answer tutorial 3.5 page 227 (MAT235 Chapter 3)

MAT235- CALCULUS II FOR ENGINEERS / Chapter 3 : Functions of Two and Three Variables

Tutorial 3.5 : Extremum of Functions

1. Jun 2019/Q3c/ 8 marks

Find the critical points of ( , ) = 3 + 2 + 2 and classify them as relative maximum,

relative minimum or saddle point(s). 1. maximum point : (2 , − 2)

33

saddle point : (0, 0)

2. Dec 2018/Q3c/ 7 marks (refer page 225-256)

Find and classify the critical point(s) of the function f (x, y ) = x3 − 3xy + y 2 − 3x .

2. minimum point : (2, 3)

saddle point : (− 1 , − 3)
3. Jun 2018/Q3c/ 8 marks (refer page 223 - 224)
24

Find and classify the critical point(s) of the function f (x, y ) = 3xy − x 3 − y 2 .

3. maximum point : (3 , 9)

24

saddle point : (0, 0)

4. Jan 2018/Q3c/ 8 marks (refer page 221-222)

Locate and classify all the critical points for f (x, y ) = x 3 − 2y 2 + x 2 + 2xy .

4. maximum point : (−1, − 1)

2

saddle point : (0, 0)

5. Mar 2017/Q3c/ 8 marks

Locate and classify all the critical points for ( , ) = 2 + 2 + 2 − 2 .

5. minimum point : (0, 1)

saddle points : (−2, −1) and (2, −1)

6. Oct 2016/Q3a/ 8 marks

Find the critical points of the function ( , ) = 3 3 − 3 2 + 2 + 2 − 1 and classify

each point as a relative maximum, relative minimum or saddle point.

6. minimum point : (8 , − 8)

99

saddle point : (0, 0)
7. Mar 2016/Q3c/ 10 marks

Find and classify the critical point(s) of the function ( , ) = 3 + 3 − 3 .

33

7. minimum point : (3, 3)
saddle point : (0, 0)

8. Dec 2019/Q3c/ 8 marks
Find all the critical points of the function ( , ) = 12 + 3 + 3 − 3 . Hence, determine

the nature of each point. 8. minimum point : (1, 1)
saddle point : (0, 0)

© Amirah 2022 227

MAT235- CALCULUS II FOR ENGINEERS / Chapter 3 : Functions of Two and Three Variables

9. Mar 2015/Q3d/ 5 marks

Given ( , ) = 2 2 + 2. Find the critical point(s) of ( , ) and determine the nature of

the point(s). 9. minimum point : (0, 0)

10. Sep 2014/Q2b/ 8 marks
Find all the critical points of ( , ) = 2 3 + 2 3 − 24 + 1 and determine the relative
maximum, relative minimum or saddle point(s), if any. 10. minimum point : (4, 4)
saddle point : (0, 0)

11. Mar 2014/Q1b/ 11 marks

Given that ( , ) = 8 2 + 2 + 4 + 2, find the critical points and determine the

relative maximum, relative minimum or saddle point. 11 minimum point : (− 1 , 1)

44

saddle points : (0, 0) and (− 1 , 0)

2

12. Sep 2013/Q1b/ 8½ marks
Find all the critical points of the function ( , ) = 2 4 − 2 + 2 and classify each

point as a relative maximum, relative minimum or saddle point.

12. minimum point : (− 1 , 1)

2

saddle points : (0, 0) and (0, 2)

13. Mar 2013/Q1b/ 8½ marks

Find all the critical points of the function ( , ) = 6 2 − 2 3 + 3 2 + 6 and classify

each point as a relative maximum, relative minimum or saddle point.
13. minimum point : (0, 0)
saddle point : (1, −1)

14. Mar 2012/Q2c/ 9 marks
Find all the extremum points of ( , ) = 2 2 − 4 + 4 + 2 and determine the nature

of the points. 14. Minimum points : (−1, −1) (1, 1)
saddle point : (0, 0)

© Amirah 2022 228

MAT235- CALCULUS II FOR ENGINEERS / Chapter 3 : Functions of Two and Three Variables

Answer Tutorial 3.5 : Extremum of Functions

1. Jun 2019/Q3c/ 8 marks
Find the critical points of ( , ) = 3 + 2 + 2 and classify them as relative maximum,
relative minimum or saddle point(s).

