a


ABC

112. (1 pt) Each of the following functions has at most onecritical point. Graph a few level curves and a few gradiants and,on this basis alone, decide whether the critical point is a localmaximum (MA), a local minimum (MI), or a saddle point (S).Enter the appropriate abbreviation for each question, or N if the-re is no critical point.

(A) f (x,y) = e−1x2−4y2

Type of critical point:

(B) f (x,y) = e1x2−4y2

Type of critical point:(C) f (x,y) = 1x2 +4y2 +4

Type of critical point:(D) f (x,y) = 1x+4y+4

Type of critical point:Answer(s) submitted:

MASMIN


MASMIN

113. (1 pt) Find the maximum and minimum values off (x,y) = 2x2 +3y2 on the disk D: x2 + y2 ≤ 1.maximum value:minimum value:



30

114. (1 pt) For each of the following functions, find the ma-ximum and minimum values of the function on the rectanglarregion: −4≤ x≤ 4,−5≤ y≤ 5.Do this by looking at level curves and gradiants.



(C) = f (x,y) = (5)2x2− (4)2y2:maximum value =minimum value =



12-6148

31

0400-400

115. (1 pt) Consider the function f (x,y) = xsin(y).In the following questions, enter an integer value or type INFfor infinity.

(A) How many local minima does f have in R2?

(B) How many local maxima does f have in R2?

(C) How many saddle points does f have in R2?



00INF

116. (1 pt) Suppose f (x,y) = xy−ax−by.(A) How many local minimum points does f have in R2?

(The answer is an integer).

(B) How many local maximum points does f have in R2?

(C) How many saddle points does f have in R2?



001

117. (1 pt) Suppose f (x,y) = x2 + y2−4x−10y+1(A) How many critical points does f have in R2?

(B) If there is a local minimum, what is the value of the dis-criminant D at that point? If there is none, type N.

(C) If there is a local maximum, what is the value of the dis-criminant D at that point? If there is none, type N.

(D) If there is a saddle point, what is the value of the discri-minant D at that point? If there is none, type N.

(E) What is the maximum value of f on R2? If there is none,type N.

(F) What is the minimum value of f on R2? If there is none,type N.



14NNN1 - 2**2 - 5**2

118. (1 pt) Find the maximum and minimum values off (x,y) = xy on the ellipse 3x2 + y2 = 2.maximum value =minimum value =



0.577350269189626-0.577350269189626

119. (1 pt) Consider the function f (x,y) = x2y+ y3−48y.

f has ? at (0,4).f has ? at (0,0).f has ? at

(−4√

3,0).

f has ? at (0,−4).f has ? at

(4√

3,0).



a minimumno critical pointa saddlea maximuma saddle

32

120. (1 pt) Find the critical points for the function

f (x,y) = 128xy− (x+ y)4

and classify each as a local maximum, local minimum, saddlepoint, or none of these.

critical points:(give your points as a comma separated list of (x,y) coordina-tes.)

classifications:(give your answers in a comma separated list, specifying maxi-mum, minimum, saddle point, or none for each, in the sameorder as you entered your critical points)

SOLUTIONAt a critical point, (fx = 0, fy = 0. Thus

fx = 128y−4(x+ y)3 = 0

andfy = 128x−4(x+ y)3 = 0.

Thus128y = 4(x+ y)3 = 128x,

and x = y. Thus 128x + 32x3 = 0, and solving gives x = 0,±2.Thus the critical points are (0,0), (2,2) and (−2,−2).

Next, fyy = fxx = −12(x + y)2, and fxy = 128− 12(x + y)2.The discriminant is D(x,y) = fxx fyy− f 2

xy, so

D = 144(x+ y)4−(128−12(x+ y)2)2

.

Thus D(0,0) =−16384 < 0, so (0,0) is a saddle point.D(2,2) = 32768 > 0 and fxx(2,2) = −192 < 0, so (2,2) is alocal maximum.D(−2,−2) = 32768 > 0 and fxx(−2,−2) = −192 < 0, so(−2,−2) is a local maximum.



