BBMP1103 - Sept 2011 exam workshop - part 7
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Transcript of BBMP1103 - Sept 2011 exam workshop - part 7
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BBMP 1103Mathematic Management
Exam Preparation Workshop Sept 2011Part 7 - Application of Partial Differentiation
Presented By: Dr Richard Ng
26 Nov 20112ptg – 4ptg
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Prepared by Dr Richard Ng (2011) Page 2
Question: 15 (May 2010)
7. Focus on Application of Partial Differentiation
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Prepared by Dr Richard Ng (2011) Page 3
Suggested Answers:
i) xyyxyxyxf 223225),(
023 xyf y … (ii)
022 yxf x … (i)
(i) x 2: 0244 yx … (iii)
(iii) - (ii): 073 x
3
7x
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Prepared by Dr Richard Ng (2011) Page 4
Substitute into (i):
03
722
y
03
142 y
03
8 y
3
8y
Hence, the critical point is
3
8,3
7
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ii) To determine maximum or minimum use M Test:
2)][()])([( xyyyxx fffM 2]1[)]2)(2[(
]1[]4[
yxf x 22 xyf y 23
2xxf
1xyf
2yyf
3
Since M > 0 and fxx > 0, hence the critical point is minimum
Prepared by Dr Richard Ng (2011) Page 5
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Prepared by Dr Richard Ng (2011) Page 6
Question: 16 (Sept 2009)
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Suggested Answers:
i) 548888),( 22 xxyyxyxf
048816 yxf x … (i)
0816 xyf y … (ii)
Equation (i) x 2:
0961632 yx … (iii)
(iii) – (ii) : 09624 x
4x
Prepared by Dr Richard Ng (2011) Page 7
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Substitute x = -4 into (ii): 0)4(816 y
3216 y
2y
Hence, the critical point is = (-4, 2)
ii) To determine maximum or minimum use M Test:
16xxf
8xyf
16yyf
48816 yxf x xyf y 816
Prepared by Dr Richard Ng (2011) Page 8
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2)][()])([( xyyyxx fffM 2]8[)]16)(16[(
]64[]256[
192
Since M > 0 and fxx > 0, hence the critical point is minimum
iii) When x = - 4, y = 2:
5)4(48)2)(4(8)2(8)4(8),( 22 yxf
51926432128
101
Hence, the minimum value is = -101
Prepared by Dr Richard Ng (2011) Page 9
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End ofPart 7