Solution :

Step 1 : find the all the first and second partial derivatives

= 3 + 2 + 2 = 3 + 2 + 2
= 3 2 + 0 + 2 (1)
= 3 2 + 2 ( ) = 0 + 2 + 2 (1)
= 2 + 2

= 3 2 + 2 = 3 2 + 2 = 2 + 2 = 2 + 2
= 3(2 ) + 0 = 0 + 2(1) = 0 + 2(1) = 2(1) + 0
= 6 = 2 = 2 = 2

Step 2 : find the critical points when = and =

When = 0 When = 0

3 2 + 2 = 0 ① 2 + 2 = 0
+ = 0
= − ②

Substitute ② into ① : = 0 ② ∶ = 0
22
3 2 + 2(− ) = 0
3 2 − 2 = 0 = 3 ② ∶ = − 3
(3 − 2) = 0
2
= 0, = 3

 critical points are : (0, 0) and (2 , − 2)

33

© Amirah 2022 229

MAT235- CALCULUS II FOR ENGINEERS / Chapter 3 : Functions of Two and Three Variables

Step 3 : find = ∙ − ( )
= ∙ − ( ) = (6 )(2) − (2)2 = 12 − 4

Step 4 : Table of sign for and

Critical points = 12 − 4 = 6 Conclusion
(0, 0) −4 (−) 0 Saddle point
Maximum point
(2 , − 2) 4 (+) −

33

 maximum point : (2 , − 2)

33

saddle point : (0, 0)

2. Dec 2018/Q3c/ 7 marks (refer page 58)
Find and classify the critical point(s) of the function ( , ) = 3 − 3 + 2 − 3 .

Solution :

Step 1 : find the all the first and second partial derivatives

= 3 − 3 + 2 − 3 = 3 − 3 + 2 − 3
= 3 2 − 3 (1) + 0 − 3(1)
= 3 2 − 3 − 3 ( ) = 0 − 3 (1) + 2 − 0
= 2 − 3

= 3 2 − 3 − 3 = 3 2 − 3 − 3 = 2 − 3 = 2 − 3
= 3(2 ) − 0 − 0 = 0 − 3(1) − 0 = 0 − 3(1) = 2(1) − 0
= 6 = −3 = −3 = 2

© Amirah 2022 230

MAT235- CALCULUS II FOR ENGINEERS / Chapter 3 : Functions of Two and Three Variables

Step 2 : find the critical points when = and =

When = 0 When = 0

3 2 − 3 − 3 = 0 ① 2 − 3 = 0
2 − − 1 = 0
2 = 3
3 ②

= 2

Substitute ② into ① :

3 = 2 ② ∶ = 3 (2) = 3
(2
2

2 − ) − 1 = 0 = − 1 ② ∶ = 3 (− 1) = − 3

2 2 − 3 − 2 = 0 2 22 4

( − 2)(2 + 1) = 0

1
= 2, = − 2

 critical points are : (2, 3) and (− 1 , − 3)

24

Step 3 : find = ∙ − ( )
= ∙ − ( ) = (6 )(2) − (−3)2 = 12 − 9

Step 4 : Table of sign for and

Critical points = 12 − = 6 Conclusion
(2, 3) 15 (+) 12 (+) Minimum point
Saddle point
(− 1 , − 3) −15 (−) −3 (−)

24

 minimum point : (2, 3)
saddle point : (− 1 , − 3)

