Test 2
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,
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f
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IALS: :
STUDENT NO:~Q..s7qd,CjlIBRANCH: ENGINEERING
,
...C;.~//..l;~j
MATHEMATICS
8z
8X
L If x2z + 3yz2 - 4y2 = 0, find (8x )y' (8y )
z and
-
2A: Test 2
Total Marks: 34
Time: 60 minutes
13 April 2005
~ .... (jjj
8Y
( ) 8z
x' Hence show that
(4)
Cb1 2. Consider the function f(x, y) = -5 + xy - 3x - 2y on the closed triangular region R bounded by the lines x = 1, y = 4 and y = x-I in the first quadrant. (a) Locate the critical points of f in R. (b) Find the points on the boundary of R where f may have extrema. (c) Using the points found in parts (a) and (b) determine the absolute maximum and minimum values of f on region R. (8)
l~
t
~
J{I;])
=-
I
~
~.\
~
t
~
:C-
I
I(~x-.~:= ~ .
-'5
-f-- {x-V-x..
::::
-5
+ x.~:c
:::-
~:c...
b X
-:3
-""3,X- -2.&:-0 - "SX -"'l-:L -t- 2-
V
C£)
3. Use the method of Lagrange multipliers.to find the extreme values of the function f(x, y, z) = 2x+y-2z
= 4.
on the sphere x2 + y2 + z2
~-.
-
(5)
fiJ J~ /Lld-
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I
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1 I
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- - - - cD
1-=-.2 ~J 7' - -
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.
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r ~
=-
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:
-2:1
=-. ~.,Lx.:J 2-
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(a)
.c[t3
- t] =
~-~D 85 82
(b)
NAD
.c[cos3t] =
~D 82 + 9
(c)
.c[sinh3t]
~D 82 + 9
~D 82 - 9
~D 82+9
82
=
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~~ 82 - 9
~
~-l!D 8
-4s
-
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2e-2s
8
1
.:8-4 - -8-2 + -8 D
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.c[sin2t]
8
82 - 9 0
8
~D -9
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. -~+lD
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~ 1 2 2 + -4 / 85 8 ~, 2" ~+-D 8- 2
NAD
s(S2- 4) 0
1 8
D
1
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(8 - ~)2 + 8 - 2~ + 1 C NAD
N1AiliNlDELA)ZL f4~\ ) ( /0/&
/1' .
-
}
f
,
~)
(dtG
)
IALS: :
STUDENT NO:~Q..s7qd,CjlIBRANCH: ENGINEERING
,
...C;.~//..l;~j
MATHEMATICS
8z
8X
L If x2z + 3yz2 - 4y2 = 0, find (8x )y' (8y )
z and
-
2A: Test 2
Total Marks: 34
Time: 60 minutes
13 April 2005
~ .... (jjj
8Y
( ) 8z
x' Hence show that
(4)
Cb1 2. Consider the function f(x, y) = -5 + xy - 3x - 2y on the closed triangular region R bounded by the lines x = 1, y = 4 and y = x-I in the first quadrant. (a) Locate the critical points of f in R. (b) Find the points on the boundary of R where f may have extrema. (c) Using the points found in parts (a) and (b) determine the absolute maximum and minimum values of f on region R. (8)
l~
t
~
J{I;])
=-
I
~
~.\
~
t
~
:C-
I
I(~x-.~:= ~ .
-'5
-f-- {x-V-x..
::::
-5
+ x.~:c
:::-
~:c...
b X
-:3
-""3,X- -2.&:-0 - "SX -"'l-:L -t- 2-
V
C£)
3. Use the method of Lagrange multipliers.to find the extreme values of the function f(x, y, z) = 2x+y-2z
= 4.
on the sphere x2 + y2 + z2
~-.
-
(5)
fiJ J~ /Lld-
. . tUWX
I
S(i/'JI:C)
J)(
3)(:::
I
1j
"-x
/
~
!J j~ -
r:/~)
~
A.i:
f
'j
~
t-
=A:~ ;L
.
h
<.y,
t~~:
(!)
2-kt
:
= -1
)~~
2..
.
'-z
1<-
+2~~
1 I
-=-
=
V'
- - - - cD
1-=-.2 ~J 7' - -
j2~ 0):
2:( t-J-~
LA~~v-.s~ ~th'fLi€.<
L =-~
-2.~
J- -
:l.X:~ -r:J~
~
Y't~)
~
"rh-
I
~
k-~
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fA-t1A
Ihj/Z)~
- If:-O
/llj Cr; J<)~
~
=
XL -IJ2. T~">-
-=-
jer<~h~
P
.
-'-
-
@
- - -@ UX
2-.-£'J
~
(j)X7r : 2/ (§) X
-X-
Cl:
~
r ~
=-
Ux.:r %:
-=- ?...L x:J
~
:
-2:1
=-. ~.,Lx.:J 2-
\
(a)
.c[t3
- t] =
~-~D 85 82
(b)
NAD
.c[cos3t] =
~D 82 + 9
(c)
.c[sinh3t]
~D 82 + 9
~D 82 - 9
~D 82+9
82
=
~D 82 + 9
~~ 82 - 9
~
~-l!D 8
-4s
-
2
2e-2s
8
1
.:8-4 - -8-2 + -8 D
(e)
.c[sin2t]
8
82 - 9 0
8
~D -9
NA!
s2 - 9 0
2
. -~+lD
~
2
8-4-8-2+!~
8
=
~D 82+4
2
~D
82- 4
3 -+ 84 3! -+ 8-3
~ 1 2 2 + -4 / 85 8 ~, 2" ~+-D 8- 2
NAD
s(S2- 4) 0
1 8
D
1
2
(8 - ~)2 + 8 - 2~ + 1 C NAD
