This document summarizes an exam preparation workshop on the application of partial differentiation presented by Dr. Richard Ng. It provides the answers to two sample exam questions, finding the critical points and determining whether they are maxima or minima for multivariate functions. For the first question, the critical point is found to be (7/3, 8/3) which is a minimum. For the second question, the critical point is (-4, 2) which is also determined to be a minimum using the second derivative test.

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