24

© Amirah 2022 231

MAT235- CALCULUS II FOR ENGINEERS / Chapter 3 : Functions of Two and Three Variables

3. Jun 2018/Q3c/ 8 marks (refer page 56)
Find and classify the critical point(s) of the function ( , ) = 3 − 3 − 2.

Solution :

Step 1 : find the all the first and second partial derivatives

= 3 − 3 − 2 = 3 − 3 − 2
= 3 (1) − 3 2 − 0
= 3 − 3 2 ( ) = 3 (1) − 0 − 2
= 3 − 2

= 3 − 3 2 = 3 − 3 2 = 3 − 2 = 3 − 2
= 0 − 3(2 ) = 3(1) − 0 = 3(1) − 0 = 0 − 2(1)
= −6 = 3 = 3 = −2

Step 2 : find the critical points when = and =

When = 0 When = 0

3 − 3 2 = 0 3 − 2 = 0 ②
− 2 = 0
= 2 ①

Substitute ① into ② : = 0 ① ∶ = (0)2 = 0
= 3 ① ∶ = (3)2 = 9
3 − 2( 2) = 0
3 − 2 2 = 0 2 24
(3 − 2 ) = 0

3
= 0, = 2

 critical points are : (0, 0) and (3 , 9)

24

© Amirah 2022 232

MAT235- CALCULUS II FOR ENGINEERS / Chapter 3 : Functions of Two and Three Variables

Step 3 : find = ∙ − ( )
= ∙ − ( ) = (−6 )(−2) − (3)2 = 12 − 9

Step 4 : Table of sign for and

Critical points = 12 − = −6 Conclusion
(0, 0) −9 (−) 0 Saddle point
Maximum point
(3 , 9) 9 (+) −9 (−)

24

 maximum point : (3 , 9)

24

saddle point : (0, 0)

4. Jan 2018/Q3c/ 8 marks (refer page 54)
Locate and classify all the critical points for ( , ) = 3 − 2 2 + 2 + 2 .

Solution :

Step 1 : find the all the first and second partial derivatives

= 3 − 2 2 + 2 + 2 = 3 − 2 2 + 2 + 2
= 3 2 − 0 + 2 + 2 (1)
= 3 2 + 2 + 2 ( ) = 0 − 2(2 ) + 0 + 2 (1)
= 2 − 4

= 3 2 + 2 + 2 = 3 2 + 2 + 2 = 2 − 4 = 2 − 4
= 3(2 ) + 2(1) + 0 = 0 + 0 + 2(1) = 2(1) − 0 = 0 − 4(1)
= 6 + 2 = 2 = 2 = −4

Step 2 : find the critical points when = and =

When = 0 When = 0

3 2 + 2 + 2 = 0 ① 2 − 4 = 0
− 2 = 0
= 2 ②

© Amirah 2022 233

MAT235- CALCULUS II FOR ENGINEERS / Chapter 3 : Functions of Two and Three Variables

Substitute ② into ① : = 0 ② ∶ = 2(0) = 0
= − 1 ② ∶ = 2 (− 1) = −1
3(2 )2 + 2(2 ) + 2 = 0
3(4 2) + 4 + 2 = 0 22

12 2 + 6 = 0
2 2 + = 0
(2 + 1) = 0

1
= 0, = − 2

 critical points are : (0, 0) and (−1, − 1)

2

Step 3 : find = ∙ − ( )
= ∙ − ( ) = (6 + 2)(−4) − (2)2 = −24 − 8 − 4 = −24 − 12

Step 4 : Table of sign for and

Critical points = −24 − 12 = 6 + 2 Conclusion
(0, 0) −12 (−) 2 (+) Saddle point

(−1, − 1) 12 (+) −4 (−) Maximum point

2

 maximum point : (−1, − 1)

2

saddle point : (0, 0)

© Amirah 2022 234

MAT235- CALCULUS II FOR ENGINEERS / Chapter 3 : Functions of Two and Three Variables

