Optimum Design
153
One of the advantages of Lagrange’s method is that it does not require
one to make a choice of independent variables. This is sometimes important
in a complex problem. The Lagrange multipliers are often used to verify
Kuhn-Tucker necessary conditions [3.22,3.25] for more complex computer
optimization. The conditions are incorporated into the computer program
and most users are unaware of them. The user’s main concern is how to
formulate the criterion function, function constraints, and the regional constraints that are bounded to obtain a reasonable solution. Also, are all
functions continuous and in a form a computer will work with?
A simple Lagrangian multiplier example is presented.
EXAMPLE 3.3.13.221.
volume which can be filled inside an ellipsoid;
x 2 y2
22
a2 ~-~5+~5= 1
The sides of the box are to be
2x, 2y, 2z (for the sake of symmetry)
so the criterion function C is
V = 8xyz
and the functional constraint is
X 2 y2 z2
F1 = ~5+~5+~-- 1 =0
Using Eq. (3.8) there is only one 2 yielding
8c
OFl
2x
o-~ + ,~1 ~x = 8yz + ,h ~ = 0
Oc
OF1
2y
oy ~- ~tl -@--y = 8x + ;~l ~ = 0
Now
Find the dimensions of the box of largest
(3.9)
(3.1o)
(3.11)
(3.12)
(3.13)
Oc
OF1
21 2z
O~-21~z=8xy-~
-
C2
=0
(3.14)
divide by 2 and multiply in order the Eqs. (3.12)-(3.14) by x, y,
X 2
4xyz + 21 ~ = 0
(3.15)
y2
4xyz + 21~ = 0
(3.16)