Math 128 Post Exam 2 Review Questions:

1.) Let 𝑓(𝑥, 𝑦) = 𝑥2 + 2𝑥𝑦 + 2𝑦2 − 4𝑥 − 6𝑦 + 12. Find the critical value(s).

(A) (−1,1) (B) (−2,1) (C) (1,1) (D) No critical values

2.) Calculate the Determinant for the function in #1.

(A) (𝑥 + 𝑦)(𝑥 + 2𝑦) − 4 (B) 8 − (𝑥 + 𝑦)2 (C) 4 (D) 0

3.) The point (1,1) determines which of the following for the function in #1:

(A) Local minimum

(B) Local maximum

(C) Saddle point

(D) None of the above.

4.) Find and classify the critical value(s) of 𝑓(𝑥, 𝑦) = 𝑥3 − 12𝑥𝑦 + 8𝑦3

(A) Local minimum at (2,1) and a saddle point at (0,0).

(B) Saddle point a (2,1) and local minimum at (0,0).

(C) Local minimum at (2,1) and a local maximum at (0,0).

(D) Local maximum at (1,2) and local minimum at (2,1).

5.) Find and classify the critical value(s) of 𝑓(𝑥, 𝑦) = 1𝑥+ 1

𝑦.

(A) Saddle point at (0,0).

(B) Local maximum at (0,0)

(C) Local minimum at (0,0)

(D) No local maximum, local minimum, nor saddle point.

6.) Which of the following is true about 𝑓(𝑥, 𝑦) = 𝑥3 + 𝑦3 − 192𝑥 − 75𝑦 + 2.

(A) (8,−5) is a saddle point.

(B) (8,−5) is a local maximum.

(C) (8,5) is a local maximum.

(D) (−8,−5) is a local minimum.

⇒

O

o ⇒

① fxx = 2 SO

Dla, , ,

= 4 so⇒ ( till is a local min

o ⇒

⇒

O

① Fx = 3×2-192=0FY =3 y

2- 75

so

⇒ ( Xy ) = ( I 8,

I 5)

Fxx = GX, fyy = Gy ,

fxy = O

Dl is ,-5g

= E8) Cs . C-51 ) LO SADDLE

I . fx = 2X t2y - 4=0 2X It 2y=4 X=4-⇒

- ( 2x +4 y =6) = 4--1=1Fy = 2x t4y - 6=0 o×2⇒1y-I'D2

. Dla, ,,= fxxifyy - thy = 2.4 - ( 25--8-4=4

. . . - - - --

-- - - - e

--

--

--

--

- -- - - - - . -

jIo.

f-I-

-- - .

4' If!:

⇒

Y='⇒x=ay=,

⇒- 12×+244455=0 ⇒ - 12×+7×4=0

Fxx-

- Gx fyy-

- 48 y ⇒ - lax ( I - gt B) = O ⇒ x = o or X=F8= 2

fxy = - 12

:÷÷::::is" 't⇒:=a④Local min

-5

.fix

, y) = x''

t y' '

f ,=

-

4×2=0⇒ no values of x andy exist

.

fy =

- Yy2=0

-

7.) A company makes two types of calculators, x and y, where x and y are measured in dollars with revenue function: 𝑅(𝑥, 𝑦) = 80𝑥 − 6𝑥2 − 4𝑥𝑦 + 68𝑦 − 2𝑦2. What prices should be charged to maximize the revenue?

(A) $1.50 and $15.50

(B) $4.00 and $17.00

(C) $5.00 and $9.00

(D) $2.00 and $12.00

8.) Find the critical value(s) of 𝑓(𝑥, 𝑦) = 2𝑥3 + 6𝑥𝑦 + 𝑦2.

(A) (3, −9), (0,0)

(B) (0, −1)

(C) (−1, − 12)

(D) (0, −9), (3,0)

9.) 𝑓(𝑥, 𝑦) is a function with critical point (𝑎, 𝑏). We have the following information: 𝑓𝑥𝑥(𝑎, 𝑏) = 3; 𝑓𝑦𝑦 (𝑎, 𝑏) = 2; 𝑎𝑛𝑑 𝑓𝑥𝑦(𝑎, 𝑏) = −1. Then what can we conclude about 𝑓(𝑎, 𝑏)?

(A) 𝑓(𝑎, 𝑏) is a local maximum.

(B) 𝑓(𝑎, 𝑏) is a local minimum.

(C) 𝑓(𝑎, 𝑏) is a saddle point.

(D) We cannot determine anything about 𝑓(𝑎, 𝑏) with the information given.

10.) If 𝑓(𝑥, 𝑦) = 𝑥𝑦 with constraint 1 − 𝑥 − 𝑦 = 0. What is the maximum value?

(A) 1 (B) 2 (C) 12 (D) 1

4

11.) Find the absolute minimum of 𝑓(𝑥, 𝑦) = 𝑥2 + 𝑦2 − 𝑥𝑦 with constraint 𝑥 − 𝑦 = 10.