5. Mar 2017/Q3c/ 8 marks
Locate and classify all the critical points for ( , ) = 2 + 2 + 2 − 2 .

Solution :

Step 1 : find the all the first and second partial derivatives

= 2 + 2 + 2 − 2 = 2 + 2 + 2 − 2
= 0 + (2 ) + 2 − 0
= 2 + 2 ( ) = 2 + 2(1) + 0 − 2(1)
= 2 + 2 − 2

= 2 + 2 = 2 + 2 = 2 + 2 − 2 = 2 + 2 − 2
= 2 (1) + 2(1) = 2 (1) + 0 = 0 + 2 − 0 = 2(1) + 0 − 0
= 2 + 2 = 2 = 2 = 2

Step 2 : find the critical points when = and =

When = 0 When = 0

2 + 2 = 0 2 + 2 − 2 = 0 ①
+ = 0 2 = 2 − 2

( + 1) = 0 = −1
= 0,

= 0 ① ∶ 2 = 2 − (0)2
2 = 2
= 1

= −1 ① ∶ 2(−1) = 2 − 2
−2 = 2 − 2
2 = 4
= ±2

 critical points are : (0, 1) , (−2, −1) and (2, −1)

© Amirah 2022 235

MAT235- CALCULUS II FOR ENGINEERS / Chapter 3 : Functions of Two and Three Variables

Step 3 : find = ∙ − ( )
= ∙ − ( ) = (2 + 2)(2) − (2 )2 = 4 + 4 − 4 2

Step 4 : Table of sign for and

Critical points = 4 + 4 − 4 2 = 2 + 2 Conclusion
(0, 1) 8 (+) 4 (+) Minimum point
Saddle point
(−2, −1) 4(−1) + 4 − 4(−2)2 = −16 (−) 2(−1) + 2 = 0 Saddle point
2(−1) + 2 = 0
(2, −1) 4(−1) + 4 − 4(2)2 = −16 (−)

 minimum point : (0, 1)
saddle points : (−2, −1) and (2, −1)

6. Oct 2016/Q3a/ 8 marks
Find the critical points of the function ( , ) = 3 3 − 3 2 + 2 + 2 − 1 and classify
each point as a relative maximum, relative minimum or saddle point.

Solution :

Step 1 : find the all the first and second partial derivatives

= 3 3 − 3 2 + 2 + 2 − 1 = 3 3 − 3 2 + 2 + 2 − 1
= 3(3 2) − 3(2 ) + 0 + 2 (1) − 0
= 9 2 − 6 + 2 ( ) = 0 − 0 + 2 + 2 (1) − 0
= 2 + 2

= 9 2 − 6 + 2 = 9 2 − 6 + 2 = 2 + 2 = 2 + 2
= 9(2 ) − 6(1) + 0 = 0 − 0 + 2(1) = 2(1) + 0 = 0 + 2(1)
= 18 − 6 = 2 = 2 = 2

Step 2 : find the critical points when = and =

When = 0 When = 0

9 = 2 −− 6 + 2 = 0 ① 2 + 2 = 0
+ = 0
= − ②

© Amirah 2022 236

MAT235- CALCULUS II FOR ENGINEERS / Chapter 3 : Functions of Two and Three Variables

Substitute ② into ① : = 0 ② ∶ = −(0) = 0
= 8 ② ∶ = − (8) = − 8
9 2 − 6 + 2(− ) = 0
9 2 − 6 − 2 = 0 9 99

9 2 − 8 = 0
(9 − 8) = 0

8
= 0, = 9

 critical points are : (0, 0) and (8 , − 8)

99

Step 3 : find = ∙ − ( )
= ∙ − ( ) = (18 − 6)(2) − (2)2 = 36 − 12 − 4 = 36 − 16

Step 4 : Table of sign for and

Critical points = 36 − 16 = 18 − 6 Conclusion
(0, 0) −16 (−) −6 (−) Saddle point

(8 , − 8) 8 (+) 10 (+) Minimum point

99

 minimum point : (8 , − 8)

