MAT 152 – 001
Name:____KEY____________
Exam 2
Grade:___110_______
1) Find the indicated partial derivatives for the functions below:
a) f ( x, y)  8x3  6 x 2 y  6 xy 2  7 y 2  2 y  1
f x  24 x 2  12 xy  6 y 2
f y  6 x2  12 xy  14 y  2
f xy  12 x  12 y
b) g ( x, y )  ln  x 4  x 2 y 2  y 4 
gx 
4 x3  2 xy 2
x4  x2 y 2  y 4
gy 
2 x2 y  4 y3
x4  x2 y 2  y 4
g xx
12 x

2
 2 y 2  x 4  x 2 y 2  y 4    4 x 3  2 xy 2 
x
4
 x2 y 2  y 4 
2
2
2) Find all points where f ( x, y)  3x 2  7 y 3  42 xy  5 has any relative extrema.
Identify any saddle points.
f x  6 x  42 y  0
f y  21 y 2  42 x  0
y ( y  14)  0
y  0 or y  14
(0, 0) and (98,14)
D ( x, y )  6(42 y )  42 2
D (0, 0)  0
D (98,14)  0
So, (0,0) is a saddle point, and (98,14) is a relative minimum.
3) A company finds that the total cost to manufacture two products (x and y) is
given by
C ( x, y)  2 x 2  6 y 2  4 xy  10 .
What amount of each product should be made to minimize the cost if a total of 10
must be produced (i.e. subject to x  y  10 ).
F ( x, y,  )  2 x 2  6 y 2  4 xy  10   ( x  y  10)
Fx  4 x  4 y    0
Fy  12 y  4 x    0
F  x  y  10  0
Solving these gives
x  10, y  0
4) Consider the graph
z  x ln  xy  .
a) Find the total differential (dz).
dz  (ln xy  1)dx 
x
dy
y
b) If x = 2, y = 3, dx = 0.1 and dy = 0.2, evaluate dz.
dz  (ln 6  1)(0.1) 
2
0.2  0.4125
3
5) Evaluate
2 3
 x
1 0
x 3
1 4

1 0  x y  y  dxdy  1  4 x y  xy  dy
x 0
2 3
2
3
2
93
ydy
4
1

93 y 2

4 2

279
8
2

y 1
93
 4  1
8
3
y  y  dxdy
6) Find the volume under the curve
z  xy x 2  y 2
and over the rectangular region given by 0  x  4, 0  y  1 .
1 4


2
2
2
xy
x

y
dxdy

0 0
0  0 x  y xdx 
1 4
1
4

1
    x 2  y 2 2 xdx  ydy
2 0
0

x4
1

3
1 2 2
x  y 2  2  ydy



20 3
 x 0
1
3
1
2
3
1
1
   y 2  16  ydy   ( y 2 ) 2 ydy
30
30
1
3
1

2
1 1 2
1
y  16  2 ydy   y 4 dy


302
30

5
12 2
y  16  2

65
y 1

y 0
11 5
y
35
y 1
y 0
1 5
1 5
1
 17 2  16 2 
15
15
15
289
205

17 
15
3
7) Evaluate
dydx
x
R
where R is the region given by 1  x  2, 0  y  x  1.

2 x 1
dydx
R x  1

0
1
dydx
x
1
 
x
1
2
x 1

0
2

1 x 1
dy dx   y y 0 dx
x
1


x 1
dx
 x
2
1
 x  ln x 1  1  ln 2
2

8) Find the general solution to
dy x 2  1

dx
y
(you may express answers as powers of y).
2
 ydy    x  1 dx
y 2 x3
  xC
2
3
2
y 2  x3  2 x  C
3
9) Consider the differential equation
dy
x
 x 2  2 , x > 0.
dx
a) Find the general solution.
dy
2
 x
dx
x
dy
2

 dx dx    x  x dx
1
y  x 2  2 ln x  C
2
b) Find the particular solution if y 
8 1 2
 (1 )  2 ln1  C
9 2
8 1 7
C  
9 2 18
So,
y
1 2
7
x  2 ln x 
2
18
8
when x  1.
9
10) Consider the differential equation
dy
x2
 xy  x3  2 x 2 , x  0 .
dx
a) Find the Integrating Factor for the equation.
dy 1
 y  x2
dx x
So,
1
I ( x)  e  x  e ln x  x
dx
b) Find the general solution of this differential equation.
1 3
2
x  x  2  dx  C 3 x  x  C 1 2
C

y

 x x
x
x
3
x
BONUS (10 pts)
Design a cylinder (by finding the appropriate height h and radius r) so that the
volume of the cylinder is 4 and the surface area is a minimum.
F (r , h,  )  2 r 2  2 rh   ( r 2 h  4 )
Fr  4 r  2 h  2 rh  0
Fh  2 r   r 2  0
F   r 2 h  4  0
Solving ,
r32
h  23 2