(0,0), (2,2), (-2,-2)saddle point, maximum, maximum

121. (1 pt) Find the critical points for the function

f (x,y) = x3 + y3−3x2−12y−4

and classify each as a local maximum, local minimum, saddlepoint, or none of these.

critical points:(give your points as a comma separated list of (x,y) coordina-tes.)


SOLUTION

To find the critical points, we solve fx = 0 and fy = 0 for xand y. Solving

fx = 3x2−6x = 0,

we have x = 0 or x = 2. Then

fy = 3y2−12 = 0,

so that y = ±2. There are four critical points: (0,2), (0,−2),(2,2), and (2,−2).

Then we have

D = ( fxx)( fyy)− ( fxy)2 = (6x−6)(6y)− (0)2 = (6x−6)(6y).

At the point (0,2), we have D < 0, so f has saddle point.At the point (0,−2), we have D > 0 and fxx < 0, so f has a localmaximum.At the point (2,2), we have D > 0 and fxx > 0, so f has a localminimum.At the point (2,−2), we have D < 0, so f has a saddle point.



(0,2), (0,-2), (2,2), (2,-2)saddle point, maximum, minimum, saddle point

122. (1 pt) Consider the three points (2,1), (1,−1), and(−2,−3).

(a) Supposed that at (2,1), we know that fx = fy = 0 andfxx < 0, fyy > 0, and fxy = 0. What can we conclude about thebehavior of this function near the point (2,1)? ?

(b) Supposed that at (1,−1), we know that fx = fy = 0 andfxx < 0, fyy = 0, and fxy > 0. What can we conclude about thebehavior of this function near the point (1,−1)? ?

(c) Supposed that at (−2,−3), we know that fx = fy = 0 andfxx = 0, fyy > 0, and fxy > 0. What can we conclude about thebehavior of this function near the point (−2,−3)? ?

Using this information, on a separate sheet of paper sketch apossible contour diagram for f .

SOLUTIONAt each point we have fx = fy = 0, so each of the three points

is a critical point. Then we have D = fxx fyy−( fxy)2, so we knowthat

(a) At (2,1), D < 0, so (2,1) is a saddle point.(b) At (1,−1), D < 0, so (1,-1) is a saddle point.(c) At (−2,−3), D < 0, so (-2,-3) is a saddle point.Answer(s) submitted:


(2,1) is a saddle point(1,-1) is a saddle point(-2,-3) is a saddle point

33

123. (1 pt) Let

f (x,y) = 1+3x2− cos(4y) .

Find all critical points and classify them as local maxima, localminima, saddle points, or none of these.

critical points:(give your points as a comma separated list of (x,y) coordina-tes. if your answer includes points that occur at a sequence ofvalues, e.g., at every odd integer, or at any constant multiple ofanother value, use m for any non-zero even integer, n for anynon-zero odd integer, and/or k for other arbitrary constants.)


SOLUTIONTo find the critical points, we must solve the equations fx =

6x = 0 and fy = 4sin(4y) = 0. The first equation has solutionx = 0, and the second equation, y = nπ/4 (where n is any in-teger). Thus our critical points are (0,0),(0, mπ

4 ), and (0, nπ

4 )(where m is any non-zero even integer and n is any non-zeroodd integer).

We calculate

D = ( fxx)( fyy)− ( fxy)2 = (6)(16cos(4y))− (0)2.

Thus, at (0,0), D > 0 and fxx > 0, so this is a minimum. Simi-larly, at (0, mπ

4 ), D > 0 and fxx > 0, so this is a minimum, and at(0, nπ

4 ), D < 0, so this is a saddlepoint,Answer(s) submitted:


(0,0), (0,4.71239), (0,5.49779)minimum, minimum, saddle point

124. (1 pt) Does the function

f (x,y) =x2

2+4y4 +9y2−7x

have a global maximum and global minimum? If it does, iden-tify the value of the maximum and minimum. If it does not, besure that you are able to explain why.