99

saddle point : (0, 0)

© Amirah 2022 237

MAT235- CALCULUS II FOR ENGINEERS / Chapter 3 : Functions of Two and Three Variables

7. Mar 2016/Q3c/ 10 marks
Find and classify the critical point(s) of the function ( , ) = 3 + 3 − 3 .

33

Solution :

Step 1 : find the all the first and second partial derivatives

= 1 3 + 1 3 − 3 = 1 3 + 1 3 − 3
3 3 3 3
()
1 1
= 3 (3 2) + 0 − 3 (1) = 0 + 3 (3 2) − 3 (1)

= 2 − 3 = 2 − 3

= 2 − 3 = 2 − 3 = 2 − 3 = 2 − 3
= 2 − 0 = 0 − 3(1) = 0 − 3(1) = 2 − 0
= 2 = −3 = −3 = 2

Step 2 : find the critical points when = and =

When = 0 When = 0

2 − 3 = 0 2 − 3 = 0 ②
3 = 2
2 ①
= 3

Substitute ① into ② :

2 2 = 0 ① ∶ = (0)2 = 0
( 3 ) − 3 = 0
3
4
9 − 3 = 0 = 3 ① ∶ = (3)2 = 3
4 − 27 = 0
( 3 − 27) = 0 3

= 0, 3 − 27 = 0
3 = 27
3 = 33

= 3

 critical points are : (0, 0) and (3, 3)

© Amirah 2022 238

MAT235- CALCULUS II FOR ENGINEERS / Chapter 3 : Functions of Two and Three Variables

Step 3 : find = ∙ − ( )
= ∙ − ( ) = (2 )(2 ) − (−3)2 = 4 − 9

Step 4 : Table of sign for and

Critical points = 4 − 9 = Conclusion
(0, 0) −9 (−) 0 Saddle point
Minimum point
(3, 3) 27 (+) 6 (+)

 minimum point : (3, 3)
saddle point : (0, 0)

8. Dec 2019/Q3c/ 8 marks
Find all the critical points of the function ( , ) = 12 + 3 + 3 − 3 . Hence, determine
the nature of each point.

Solution :

Step 1 : find the all the first and second partial derivatives

= 12 + 3 + 3 − 3 = 12 + 3 + 3 − 3
= 0 + 3 2 + 0 − 3 (1)
= 3 2 − 3 ( ) = 0 + 0 + 3 2 − 3 (1)
= 3 2 − 3

= 3 2 − 3 = 3 2 − 3 = 3 2 − 3 = 3 2 − 3
= 3(2 ) − 0 = 0 − 3(1) = 0 − 3(1) = 3(2 ) − 0
= 6 = −3 = −3 = 6

Step 2 : find the critical points when = and =

When = 0 When = 0

3 2 − 3 = 0 3 2 − 3 = 0 ②
2 − = 0 2 − = 0
= 2
①

© Amirah 2022 239

MAT235- CALCULUS II FOR ENGINEERS / Chapter 3 : Functions of Two and Three Variables

Substitute ① into ② : = 0 ① ∶ = (0)2 = 0
= 1 ① ∶ = (1)2 = 1
( 2)2 − = 0
4 − = 0

( 3 − 1) = 0
= 0, 3 − 1 = 0
3 = 1
3 = 13
= 1

 critical points are : (0, 0) and (1, 1)

Step 3 : find = ∙ − ( )
= ∙ − ( ) = (6 )(6 ) − (−3)2 = 36 − 9

Step 4 : Table of sign for and

Critical points = − = Conclusion
(0, 0) −9 (−) 0 Saddle point
Minimum point
(1, 1) 27 (+) 6 (+)

 minimum point : (1, 1)
saddle point : (0, 0)

© Amirah 2022 240

MAT235- CALCULUS II FOR ENGINEERS / Chapter 3 : Functions of Two and Three Variables

9. Mar 2015/Q3d/ 5 marks
Given ( , ) = 2 2 + 2. Find the critical point(s) of ( , ) and determine the nature of
the point(s).

