Engineering Optimization: Theory and Practice, Fourth Edition

8.4 Solution Using Differential Calculus 499

It can be seen that the minimum total cost has been obtained before finding the
optimal size of the box. To find the optimal values of the design variables, let us write
Eqs. (8.14) as

U 1 ∗= 08 x 1 ∗x 2 ∗=∗ 1 f∗=

1

5

( 200 )= 40 (E 8 )

U 2 ∗= 04 x 2 ∗x 3 ∗=∗ 2 f∗=

1

5

( 200 )= 40 (E 9 )

U 3 ∗= 02 x 1 ∗x 3 ∗=∗ 3 f∗=

1

5

( 200 )= 40 (E 10 )

U 4 ∗=

80

x 1 ∗x 2 ∗x 3 ∗

=∗ 4 f∗=

2

5

( 200 )= 80 (E 11 )

From these equations, we obtain

x 2 ∗=^12

1

x∗ 1

=

1

x∗ 3

, x 1 ∗=

x 3 ∗

2

, x∗ 2 =

1

x∗ 3

1

x∗ 1 x∗ 2 x∗ 3

= 1 =

2 x 3 ∗

x∗ 3 x∗ 3

, x 3 ∗= 2

Therefore,

x 1 ∗ m= 1 , x∗ 2 =^12 m , x 3 ∗= m 2 (E 12 )

Itis to be noticed that there is one redundant equation among Eqs. (E 8 ) o (Et 11 ) which,
is not needed for the solution ofx∗i ( i= 1 ton).
The solution given in Eq. (E 12 ) canalso be obtained using Eqs. (8.18). In the
present case, Eqs. (8.18) lead to

1 w 1 + 1 w 2 + 0 w 3 = nl

200 ×^15

80

=ln

1

2

(E 13 )

0 w 1 + 1 w 2 + 1 w 3 = nl

200 ×^15

40

=ln 1 (E 14 )

1 w 1 + 0 w 2 + 1 w 3 = nl

200 ×^15

20

=ln 2 (E 15 )

− 1 w 1 − 1 w 2 − 1 w 3 = nl

200 ×^25

80

=ln 1 (E 16 )

By adding Eqs. (E 13 ) (E, 14 ) and (E, 16 ) we obtain,

w 2 = nl^12 n 1+l +ln 1=ln(^12 · 1 · 1 )=ln^12 = nl x 2 ∗

or
x 2 ∗=^12

Similarly, by adding Eqs. (E 13 ) (E, 15 ) and (E, 16 ) we get,

w 1 = nl^12 + n 2l +ln 1=ln 1=lnx∗ 1