Global maximum?(Enter the value of the global maximum, or none if there is noglobal maximum.)Global minimum?(Enter the value of the global minimum, or none if there is noglobal minimum.)

SOLUTIONSuppose x is fixed. Then for large values of y the sign of

f is determined by the highest power of y, namely y4. Thus,f (x,y)→ ∞ as y→ ∞, and there can be no global maximum.. For y → −∞, and as x → ±∞, we have the same behavior.So we could have a global minimum. We can check this by

finding the critical points and testing them: fx = x− 7, andfy = 16y3 +18y = 2y(8y2 +9), so the only critical point is (7,0).At this point, D = fxx fyy−( fxy)2 = (1)(18) > 0, and fxx > 0, sothis a local and the global minimum. Thus the global minimumis f (7,0) =−24,5.



none-7*7/2

125. (1 pt) Let f (x,y) = 3/x +4/y +5x +6y in the region Rwhere x,y > 0.

Explain why f must have a global minimum at some pointin R (note that R is unbounded—how does this influence yourexplanation?). Then find the global minimum.minimum =

SOLUTIONThe function f is continuous in the region R, but R is not

closed and bounded so a special analysis is required.Notice that f (x,y) tends to ∞ as (x,y) tends farther and fart-

her from the origin or tends toward any point on the x or y axis.This suggests that a minimum for f , if it exists, can not be too farfrom the origin or too close to the axes. For example, if x > 10then f (x,y) > 5x > 50, and if y > 10 then f (x,y) > 6y > 60. If0 < x < 0,1 then f (x,y) > 3/x > 30, and if 0 < y < 0,1 thenf (x,y) > 4/y > 40.

Since f (1,1) = 18, a global minimum for f if it exists mustbe in the smaller region R′ : 0,1 ≤ x ≤ 10, 0,1 ≤ y ≤ 10. Theregion R′ is closed and bounded and so have a minimum valueat some point in R′, and since that value is at most 18, it is alsoa global minimum for all of R.

Since the region R has no boundary, the minimum value mustoccur at a critical point of f . At a critical point we have

fx =− 3x2 +5 = 0 = fy =− 4

y2 +6 = 0.

The only critical point is (√

35 ,√

23 ), at which f achieves the

minimum value

f (

√35,

√23) = 2

√15+2

√24



2*sqrt(5*3) + 2*sqrt(6*4)

34

126. (1 pt) What is the shortest distance from the surfacexy+6x+ z2 = 36 to the origin?

distance =SOLUTIONWe minimize the square of the distance from the point (x,y,z)

to the origin:S = x2 + y2 + z2.

Since z2 = 36− xy−6x, we have

S = x2 + y2 +36− xy−6x.

At a critical point∂S∂x

= 2x− y−6 = 0,

and∂S∂y

= 2y− x = 0,

so x = 2y, and2(2y)− y−6 = 0

giving y = 2, so x = 4 and z2 = 36− (4)(2)− 6(4), so z = ±2.We have

D = SxxSyy− (Sxy)2 = 2 ·2− (−1)2 = 4−1 > 0,

so, since D > 0 and Sxx > 0, the critical points are local minima.Since S→∞ as x,y→±∞, the local minima are global minima.

If x = 4, y = 2, z = ±2, we have S = 42 + 22 + 22 = 24, sothe shortest distance to the origin is

√24.



sqrt(24)

127. (1 pt) For each of the following functions, find the maxi-mum and mimimum values of the function on the circular disk:x2 +y2 ≤ 1. Do this by looking at the level curves and gradients.



(C) f (x,y) = 4x2−5y2:maximum value =minimum value =



5.41421356237309

2.58578643762691504-5

128. (1 pt) Find the absolute maximum and absolute mini-mum of the function f (x,y) = 2x3 +y4 on the region {(x,y)|x2 +y2 ≤ 64}

As usual, ignore unneeded answer blanks, and list points le-xicographically.


Absolute maximum value:attained at ( , ), ( , ) and ( , ).



