Engineering Optimization: Theory and Practice, Fourth Edition

498 Geometric Programming

Figure 8.1 Open rectangular box.

that is,

 1 + 3 − 4 = 0 (E 2 )

 1 + 2 − 4 = 0 (E 3 )

 2 + 3 − 4 = 0 (E 4 )

 1 + 2 + 3 + 4 = 1 (E 5 )

From Eqs. (E 2 ) nd (Ea 3 ) we obtain,

 4 = 1 + 3 = 1 + 2 or  2 = 3 (E 6 )

Similarly, Eqs. (E 3 ) nd (Ea 4 ) ive usg

 4 = 1 + 2 = 2 + 3 or  1 = 3 (E 7 )

Equations(E 6 ) nd (Ea 7 ) ieldy

 1 = 2 = 3

while Eq. (E 6 ) ivesg

 4 = 1 + 3 = 2  1

Finally, Eq. (E 5 ) eads to the unique solutionl

∗ 1 =∗ 2 =∗ 3 =^15 and ∗ 4 =^25

Thus the optimal value of the objective function can be found from Eq. (8.13) as

f∗=

(

80

1 / 5

) 1 / 5 (

40

1 / 5

) 1 / 5 (

20

1 / 5

) 1 / 5 (

80

2 / 5

) 2 / 5

= ( 4 × 102 )^1 /^5 ( 2 × 102 )^1 /^5 ( 1 × 102 )^1 /^5 ( 4 × 104 )^1 /^5

= 2 ( 3 × 1010 )^1 /^5 = 200 $