(A) 75

(B) 25

(C) 100

(D) 0

① next ⇒

page

① fi ,= 6×2+6,1=0 ⇒

fy=y=e⇒ 6×+21×4=0 ⇒ 2×13 - x) = O

( o,

-07=0,0 ) ⇒ X=o↳ix -5

Dha,lol = 3.2 - f- 1)

t

= 6 - I =

520O Fxx ha,

=3 SO location

next ⇒O page

o ¥::as: ⇒¥#' " "

constraint '

to -- x - y = f yl - y = - zy ⇒ y = -5

-

'

and

FCT,-51=25+25-1-25) = 75 ×=5

7.

Max Rex ,y , noting that by Io since the representthe number of . things.

R×= 80 - 12×-44=0 ⇒ 12×+4,1=80

Ry =- 4×+68 - 4y=o ⇒-(4xt4)

SIX toy = 12

⇒ X=Ig= 3/2

I 12171+4 y= 80

4y= 80 - 18=62( 312,342 ) = ( 1.5

,15.5 )

y == 312

-10

.

fx = y = Xgx = Xfl )

fy=x=xgy= , .fi ,⇒ /X

0=1 - x - y = I - Cy) - y= I - zy

constraint -

-

⇒ y-

- in ⇒ GEL)and fftntl-I.fi/fJ

12.) If the production of items is measured by 𝑓(𝑥, 𝑦) = 𝑥1/2𝑦1/2 where 𝑥 is the number of labor hours scheduled and 𝑦 is the number of units of a material to produce the item. The labor costs $2300 per hour and materials cost $700 per unit, and the maximum budget is $1,600,000. Find the optimal production total.

(A) 631 units

(B) 1262 units

(C) 577 units

(D) 843 units

13. Consider 𝑓(𝑥, 𝑦) = 𝑥𝑦 with constraint 𝑥2 + 𝑦2 = 32. Using Lagrange Multipliers, which of the following is false?

(A) There is a maximum value at (4,4) and one other point. (B) There is a minimum value at (-4,4) and one other point. (C) The maximum value is 16 and the minimum value is 0. (D) An endpoint on the constraint is (0, √32), which does not represent the maximum nor minimum.

14.) Suppose we have a rectangle inscribed under the curve 𝑥2

16+ 𝑦2

4= 1, 𝑥, 𝑦 ≥ 0 as

illustrated. Find the values of 𝑥 and 𝑦 which make the area 𝑥𝑦 of the rectangle maximal.

2

y

x 4

(A) (2,2) (B) (2√2, √2)

(C) (√2, √142

) (D) (4,2)

① gfx , y ) = 2300×+700 y= I

,600

,000

FC , 7=(8%-5)" ? (8%0)

"

neat ⇒= = 630 .

488 .. . . page

next page

The max is 16⇒

and the min - 16 !O

Max AH,µ=xy subject to gcx , y ) = YI t If =/

::: ::::::¥±⇒¥=¥⇒ H=4

•

1=94×1--4,85+17 = IT t IT = Ia⇒ y ? 2 or f- IF and we need y > O

so 1*7× ? x ? 45=4 ( ru ) ! 8 ⇒ x = It ± 252 and

X > O So X = 252

and Cara,

ru )

12 .

Maximize f subject to g= 1600000

¥:::÷:::::. ¥ooI÷=¥oo¥I ⇒y.iq#ox--FDapply constraint : 1,600,000 = 2300 X t 70002,3 x = 4600 X

' KEI -

- soft coast,soot )⇒y=2⇒.8g=8#"¥tyS¥;⇒⇒

¥=¥=I⇒ ¥ =¥ ⇒ xEF

constraint : 32 =x7yI 2×2 ⇒ x=I4 ⇒ y = I 4

so there are 4 pts that result : ftp.%EfY.fi/IE@4!

15.) Solve the differential equation: 𝑦′ = 2𝑥 + 3, 𝑦(1) = 2.

(A) 𝑥2 + 3𝑥 + 2

(B) 2

(C) 𝑥2 + 3𝑥 − 2

(D) 13

𝑥3 + 32

𝑥2 + 2

16.) Solve the differential equation. 𝑦′ = 𝑥𝑒𝑥−1, 𝑦(1) = 1.

(A) 𝑒𝑥 + (1 − 𝑒)

(B) (𝑥 − 1)𝑒𝑥−1 + 1

(C) 𝑥2

2𝑒𝑥 + (1 − 𝑒

2)

(D) 𝑒𝑥−1 + 𝑒

17.) Solve the differential equation. 𝑥 𝑑𝑦𝑑𝑥

= 𝑥2 + 1.

(A) (𝑥2+1)2

2𝑥2 + 𝐶

(B) 𝑥2 − 2 ln 𝑥 + 𝐶

(C) 𝑥2

2+ ln 𝑥 + 𝐶

(D) 𝑥2+𝑥𝑥2 + 𝐶

18.) Solve the differential equation. 𝑦′ = 𝑥𝑦, 𝑦(0) = 5.