-1024-80

40960-808

129. (1 pt)

Find the maximum and minimum values of the functionf (x,y,z) = yz + xy subject to the constraints y2 + z2 = 324 andxy = 6.

Maximum value is .Minimum value is .Answer(s) submitted:


35

168-156

130. (1 pt)

Find the absolute maximum and minimum of the functionf (x,y) = x2− y2 subject to the constraint x2 + y2 = 256.

As usual, ignore unneeded answer blanks, and list points inlexicographic order.


Absolute maximum value:attained at ( , ) and ( , ).



-2560-16016256-160160

131. (1 pt) Find the maximum and minimum values of thefunction f (x,y) = 2x2 + 3y2− 4x− 5 on the domain x2 + y2 ≤256. As usual, ignore unneeded answer blanks, and list pointslexicographically.

Maximum value is , occuring at ( , ),and ( , ).

Minimum value is , occuring at ( , ) and( , ).


(score 0.2)

Correct Answers:767-2-15.8745078663875-215.8745078663875-710

132. (1 pt) Find the maximum and minimum values of thefunction f (x,y,z) = x + 2y subject to the constraints y2 + z2 =225 and x+ y+ z = 5.

Maximum value is , occuring at ( ,, ).

Minimum value is , occuring at ( ,, ).



26.2132034355964510.6066017177982-10.6066017177982-16.21320343559645-10.606601717798210.6066017177982

133. (1 pt) Find the maximum and minimum values of thefunction f (x,y,z) = x2y2z2 subject to the constraint x2 + y2 +z2 = 9.

Maximum value is , occuring atpoints (positive integer or ”infinitely many”).

Minimum value is , occuring at points(positive integer or ”infinitely many”).



2780infinitely many

36

134. (1 pt) For each value of λ the function h(x,y) = x2 +y2−λ(2x+4y−16) has a minimum value m(λ).

(a) Find m(λ)m(λ) =(Use the letter L for λ in your expression.)

(b) For which value of λ is m(λ) the largest, and what is thatmaximum value?λ =maximum m(λ) =

(c) Find the minimum value of f (x,y) = x2 + y2 subject tothe constraint 2x +4y = 16 using the method of Lagrange mul-tipliers and evaluate λ.minimum f =λ =

(How are these results related to your result in part (b)?)SOLUTION(a) The critical points of h(x,y) occur where

hx(x,y) = 2x−2λ = 0

andhy(x,y) = 2y−4λ = 0.

The only critical point is (x,y) = (λ,2λ) and it gives a minimumvalue for h(x,y). That minimum value is

m(λ) = h(λ,2λ) = λ2 +4λ

2−λ(2λ+8λ−16)

orm(λ) =−5λ

2 +16λ.

(b) The maximum value of m(λ) = −5λ2 + 16λ occurs ata critical point, where m′(λ) = −10λ + 16 = 0. At this point,λ = 1,6 and m(1,6) = 12,8.

(c) We want to minimize f (x,y) = x2 +y2 subject to the cons-traint g(x,y) = 16, where g(x,y) = 2x + 4y. The method of La-grange multipliers has us solve

2x = 2λ,

2y = 4λ,

and2x+4y = 16.

We can see that this gives the same solutions for x and y as weobtained in part (a), and then, plugging in to the third equation,we have

2λ+8λ = 16.

Thus we get the same λ as before: λ = 1,6, so that the minimumvalue of f is f (λ,2λ), where λ = 1,6, or fmin = 12,8.

Note that the two question have the same answer. This makessense, because in the first problem the largest minimum valuewill clearly occur when 2x+4y = 16, which is the constraint inthe third equation.



16*L - (2*2 + 4*4)*Lˆ2/42*16/(2*2 + 4*4)16*16/(2*2 + 4*4)16*16/(2*2 + 4*4)2*16/(2*2 + 4*4)

135. (1 pt) Use Lagrange multipliers to find the maximumand minimum values of f (x,y) = 3x− 4y subject to the cons-traint x2 +3y2 = 129, if such values exist.

maximum =minimum =(For either value, enter DNE if there is no such value.)

