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Module-6: Nonlinear Programming Problems

Syllabus:

NLPP with One Equality Constraint (Two or Three Variables) using the Method of
Lagrange’s Multipliers. NLPP with Two Equality Constraints. NLPP with Inequality
Constraint: Kuhn-Tucker Conditions.

Type-I: Nonlinear Programming Problems with no constraints

1. Find the relative maximum or minimum of the function

(i) 𝑧 = 𝑥1 2 + 𝑥2 2 + 𝑥3 2 − 4𝑥1 − 8𝑥2 − 12𝑥3 + 100

[𝐴𝑛𝑠: 𝑧 𝑖𝑠 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑎𝑡 (2,4,6), 𝑧𝑚𝑖𝑛 = 44]

(ii) 𝑧 = 𝑥1 2 + 𝑥2 2 + 𝑥3 2 − 6𝑥1 − 8𝑥2 − 10𝑥3

[𝐴𝑛𝑠: 𝑧 𝑖𝑠 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑎𝑡 (3,4,5), 𝑧𝑚𝑖𝑛 = −50 ]

(iii) 𝑧 = −𝑥1 2 - 𝑥2 2 - 𝑥3 2 + 𝑥1 + 2𝑥3 + 𝑥2 𝑥3

1 2 4 19
[𝐴𝑛𝑠: 𝑧 𝑖𝑠 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑎𝑡 ( , , ) , 𝑧𝑚𝑎𝑥 = ]
2 3 3 12

(iv) 𝑧 = 𝑥1 2 + 𝑥2 2 + 𝑥3 2 − 6𝑥1 − 8𝑥2 − 10𝑥3

[𝐴𝑛𝑠: 𝑧 𝑖𝑠 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑎𝑡 (3,4,5), 𝑧𝑚𝑖𝑛 = −50 ]

(v) 𝑧 = 2𝑥1 + 𝑥3 + 3𝑥2 𝑥3 −𝑥1 2 - 3𝑥2 2 - 3𝑥3 2 + 17

1 2
[𝐴𝑛𝑠: 𝑧 𝑖𝑠 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑎𝑡 (1, , ) , 𝑧𝑚𝑎𝑥 = 18 ]
9 9

Type-II: Nonlinear Programming Problems with equality constraints

A. Objective function is function of two variables and with one equality

constraint
Optimal condition: (i) If ∆𝟑 is positive z has maximum value at
stationary point.
(ii) If ∆𝟑 is negative z has minimum value at stationary point.

Problems:

2. Using Lagrange’s multipliers, solve the following NLPP

(i) Optimise 𝑧 = 4𝑥1 + 6𝑥2 − 2𝑥1 𝑥2 −2𝑥1 2 - 2𝑥2 2

Subject to 𝑥1 + 2𝑥2 = 2 , 𝑥1 , 𝑥2 ≥ 0.
1 5 77
[Ans: At (3 , 6) , 𝑧𝑚𝑎𝑥 = 18 ]
(ii) Optimise 𝑧 = 5𝑥1 + 𝑥2 − 2𝑥1 𝑥2 −𝑥1 2 - 𝑥2 2
Subject to 𝑥1 + 𝑥2 = 4 , 𝑥1 , 𝑥2 ≥ 0.
5 3
[Ans: At (2 , 2) , 𝑧𝑚𝑎𝑥 = 13 ]

Prof. Divesh Singh KJSIT

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(iii) Optimise 𝑧 = 7𝑥1 2 +5 𝑥2 2

Subject to 2𝑥1 + 5𝑥2 = 7 , 𝑥1 , 𝑥2 ≥ 0.
14 49 343
[Ans: At (39 , 39) , 𝑧𝑚𝑖𝑛 = 39
]
(iv) Optimise 𝑧 = 6𝑥1 +5 𝑥22 2

