Find the indicated function values, if possible
1.
a. f(-2,1)
b. f(3,-2)
g ( x, y )  2 xy  5 x 2 y  y 3
g (2, 2)  2  2  2  5  (2) 2  2  23  8  40  8  40
g (1, 1)  2  1 1  5  (1) 2  1  (1) 3  2  5  1  4
2.
a.
b.
f ( x, y ) 
5x2  y3
4x  y2
 f (1,3) 
5(1)2  33 32

4  9
5
5(3)2  23 53
f (3, 2) 

 13.25
43  4 4
3.
a.
b.
f ( x, y)  2 x3 y  ln y  f (2, e)  2(2)3 e  ln e  16e  1
f (3,1)  2(3)3 1  ln1  54
Draw a three-dimensional coordinate system, and graph and label the given points.
4. A(-3,0,2); B(0,0,2); C(2,-2,2); D(0,-2,2)
The given points will be represented as given below.
Find
for each of the functions
5.
f ( x, y )  16  2 x 2  y 2 
3

f 1
 (16  2 x 2  y 2 ) 2  4 x 
x 2
3

f 1
 (16  2 x 2  y 2 ) 2  2 y 
y 2
6.
1

y
f
y
x 
f ( x, y) 

y
2
ln x
x
(ln x)
x(ln x) 2
f
1

y ln x
7.
f ( x, y )  xe y  y 2e x 
3
3
f
 e y  y 2e x
x
3
f
 3xy 2e y  2 ye x
y
2x
(16  2 x  y )
2
2
y
3
(16  2 x 2  y 2 ) 2
3
2
Find all second-order partial derivatives
8.
ex
f 3x 2e x
f
12 x 2e x
2 x3
5
f ( x, y )  4 



3
x
e


4
y


y
x
y4
xy
y5
3
3
3
f 3x 2e x
 2 f 3xe x


 4 3x3  2
x
y4
x 2
y
3
3
ex
f 4e x
 2 f 20e x
f ( x, y )  4 
 5  2  6
y
y
y
y
y
3
3
3
9.
1

f 1
21
3
2
f ( x, y )  (7 x  y ) 
 (7 x  y ) 2  21x 2 
x 2
2
3
1
2 2
x2
1
2 2
(7 x  y )
3
1
1



3
2 2
2 1
3
2
2
2
x
(7
x

y
)

x
(7
x

y
)
21
x


2
 2 f 21 


x 2
2
(7 x3  y 2 )
3
2
21 3.5 x  2 xy 

3
2
3
2 2
(7 x  y )
f 21

x 2
x2
1
2 2

(7 x  y )
3
1
3

f
21
1
21
 x 2   (7 x3  y 2 ) 2 2 y  
xy 2
2
2
f ( x, y )  (7 x 3  y 2 ) 2 
1

f 1
 (7 x 3  y 2 ) 2  2 y 
y 2
1
f

y
y
1
2 2
(7 x  y )
3
(7 x  y )
3
2
y
1
(7 x3  y 2 ) 2
1

1
(7 x3  y 2 ) 2  y (7 x3  y 2 ) 2 2 y
 f
2
 2 

3
y
(7 x  y 2 )
2
x2 y
7 x3
3
2 2
(7 x  y )
3

3
2
Find
10.
S
 16(8m  b  17)  18(9m  b  23)  20(10m  b  28)  2(245m  27b  623)
m
S
 2(8m  b  17)  2(9m  b  23)  2(10m  b  28)  6(9m  b  13)
b
11.
S (m, b)  (12m  b  81)2  (13m  b  88) 2  (14m  b  96) 2  (15m  b  101) 2
S
 2 12(12m  b  81)  13(13m  b  88)  14(14m  b  96)  15(15m  b  101) 
m
S
 2  (12m  b  81)  (13m  b  88)  (14m  b  96)  (15m  b  101) 
b
Find
12.
n is greek for lambda
F ( x, y, )  3x 2  12 y 2   ( x  2 y  84)
F
 6x 
x
F
 24 y  2
y
F
 x  2 y  84

