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MATH
32A
FINAL
EXAM
December
13,
2007
LASTNAME
FIRST NAME
IDNO.
YourTA:
To
receive
credit,
you
must
write
your
answer
in
the
space
provided.
DO NOT WRITE BELOW THIS LINE
1
(20 pts)
4
(20 pts)
2
(20 pts)
5
(20 pts)
3
(20 pts)
6
(20 pts)
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TOTAL
FOR
WRITTEN
PROBLEMS
1
PROBLEM
1
(20
Points)
The plane
xz
+ y+ =1
23
intersects the x, y, and z axes in points P, Q, and R.
(A)Determine P, Q, and R.
(B)Find the area of the triangle .PQR.
Solution:
(A)The points are
P =(2,0,0),Q=(0,1,0),R =(0,0,3)
(B)The area of .PQR is onehalf the length of the cross
product:
..
PQ
PR =(j
2i)
(3k
2i)=3i
+2k
+6j
Err:509
Thus, the area is:
.. p
11 7
 PQ
PR=32
62
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22
Err:520
22 2
2
PROBLEM
2
(20
Points)
Thetemperature atposition(x,y,z)in a room is
oC
f(x,y,z)= xy +2z +17
A woman walksdown a spiral staircasein the middle of the
room. She holds a thermometer whose position at time t
(seconds)is
tt t
c(t)= hcos
,sin
,8
333
(t in seconds). How fast is temperature reading on the
thermometer changing at t =6 s.
Solution:
We have
.f = hy,x,2i
1 t 1 t 1
c
(t)= h
sin
,
cos
,
3 33 33
At t = 6,
c(6)= (x,y,z)= h1,0,82i
.f = h0,1,2i
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c
(6)= h0,
11
,
33
By the ChainRule forpaths, the rate of change of tempera
ture is
d
(t)
dt
The temperature reading at t =6 is changing at the rate
11 1
,
i = 
oC/s
33 3
3
4
PROBLEM
3
(20
Points)
Findthe maximum value of f(x,y,z)= xyz, subject to the
constraint
222
g(x,y,z)=9x + y + z =243
Solution:
Use the method of Lagrange multipliers:
.f = hyz,xz,xyi = .g = h18x, 2y,2zi
Case 1: x,y,z are all nonzero. Then
yz xz xy
f(c(t))= .f h
c
.f h
c ' (6)= h0,1,2ihh0,
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Err:520
18x 2y 2z
This yields
2
22
2
y =9x, z =9x
Plug into the constraint:
222
22
9x + y + z =9x 2
+9x 2
+9x =243 . x =243/27 =9
f(3,9,9)=3(9)(9)=243
Case2: Atleastone of x,y,z is zero. Inthis casef(x,y,z)=
xyz =0.
We conclude that the maximum value of f = xyz subject
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This test prep was uploaded on 04/21/2008 for the course MATH 31A taught by Professor Jonathanrogawski during the Fall '07 term at UCLA.
 Fall '07
 JonathanRogawski
 Math

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