Prepared by Dr. Richard Ng

1. BBMP 1103
Mathematic Management
Exam Preparation Workshop Sept 2011
Part 7 - Application of Partial Differentiation
Presented By: Dr Richard Ng
26 Nov 2011
2ptg – 4ptg

2. 7. Focus on Application of Partial Differentiation
Question: 15 (May 2010)
Prepared by Dr Richard Ng (2011) Page 2

3. Suggested Answers:
i) f ( x, y) 25 2 x 3 y x 2 y2 xy
fx 2 2x y 0 … (i)
fy 3 2y x 0 … (ii)
(i) x 2: 4 4x 2 y 0 … (iii)
(iii) - (ii): 3x 7 0
7
x
3
Prepared by Dr Richard Ng (2011) Page 3

4. Substitute into (i):
7
2 2 y 0
3
14
2 y 0
3
8
y 0
3
8
y
3
7 8
Hence, the critical point is ,
3 3
Prepared by Dr Richard Ng (2011) Page 4

5. ii) To determine maximum or minimum use M Test:
fx 2 2x y fy 3 2y x
f xx 2 f yy 2
f xy 1
M [( f xx )( f yy )] [( f xy )]2
[(2)(2)] [1]2
[4] [1]
3
Since M > 0 and fxx > 0, hence the critical point is minimum
Prepared by Dr Richard Ng (2011) Page 5

6. Question: 16 (Sept 2009)
Prepared by Dr Richard Ng (2011) Page 6

7. Suggested Answers:
i) f ( x, y) 8x 2 8 y 2 8 xy 48x 5
fx 16x 8 y 48 0 … (i)
fy 16 y 8x 0 … (ii)
Equation (i) x 2:
32x 16 y 96 0 … (iii)
(iii) – (ii) : 24x 96 0
x 4
Prepared by Dr Richard Ng (2011) Page 7

8. Substitute x = -4 into (ii): 16y 8( 4) 0
16y 32
y 2
Hence, the critical point is = (-4, 2)
ii) To determine maximum or minimum use M Test:
fx 16x 8 y 48 fy 16 y 8 x
f xx 16 f yy 16
f xy 8
Prepared by Dr Richard Ng (2011) Page 8

9. M [( f xx )( f yy )] [( f xy )]2
[(16)(16)] [8]2
[256] [64]
192
Since M > 0 and fxx > 0, hence the critical point is minimum
iii) When x = - 4, y = 2:
f ( x, y) 8( 4) 2 8(2) 2 8( 4)(2) 48( 4) 5
128 32 64 192 5
101
Hence, the minimum value is = -101
Prepared by Dr Richard Ng (2011) Page 9

10. End of
Part 7