find all local maxima, local minima, and saddle points.
13.
F ( x, y)  12 x  8 y  x 2  y 2  7
Fx  12  2 x
Fy  8  2 y
Fxx  2
Fyy  2
Fxy  0
Putting Fx = 0 and Fy = 0 and solving, we get x =6, y =4.
Hence, FxxFyy-Fxy2 = 4 and (6, 4) is maximum point.
14.
F ( x, y)  x 2  2 xy  4 y 2  6 y  3
Fx  2 x  2 y
Fy  2 x  8 y  6
Fxx  2
Fyy  8
Fxy  2
Putting Fx = 0 and Fy = 0 and solving, we get x =1, y =1.
Hence, FxxFyy-Fxy2 = 16-4 = 12 and, hence, (1,1) is minimum point.
15.
F ( x, y)  3x 2  2 xy  y 2  16 x  4 y  14
Fx  6 x  2 y  16
Fxx  6
Fy  2 x  2 y  4
Fyy  2
Fxy  2
Putting Fx = 0 and Fy = 0 and solving, we get x =3, y =1.
Hence, FxxFyy-Fxy2 = 12-4 = 8 and, hence, (3,1) is minimum point.
16.
F ( x, y)  x3  3x 2  2 y 2  9 x  8 y  7
Fx  3x 2  6 x  9
Fy  4 y  8
Fxx  6 x  6
Fyy  4
Fxy  0
Putting Fx = 0 and Fy = 0 and solving, we get x =-1, 3, y =4. Hence, points of interest are (-1,-4) and (3,4).
At (3,-4), Fxx = 12, Fyy = -4, Fxy = 0.
Hence, FxxFyy-Fxy2 = -48 and, hence, (3,1) is saddle point.
At (-1,-4), Fxx = -12, Fyy = -4, Fxy = 0.
Hence, FxxFyy-Fxy2 = 48 and, hence, (-1,-4) is maximum point.
17.
F ( x, y)  9 x  xy 2  2 y 3
Fx  9  y 2
Fy  2 xy  6 y
Fxx  0
Fyy  2 x  6
Fxy  2 y
Putting Fx = 0 and Fy = 0 and solving, we get x =3, y = 3,-3. Hence, points of interest are (3, 3) and (3,-3).
At (3,3), Fxx = 0, Fyy = 0, Fxy =-6.
Hence, FxxFyy-Fxy2 = -36 and, hence, (3, 3) is saddle point.
At (3,-3), Fxx = 0, Fyy = 12, Fxy = 6.
Hence, FxxFyy-Fxy2 = -36 and, hence, (3,-3) is also saddle point.
If fxxfyy-fxy2 is NEGATIVE, then the point is neither a max nor a min, it is a SADDLE POINT.
If fxxfyy-fxy2 is POSITIVE, and fxx and fyy are both negative, the point is a MAXIMUM.
If fxxfyy-fxy2 is POSITIVE, and fxx and fyy are both positive, the point is a MINIMUM.
Use the method of Lagrange multipliers to find the minimum value of
constraint.
subject to the given
18.
F ( x, y)  2 x 2  y 2  18x
Condition
3x-y-8 = 0
F ( x, y,  )  2 x 2  y 2  18x   (3x  y  8)
F
 4 x  18  3  0
x
F
 2y   = 0
y
F
 3 x  y  8 =0

Solving these equations, we get
λ = -2, x = 3 and y = 1. Hence,
F ( x, y)min  18  1  54  35
19.
F ( x, y)  x3  y 3
Condition
x-y = 10
F ( x, y,  )  x3  y 3   ( x  y  10)
F
 3x 2    0
x
F
 3 y 2    0
y
F
 x  y  10 = 0

Solving these equations, we get
F ( x, y)min  125  (125)  250
x = 5 and y = -5. Hence,
Use the method of Lagrange multipliers to find the maximum value of
constraint.
20.
F ( x, y)  6 x 2  5xy
Condition
2x-y -8 = 0
F ( x, y,  )  6 x 2  5 xy   (2 x  y  8)
F
 12 x  5 y  2  0
x
F
 5 x    0
y
F
 2x  y  8  0

Solving these equations, we get
x = 5 and y = 2. Hence,
F ( x, y)min  6  4  5  2  5  26
21.
F ( x, y)  x 2  4 y 2  84 xy
Condition
5x+2y-18 = 0
F ( x, y,  )  x 2  4 y 2  84 xy   (5x  2 y  18)
F
 2 x  84 y  5  0
x
F
 8 y  84 x  2  0
y
F
 5 x  2 y  18  0

Solving these equations, we get
x = 2 and y = 4. Hence,
subject to the given
F ( x, y)min  4  64  84  8  612
Use the method of Lagrange multipliers to find the maximum and minimum values of
subject
to the given constraint.
22.
F ( x, y )  5 xy
Condition
9x2+y2-162 = 0
F ( x, y,  )  5 xy   (9 x 2  y 2  162)
F
 5 y  18 x  0
x
F
 5 x  2 y  0
y
F
 9 x 2  y 2  162  0