Solution :

Step 1 : find the all the first and second partial derivatives

= 2 2 + 2 = 2 2 + 2
= 2(2 ) + 0
= 4 ( ) = 0 + 2
= 2

= 4 = 4 = 2 = 2
= 4(1) = 0 = 0 = 2(1)
= 4 = 2

Step 2 : find the critical points when = and =

When = 0 When = 0

4 = 0 2 = 0
= 0 = 0

 critical points are : (0, 0)

Step 3 : find = ∙ − ( )
= ∙ − ( ) = (4)(2) − (0)2 = 8 − 8 − 4 = −24 − 12

Step 4 : Table of sign for and

Critical points = = Conclusion
(0, 0) (+) (+) Minimum point

 minimum point : (0, 0)

© Amirah 2022 241

MAT235- CALCULUS II FOR ENGINEERS / Chapter 3 : Functions of Two and Three Variables

10. Sep 2014/Q2b/ 8 marks
Find all the critical points of ( , ) = 2 3 + 2 3 − 24 + 1 and determine the relative
maximum, relative minimum or saddle point(s), if any.

Solution :

Step 1 : find the all the first and second partial derivatives

= 2 3 + 2 3 − 24 + 1 = 2 3 + 2 3 − 24 + 1
= 2(3 2) + 0 − 24 (1) + 0
= 6 2 − 24 ( ) = 0 + 2(3 2) − 24 (1) + 0
= 6 2 − 24

= 6 2 − 24 = 6 2 − 24 = 6 2 − 24 = 6 2 − 24
= 6(2 ) − 0 = 0 − 24(1) = 0 − 24(1) = 6(2 ) − 0
= 12 = −24 = −24 = 12

Step 2 : find the critical points when = and =

When = 0 When = 0

6 2 − 24 = 0 6 2 − 24 = 0 ②
2 − 4 = 0 2 − 4 = 0
2 = 4
2 ①
= 4

Substitute ① into ② :

2 2 = 0 ① ∶ = (0)2 = 0
( 4 ) − 4 = 0
4
4
16 − 4 = 0 = 4 ① ∶ = (4)2 = 4
4 − 64 = 0
( 3 − 64) = 0 4

= 0, 3 − 64 = 0
3 = 64
3 = 43

= 4

 critical points are : (0, 0) and (4, 4)

© Amirah 2022 242

MAT235- CALCULUS II FOR ENGINEERS / Chapter 3 : Functions of Two and Three Variables

Step 3 : find = ∙ − ( )
= ∙ − ( ) = (12 )(12 ) − (24)2 = 144 − 576

Step 4 : Table of sign for and

Critical points = 144 − 576 = Conclusion
(0, 0) −576 (−) 0 Saddle point
Minimum point
(4, 4) 1728 (+) 48 (+)

 minimum point : (4, 4)
saddle point : (0, 0)

11. Mar 2014/Q1b/ 11 marks
Given that ( , ) = 8 2 + 2 + 4 + 2, find the critical points and determine the
relative maximum, relative minimum or saddle point.