SOLUTIONOur objective function is f (x,y) = 3x−4y and our equation

of constraint is g(x,y) = x2 + 3y2 = 129. To optimize f (x,y)with Lagrange multipliers we solve the following system ofequations

fx(x,y) = λgx(x,y), so 3 = 2λx,

fy(x,y) = λgy(x,y), so −4 = 6λy,and

g(x,y) = 129, so x2 +3y2 = 129.

Solving for λ gives us

λ =32x

=−23y

,

which we can use to find x in terms of y:

x =−94

y.

Using this relation in our equation of constraint, we can solvefor y:

x2 +3y2 = 129,

so

(−94

y)2 +3y2 = 129,

and thus

((94)2 +3)y2 = 129.

So, solving for y, we have y =±4, and x =∓9. Thus, the criti-cal points are (−9,4) and (9,−4). Since the constraint is closedand bounded, maximum and minimum values of f subject tothe constraint exist. Evaluating f at the critical points, we findthat the maximum value is f (9,−4) = 3(9)+4(4) = 43 and theminimum value is f (−9,4) =−3(9)−4(4) =−43.



3*3*3 + 4*4-3*3*3 - 4*4

37

136. (1 pt) A company manufactures x units of one item andy units of another. The total cost in dollars, C, of producing thesetwo items is approximated by the function

C = 5x2 + xy+4y2 +700.

(a) If the production quota for the total number of items (bothtypes combined) is 240, find the minimum production cost.cost =

(b) Estimate the additional production cost or savings if theproduction quota is raised to 241 or lowered to 239.production cost or savings =

SOLUTION(a) We want to minimize C subject to g = x+y = 240. Using

Lagrange multipliers, we require

10x+1y = λ,

and1x+8y = λ.

From the constraint, y = 240− x, so that, equating these twoequations and substituting for y, we get

10x+1(240− x) = λ = 1x+8(240− x),

or 16x = 1680. Thus x = 105 and y = 240−105 = 135. There-fore C = 142900 dollars.

(b) Note that from the preceding we have

λ = 10x+1y = (10)(105)+(1)(135) = 1185.

Thus, increasing production by 1 will cause costs to increaseby approximately $1185. Similarly, decreasing production by 1will save approximately $1185.



5*105*105 + 1*105*135 + 4*135*135 + 7002*5*105 + 1*135

137. (1 pt) Use Lagrange multipliers to find the maximumand minimum values of f (x,y) = 5x+y+ z, subject to the cons-traint x2 + y2 + z2 = 4, if such values exist.

maximum =minimum =(For either value, enter DNE if there is no such value.)

SOLUTION

The objective function is f (x,y,z) = 5x+y+ z and the equa-tion of constraint is g(x,y,z) = x2 + y2 + z2 = 4. Their gradientsare

∇ f (x,y,z) = 5~i+~j +~k,

and

∇g(x,y,z) = 2x~i+2y~j +2z~k.

So the equation ∇ f = λ∇g becomes

5~i+~j +~k = λ(2x~i+2y~j +2z~k).

Solving for λ we find

λ =52x

=12y

=12z

.

Which provides us with the equations

2x = 10y and 2x = 10z.

Thus y = 15 x and z = 1

5 x. Substituting these into the equation ofconstraint, we can find x:

x2 +(15

x)2 +(15

x)2 = 4,

so that

(1+(15)2 +(

15)2)x2 = 4,

and thus

x =± 10√27

.

Since y = 15 x and z = 1

5 x, the critical points are at±( 10√

27,( 1

5 )( 10√27

),( 15 )( 10√

27)). Since the constraint is closed and

bounded, maximum and minimum values of f subject tothe constraint exist. Evaluating f at the critical points, wefind the maximum is f ( 10√

27,( 1

5 )( 10√27

),( 15 )( 10√

27)) ≈ 10,3923,

and the minimum value is f (− 10√27

,−( 15 )( 10√

27),−( 1

5 )( 10√27

) ≈−10,3923.



2*sqrt(27)-2*sqrt(27)

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