Solving these equations, we get
x = ±3 and y = ±9. Hence,
F ( x, y)min  5  3  9  135
23. Production. The management of a company has determined that x units of labor and y
units of capital are required to produce
units of product. Each unit
of labor cost $450, and each unit of capital cost $360. Find the maximum number of
units that can be produced if a total of $90,000 is available for labor and capital.
F ( x, y)  130 x0.4 y 0.6
Condition
450x+360y = 90000
F ( x, y,  )  130 x0.4 y 0.6   (450 x  360 y  90000)
F
 130 y 0.6  0.4 x 0.6  450  0
x
F
 130 x 0.4  0.6 y 0.4  360  0
y
F
 5 x  2 y  0
y
F
 450 x  360 y  90000  0

From Fx and Fy, we get  
130 y 0.6  0.4 x 0.6 130 x0.4  0.6 y 0.4

 15 x  8 y
450
360
Putting this value in Fλ = 0, we get and solving, we get x = 80 and y = 150
Hence, F ( x, y ) max  130  800.4 1500.6  15164.7097 say 15164units
a.Find the equation of the regression line for the given points, and b. draw the scatter diagram and
graph the regression line.
24. (1, 5.2), (2, 6.4), (3, 8.1), (4,9.2), (5, 10.6)
Regression line: y =1.36x+3.82
Scatter plot and line plot are given below.
Scatter plot
12
5, 10.6
10
4, 9.2
3, 8.1
y
8
2, 6.4
6
1, 5.2
4
2
0
0
1
2
3
x
4
5
6
Find the equation of the regression line for the given points
25. (1, 0.2), (2, 0.4), (3, 0.3), (4, 0.6), (5, 0.6)
Regression line: y=0.1x+0.12
26. (3.2, 0.10), (4.1, 0.15), (4.8, 0.20), (5.1, 0.23), (6.0, 0.29)
Regression line: y = 0.069x-0.12617
27. Tourism. The total number of foreign tourist visiting the United States, as reported by
the U.S. Travel and Tourism Administration, is shown in the following table.
Year (x)
Tourist (y) (in millions)
2000
25.7
2001
26.3
2002
29.7
2003
34.2
2004
38.3
a. Find the regression line.
b. Estimate the number of foreign tourists that will visit the United States during 2005.
Regression line: y= 3.31x-6595.78
y(2005) = 3.31*2005-6595.78 = 40.77 million
Evaluate the given double integral.
28.
2 3
2
2
2
0
0
2
  (4  x)dydx    4 y  xy 1 dx   12  3x  4  x  dx   8  2 x  dx  8x  x   12
2
3
0
0 1
0
29.
2
1
1
 e xy 
 e2 y
2y
y
y 
2
1


ye
dxdy

y
dy

e

e
dy

 2  e   0.5e  e  1.5
0 1
0  y 
0 


0
1
1 2
1
xy
Evaluate the double integral on the given rectangular region.
30.
6 4
x
1 1
4
6
6
3


x3 
65
65 2 
3  ydxdy    3  y  dy   3  ydy   (3  y) 2   130
3  1
3 1
3 3
1
1 
6
2
Evaluate the double integral
31.
x2
4
4
4
3
 12 
 2 52 x 2 
x
49
2
dydx   x  2 y  dx  2 ( x  x)dx  2  x   
y
2 1 5

x
5
1
1
4 x2

1 x
32.
2
2 4 y
 
0
0
4 y 2
2 3 
y xdxdy   y  x 2 
 3 0
0
2
2
dy 
0
3
2
2
2 2
y
(4

y
)
dy    t 4 dt if t 2  4  y 2  2tdt  2 ydy

30
32
2
2
2 4
2 t5 
64
  t dt    
30
3  5  0 15
Evaluate the double integral on the given region.
33.
 (3  2x  2 y)dA
R : 0  x  1;0  y  2  x
R
1 2 x
1
  (3  2 x  2 y)dydx  3 y  2 xy  y
0 0
0
2
1
2 x
 dx   3(2  x )  2 x (2  x )  (2  x ) 2  dx
0
0
1
 x3

2
2


x

6
x

2
dx

 3  3x  2 x    3
0 


0
1
2
34. Find the volume of the solid bounded above by the graph of
below by the triangle with vertices (0,0,0), (0,2,0), and (2,0,0).
V     dxdydz
4
V 
0
4 y
2 3 x  2 y
 
0
0
4
4 y
2
0
0
dzdxdy  

4 y

 2
x2
(3  x  2 y )dxdy   3 x   2 xy  dy
2
0
0 
4
and
4

9 y2 7 3 
7
94
 9y 7 2 
8


y
dy

8
y


y   32  18   64 
0  4 8  
8
24  0
24
3
4