Subject to 𝑥1 + 5𝑥2 = 3 , 𝑥1 , 𝑥2 ≥ 0.
3 18 54
[Ans: At (31 , 31) , 𝑧𝑚𝑎𝑥 = 31 ]
(v) Optimise 𝑧 = 4𝑥1 + 8𝑥2 −𝑥1 2 - 𝑥2 2
Subject to 𝑥1 + 𝑥2 = 2 , 𝑥1 , 𝑥2 ≥ 0.
[Ans: At (0,2) , 𝑧𝑚𝑎𝑥 = 12 ]
(vi) Optimise 𝑧 = 6𝑥1 + 2𝑥2 + 2𝑥1 𝑥2 +3𝑥1 2 + 𝑥2 2
Subject to 2𝑥1 + 𝑥2 = 4 , 𝑥1 , 𝑥2 ≥ 0.
[Ans: At (1,2) , 𝑧𝑚𝑖𝑛 = 28 ]
(vii) Optimise 𝑧 = 6𝑥1 2 +5 𝑥2 2
Subject to 𝑥1 + 5𝑥2 = 11 , 𝑥1 , 𝑥2 ≥ 0.
11 66 726
[Ans: At (31 , 31) , 𝑧𝑚𝑎𝑥 = 31
]
(viii) Optimise 𝑧 = 2𝑥1 + 6𝑥2 −𝑥1 2 - 𝑥2 2
+ 14
Subject to 𝑥1 + 𝑥2 = 4 , 𝑥1 , 𝑥2 ≥ 0.
[Ans: At (1,3) , 𝑧𝑚𝑎𝑥 = 24 ]
(ix) Optimise 𝑧 = 4𝑥1 + 2𝑥2 +3𝑥1 2 +2𝑥2 2
Subject to 3 𝑥1 + 5𝑥2 = 11 , 𝑥1 , 𝑥2 ≥ 0.
1 41
[Ans: At ( , 2) , 𝑧𝑚𝑖𝑛 = ]
3 3
(x) Optimise 𝑧 = 6𝑥1 2 +5 𝑥2 2
Subject to 𝑥1 + 5𝑥2 = 7 , 𝑥1 , 𝑥2 ≥ 0.
7 42 294
[Ans: At ( , ) , 𝑧𝑚𝑎𝑥 = ]
31 31 31
(xi) Optimise 𝑧 = 4𝑥1 + 8𝑥2 −𝑥1 - 𝑥2 2 Subject
2
to 𝑥1 + 𝑥2 = 4 , 𝑥1 , 𝑥2 ≥ 0.
[Ans: At (1,3) , 𝑧𝑚𝑎𝑥 = 18 ]

B. Objective function is function of three variables and with one equality

constraint
Optimal condition: (i) If ∆𝟑 𝒂𝒏𝒅 ∆𝟒 are having different sign z has
maximum value at stationary point.
(ii) ) If ∆𝟑 𝒂𝒏𝒅 ∆𝟒 are having same sign z has minimum value at
stationary point.

Problems:

3. Using Lagrange’s multipliers, solve the following NLPP

(i) Optimise 𝑧 = 2𝑥1 2 + 2𝑥2 2 +2𝑥3 2 − 24𝑥1 − 8𝑥2 − 12𝑥3 + 196

Subject to 𝑥1 + 𝑥2 + 𝑥3 = 11 , 𝑥1 , 𝑥2, 𝑥3 ≥ 0.
[Ans: At (6,2,3) , 𝑧𝑚𝑖𝑛 = 98 ]
(ii) Optimise 𝑧 = 𝑥1 2 + 𝑥2 2 +𝑥3 2 Subject to 4𝑥1 + 𝑥2 2 + 2𝑥3 − 14 = 0
, 𝑥1 , 𝑥2, 𝑥3 ≥ 0. [Ans: At (2,2,1) , 𝑧𝑚𝑖𝑛 = 9 ]

Prof. Divesh Singh KJSIT

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(iii) Optimise 𝑧 = 3𝑥1 2 + 𝑥2 2 +𝑥3 2 . Subject to 𝑥1 + 𝑥2 + 𝑥3 = 2 ,