Solution :

Step 1 : find the all the first and second partial derivatives

= 8 2 + 2 + 4 + 2 = 8 2 + 2 + 4 + 2
= 8 (2 ) + 0 + 4 (1) + 0
= 16 + 4 ( ) = 8 2(1) + 2 + 4 (1) + 0
= 8 2 + 2 + 4

= 16 + 4 = 16 + 4 = 8 2 + 2 + 4 = 8 2 + 2 + 4
= 16 (1) + 0 = 16 (1) + 4(1) = 8(2 ) + 0 + 4(1) = 0 + 2(1) + 0
= 16 = 16 + 4 = 16 + 4 = 2

© Amirah 2022 243

MAT235- CALCULUS II FOR ENGINEERS / Chapter 3 : Functions of Two and Three Variables

Step 2 : find the critical points when = and =

When = 0 When = 0

16 + 4 = 0 8 2 + 2 + 4 = 0 ①
4 + = 0 4 2 + + 2 = 0

(4 + 1) = 0 4 + 1 = 0
= 0,
4 = −1
1

= − 4

Substitute = − 1 into ① : Substitute = 0 into ① :

4 4 2 + 0 + 2 = 0
4 2 + 2 = 0
12 1 2 2 + = 0
4 (− ) + + 2 (− ) = 0 (2 + 1) = 0
1
44 = 0, = − 2

11
4 ( ) + − = 0

16 2

1 + − 1 = 0
4 2
11
= −

24

1
= 4

 critical points are : (− 1 , 1) , (0, 0) and (− 1 , 0)
44 2

Step 3 : find = ∙ − ( )
= ∙ − ( ) = (16 )(2) − (16 + 4)2 = 32 − (16 + 4)2

Step 4 : Table of sign for and

Critical points = − ( + ) = Conclusion
(− 1 , 1) 8 (+) 4 (+) Minimum point
0 Saddle point
44 −16 (−) 0 Saddle point

(0, 0)

(− 1 , 0) −16 (−)

2

 minimum point : (− 1 , 1)

44

saddle points : (0, 0) and (− 1 , 0)

2

© Amirah 2022 244

MAT235- CALCULUS II FOR ENGINEERS / Chapter 3 : Functions of Two and Three Variables

12. Sep 2013/Q1b/ 8½ marks
Find all the critical points of the function ( , ) = 2 4 − 2 + 2 and classify each
point as a relative maximum, relative minimum or saddle point.

Solution :

Step 1 : find the all the first and second partial derivatives

= 2 4 − 2 + 2 = 2 4 − 2 + 2
= 2(4 3) − 2(1) + 2 (1)
= 8 3 − 2 + 2 ( ) = 0 − (2 ) + 2 (1)
= −2 + 2

= 8 3 − 2 + 2 = 8 3 − 2 + 2 = −2 + 2 = −2 + 2
= 8(3 2) − 0 + 0 = 0 − 2 + 2(1) = −2 (1) + 2(1) = −2 (1) + 0
= 24 2 = 2 − 2 = 2 − 2 = −2

Step 2 : find the critical points when = and =

When = 0 When = 0

8 3 − 2 + 2 = 0 ① −2 + 2 = 0
− + = 0
− = 0 1 − = 0
(1 − ) = 0 = 1
= 0,

Substitute = 0 into ① : Substitute = 1 into ① :

8(0)3 − 2 + 2 = 0 8 3 − (1)2 + 2(1) = 0
2 − 2 = 0
(2 − ) = 0 8 3 − 1 + 2 = 0

= 0, 8 3 + 1 = 0

2 − = 0 8 3 = −1
= 2
3 = −1
8
(−1)3
3 = (2)3

3 = −1 3
(2)

3 = (− 13
2)

1
= − 2

 critical points are : (0, 0), (0, 2) and ( − 1 , 1)

2

© Amirah 2022 245

MAT235- CALCULUS II FOR ENGINEERS / Chapter 3 : Functions of Two and Three Variables

Step 3 : find = ∙ − ( )
= ∙ − ( ) = (24 2)(−2 ) − (2 − 2 )2 = −48 3 − (2 − 2 )2

Step 4 : Table of sign for and

Critical points = − − ( − ) = Conclusion
0 Saddle point
(0, 0) −4 (−) 0 Saddle poin
Minimum point
(0, 2) −4 (−) 6 (+)
(− 1 , 1) 6 (+)

2

 minimum point : (− 1 , 1)

2

saddle points : (0, 0) and (0, 2)

13. Mar 2013/Q1b/ 8½ marks
Find all the critical points of the function ( , ) = 6 2 − 2 3 + 3 2 + 6 and classify
each point as a relative maximum, relative minimum or saddle point.

