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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Question: 15 (May 2010)

7. Focus on Application of Partial Differentiation

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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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Substitute into (i):

03

722

y

03

142 y

03

8 y

3

8y

Hence, the critical point is

3

8,3

7

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

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Question: 16 (Sept 2009)

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

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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

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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

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End ofPart 7