, 𝑥1 , 𝑥2, 𝑥3 ≥ 0.
[Ans: At (0.81,0.35,0.28) , 𝑧𝑚𝑖𝑛 = 0.84 ]
(iv) Optimise 𝑧 = 𝑥1 2 + 𝑥2 2 +𝑥3 2 − 10𝑥1 − 6𝑥2 − 4𝑥3 . Subject to 𝑥1 +
𝑥2 + 𝑥3 = 7 , , 𝑥1 , 𝑥2, 𝑥3 ≥ 0.
[Ans: At (4,2,1) , 𝑧𝑚𝑖𝑛 = −35 ]
(v) Optimise 𝑧 = 2𝑥1 2 +2 𝑥2 2 +2𝑥3 2 − 24𝑥1 − 8𝑥2 − 12𝑥3 +260. Subject
to 𝑥1 + 𝑥2 + 𝑥3 = 11 , , 𝑥1 , 𝑥2, 𝑥3 ≥ 0.
[Ans: At (6,2,3) , 𝑧𝑚𝑖𝑛 = 162 ]
(vi) Optimise 𝑧 = 2𝑥1 2 +2 𝑥2 2 +3𝑥3 2 + 10𝑥1 + 8𝑥2 + 6𝑥3 -100. Subject to
𝑥1 + 𝑥2 + 𝑥3 = 20 , , 𝑥1 , 𝑥2, 𝑥3 ≥ 0.
[Ans: At (5,3,2) , 𝑧𝑚𝑎𝑥 = 35 ]

C. Objective function is function of two variables and with two equality

constraint
𝜕2 𝑧 𝜕2 𝑧
2
𝜕2 𝑧 𝜕𝑥1 𝜕𝑥1 𝜕𝑥2
Optimal condition: (i) If and | | have same sign then
𝜕𝑥1 2 𝜕2 𝑧 𝜕2 𝑧
2
𝜕𝑥1 𝜕𝑥2 𝜕𝑥2
z has maximum value at stationary point.
𝜕2 𝑧 𝜕2 𝑧
2
𝜕2 𝑧 𝜕𝑥1 𝜕𝑥1 𝜕𝑥2
(ii) If and | | have opposite sign then z has minimum
𝜕𝑥1 2 𝜕2 𝑧 𝜕2 𝑧
2
𝜕𝑥1 𝜕𝑥2 𝜕𝑥2
value at stationary point.

Problems:

4. Using Lagrange’s multipliers, solve the following NLPP

(i) Optimise 𝑧 = 6𝑥1 + 8𝑥2 −𝑥1 2 - 𝑥2 2 + 14
Subject to 4𝑥1 + 3𝑥2 = 16 ,3𝑥1 + 5𝑥2 = 15 𝑥1 , 𝑥2 ≥ 0.
35 12
[Ans: At (11 , 11) , 𝑧𝑚𝑎𝑥 = 16.504 ]
(ii) Optimise 𝑧 = 4𝑥1 + 9𝑥2 −𝑥1 2 - 𝑥2 2
Subject to 4𝑥1 + 3𝑥2 = 15 ,3𝑥1 + 5𝑥2 = 14 𝑥1 , 𝑥2 ≥ 0.
[Ans: At (3,1) , 𝑧𝑚𝑎𝑥 = 11 ]
(iii) Optimise 𝑧 = 8𝑥1 + 9𝑥2 −𝑥1 2 - 𝑥2 2
Subject to 2𝑥1 + 3𝑥2 = 12 ,5𝑥1 + 2𝑥2 = 19 𝑥1 , 𝑥2 ≥ 0.
[Ans: At (3,2) , 𝑧𝑚𝑎𝑥 = 29 ]

Prof. Divesh Singh KJSIT

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D. Objective function is function of three variables and with two equality

constraint
Optimal condition: (i) If minor of order 5 has negative sign then z has
maximum value at stationary point.
(ii) If minor of order 5 has positive sign then z has minimum value at
stationary point.