Solution :

Step 1 : find the all the first and second partial derivatives

= 6 2 − 2 3 + 3 2 + 6 = 6 2 − 2 3 + 3 2 + 6
= 6(2 ) − 2(3 2) + 0 + 6 (1)
= 12 − 6 2 + 6 ( ) = 0 − 0 + 3(2 ) + 6 (1)
= 6 + 6

= 12 − 6 2 + 6 = 12 − 6 2 + 6 = 6 + 6 = 6 + 6
= 12(1) − 6(2 ) + 0 = 0 − 0 + 6(1) = 0 + 6(1) = 6(1) + 0
= 12 − 12 = 6 = 6 = 6

© Amirah 2022 246

MAT235- CALCULUS II FOR ENGINEERS / Chapter 3 : Functions of Two and Three Variables

Step 2 : find the critical points when = and =

When = 0 When = 0

12 − 6 2 + 6 = 0 ① 6 + 6 = 0
2 − 2 + = 0 + = 0
= −
②

Substitute ② into ① : = 0 ② ∶ = −(0) = 0
= 1 ② ∶ = −(1) = −1
2 − 2 + (− ) = 0
2 − 2 − = 0
− 2 = 0
(1 − ) = 0
= 0, = 1

 critical points are : (0, 0) and (1, −1)

Step 3 : find = ∙ − ( )
= ∙ − ( ) = (12 − 12 )(6) − (6)2 = 72 − 72 − 36 = 36 − 72

Step 4 : Table of sign for and

Critical points = − = − Conclusion
(0, 0) 36 (+) 12 (+) Minimum point
Saddle point
(1, −1) −36 (−) 0

 minimum point : (0, 0)
saddle point : (1, −1)

© Amirah 2022 247

MAT235- CALCULUS II FOR ENGINEERS / Chapter 3 : Functions of Two and Three Variables

14. Mar 2012/Q2c/ 9 marks
Find all the extremum points of ( , ) = 2 2 − 4 + 4 + 2 and determine the nature
of the points.

Solution :

Step 1 : find the all the first and second partial derivatives

= 2 2 − 4 + 4 + 2 = 2 2 − 4 + 4 + 2
= 2(2 ) − 4 (1) + 0 + 0
= 4 − 4 ( ) = 0 − 4 (1) + 4 3 + 0
= 4 3 − 4

= 4 − 4 = 4 − 4 = 4 3 − 4 = 4 3 − 4
= 4(1) − 0 = 0 − 4(1) = 0 − 4(1) = 4(3 2) − 0
= 4 = −4 = −4 = 12 2

Step 2 : find the critical points when = and =

When = 0 When = 0

4 − 4 = 0 4 3 − 4 = 0 ②
− = 0 3 − = 0
=
①

Substitute ① into ② : = 0 ① ∶ = 0
= −1 ① ∶ = −1
( )3 − = 0 = 1 ① ∶ = 1
3 − = 0

( 2 − 1) = 0
( + 1)( − 1) = 0
= 0, = −1, = 1

 critical points are : (0, 0), (−1, −1) (1, 1)

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MAT235- CALCULUS II FOR ENGINEERS / Chapter 3 : Functions of Two and Three Variables

Step 3 : find = ∙ − ( )
= ∙ − ( ) = (4)(12 2) − (−4)2 = 48 2 − 16

Step 4 : Table of sign for and

Critical points = 48 2 − 16 = Conclusion
+ Saddle point
(0, 0) −16 (−)

(−1, −1) 32 (+) + Minimum point

(1, 1) 32 (+) + Minimum point

 Minimum points : (−1, −1) (1, 1)
saddle point : (0, 0)

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