(A) (𝑥𝑦)2

2+ 5

(B) ln ((𝑥−4)2

2) + 5

(C) 𝑒12𝑥2

+ 5

(D) 5𝑒12𝑥2

ycxl = S y'

dx = fzxtzdx = I t3xtC

OI 2=41 ) = I 't 3. Itc ⇒ C= -2

ya = 573×-2

D

OYHHSY

'dx=Sxe*'d×=Sxe×e'dx=ei/fxedx]

Sxexdx = XE - SEdx=xe× - Etc

u= x vet I=yN= e-I ( e - etc ) =L ⇒ c -

- e

du -- dxdu=e×dx ⇒ ycxl=teXe× - text I

= # Hee 't I

¥'=x¥=xt¥=*O it ⇒ y = Sxttxdx =/EIthlxltI

¥t=xy ⇒ Soft=fxdx ⇒ HIM - txtc

lyke" " cent ⇒ y - cette

O C ? 5- yes =ce÷Ecy=c ⇒ y=5e" "

'

19.) Solve the differential equation: 𝑑𝑦𝑑𝑥

= 5𝑦2, 𝑦(0) = 1.

(A) 11−5𝑥

(B) 𝑒52𝑥

(C) 1(1−5𝑥)3

(D) 5𝑒2𝑥

20.) Solve the differential equation: 𝑥2𝑦′ + 𝑦′ = 𝑥, 𝑦(0) = 4.

(A) 12

ln(𝑥2 + 1) + 4

(B) − 16

𝑥3 + 14

𝑥2 + 4

(C) −2𝑥 + 1

(D) 4𝑒𝑥2+𝑥

21.) Solve the differential equation. 𝑑𝑦𝑑𝑥

= 2𝑦 + 3, 𝑦(0) = 0.

(A) 𝑦2 + 3𝑦

(B) − 32

+ 32

𝑒2𝑥

(C) 32

𝑒2𝑥

(D) ln (𝑦 + 32)

22.) Solve the differential equation: 𝑦′′ = 10, 𝑦′(1) = 5, 𝑦(2) = 6.

(A) 10𝑥 + 11

(B) 0

(C) 5𝑥2 − 5𝑥 − 4

(D) 5𝑥2 + 5𝑥 + 6

0 Sff = Ssdx c ? I -- ya = IIE - I

- tf = 5xtC⇒ c = - I

⇒ y=⇒ xxx IET =÷⇒

3- Xtc

y'

C x' ti ) = X

O Eff -

- XII ⇒

y.es#.dx=SxItTdxu=x2--Skdu4-=Ih1altC=Ilulx2tIltCdu=rxdx

=Iei¥oc?4=40 '

separation of variable ,

Ofyf¥z=f2dx ⇒ In lytta = 2xtc

⇒ ly +314 = E " I Cem

⇒ Y = CEE 31 ,c ? o = ya

= c- 312⇒ 0=72

1441=321322

c ,y

'= Sy"dx = Slodx = 10x the

O-

5- y 'M = Otc ⇒ c= -5

y = Sy'd× = 510×-5 DX

= 5×2 - 5x to

⇐ 6=-1121--5.4-5.2 to = to to

C = 6- 10=-4

⇒ yea = 5×2-5×-4

23.) Solve the differential equation: 𝑑2𝑦𝑑𝑡2 = −10 and 𝑦(0) = 100, 𝑦′(0) = −20.

(A) 100 + 20𝑡 − 5𝑡2 (B) 100 − 20𝑡 + 10𝑡2 (C) 100 + 20𝑡 + 5𝑡2 (D) 100 − 20𝑡 − 5𝑡2

24.) A pollutant spilled on the ground decays at a rate of 8% per day. In addition, clean-up crews remove the pollutant at a rate of 30 gallons/day. Write a differential equation for the amount of pollutant, P, in gallons left after t days.

(A) 𝑑𝑃𝑑𝑡

= 0.08𝑃 + 30

(B) 𝑑𝑃𝑑𝑡

= (0.08 + 30)𝑃

(C) 𝑑𝑃𝑑𝑡

= −0.08𝑃 + 30

(D) 𝑑𝑃𝑑𝑡

= −0.08𝑃 − 30

25.) Suppose 𝑧 = 𝑓(𝑥, 𝑦) is a function with continuous second partial derivatives, 𝑓𝑥(𝑎, 𝑏) =𝑓𝑦(𝑎, 𝑏) = 0, 𝑓𝑥𝑥(𝑎, 𝑏) > 0, 𝑎𝑛𝑑 𝑓𝑦𝑦 (𝑎, 𝑏) < 0. Then, the critical value (𝑎, 𝑏) determines

(A) a relative maximum for 𝑓 (B) a relative minimum for 𝑓 (C) a possible relative maximum, but not a relative minimum (D) definitely not a relative maximum or minimum

Iff = 'S y' 'd t

-

- S - todt = - lotte

①C ? - 20 = Y' 10 ) = - to .

o t C ⇒ C = - 20

Htt = Sy 'dt= S - lot - aodt = -57-20 ttc

C ? 100--4107=2

- -0,08 P

- 30

:Fxx team > O

Dha,

,= fxxkhd.fyyla.io) - ¥5 mole: fxIN

= A . C - ) - Exy5= minus ]= negative

SADDLE