Problems:

5. Using Lagrange’s multipliers, solve the following NLPP

(i) Optimise 𝑧 = 4𝑥1 2 - 𝑥2 2 − 𝑥3 2 − 4𝑥1 𝑥2 . Subject to 𝑥1 + 𝑥2 + 𝑥3 = 15 ,
2𝑥1 − 𝑥2 + 2𝑥3 = 20, 𝑥1 , 𝑥2, 𝑥3 ≥ 0.
[Ans: At (5.95,3.33,5.71) , 𝑧𝑚𝑖𝑛 = 83.87 ]

(ii) Optimise 𝑧 = 4𝑥1 2 +2 𝑥2 2 + 𝑥3 2 − 4𝑥1 𝑥2 . Subject to 𝑥1 + 𝑥2 + 𝑥3 = 15 ,

2𝑥1 − 𝑥2 + 2𝑥3 = 20, 𝑥1 , 𝑥2, 𝑥3 ≥ 0.
11 10 820
[Ans: At ( 3 , 3
, 8) , 𝑧𝑚𝑖𝑛 = 9 ]
(iii) Optimise 𝑧 = 𝑥1 2 + 𝑥2 2 + 𝑥3 2 . Subject to 𝑥1 + 𝑥2 + 3𝑥3 = 2 , 5𝑥1 + 2𝑥2 +
𝑥3 = 5, 𝑥1 , 𝑥2, 𝑥3 ≥ 0.
[Ans: At (. 804, .348, .283) , 𝑧𝑚𝑖𝑛 = 0.8476 ]
(iv) Optimise 𝑧 = 2𝑥1 2 +3 𝑥2 2 + 𝑥3 2 . Subject to 𝑥1 + 𝑥2 + 2𝑥3 = 13 , 2𝑥1 +
𝑥2 + 𝑥3 = 10, 𝑥1 , 𝑥2, 𝑥3 ≥ 0.
[Ans: At (2,1,5) , 𝑧𝑚𝑖𝑛 = 36 ]

Type-II: Nonlinear Programming Problems with inequality constraints

A. Problems with one inequality constraint

Problems
6. Solve the following NLPP
(i) Maximize 𝑧 = 2𝑥1 2 − 7𝑥2 2 + 12𝑥1 𝑥2, subject to 2𝑥1 + 5𝑥2 ≤ 98
, 𝑥1 , 𝑥2 ≥ 0. [Ans: At (44,2) , 𝑧𝑚𝑎𝑥 = 4900]
(ii) Maximize 𝑧 = 10𝑥1 + 4𝑥2 − 2𝑥1 2 − 𝑥2 2 , subject to 2𝑥1 + 𝑥2 ≤ 5
11 4 91
, 𝑥1 , 𝑥2 ≥ 0. [Ans: At ( 6 , 3) , 𝑧𝑚𝑎𝑥 = 6
]
(iii) Maximize 𝑧 = 8𝑥1 + 10𝑥2 − 𝑥1 2 − 𝑥2 2 , subject to 3𝑥1 + 2𝑥2 ≤ 6
4 33 277
, 𝑥1 , 𝑥2 ≥ 0. [Ans: At (13 , 13) , 𝑧𝑚𝑎𝑥 = 13 ]
(iv) Maximize 𝑧 = 16𝑥1 + 6𝑥2 − 2𝑥1 2 − 𝑥2 2 − 17, subject to 3𝑥1 +
2𝑥2 ≤ 6
, 𝑥1 , 𝑥2 ≥ 0. [Ans: At (3,2) , 𝑧𝑚𝑎𝑥 = 21]

And more similar type of problems

B. Problems with two inequality constraints
7. Solve the following NLPP

Prof. Divesh Singh KJSIT

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(i) Use the Kunh-Tucker conditions to solve the following NLPP

Maximise = 7𝑥1 2 + 5 𝑥2 2 + 6𝑥1 , subject to 𝑥1 + 2𝑥2 ≤ 10
48 1
𝑥1 − 3𝑥2 ≤ 9, 𝑥1 , 𝑥2 ≥ 0. [Ans: At ( 5 , 5) , 𝑧𝑚𝑖𝑛 = 702.92]
(ii) Use the Kunh-Tucker conditions to solve the following NLPP
Minimise = 7𝑥1 2 + 5 𝑥2 2 − 6𝑥1 , subject to 𝑥1 + 2𝑥2 ≤ 10
3 9
𝑥1 + 3𝑥2 ≤ 9, 𝑥1 , 𝑥2 ≥ 0. [Ans: At ( , 0) , 𝑧𝑚𝑖𝑛 = − ]
7 7

And more similar type of problems

Prof. Divesh Singh KJSIT