2.4 Multivariable Optimization with Equality Constraints 77
The necessary conditions for the maximum off give
∂f
∂x 1
= 8 x 2
[
( 1 −x 12 −x^22 )^1 /^2 −
x 12
( 1 −x 12 −x 22 )^1 /^2
]
= 0 (E 5 )
∂f
∂x 2
= 8 x 1
[
( 1 −x 12 −x^22 )^1 /^2 −
x 22
( 1 −x 12 −x 22 )^1 /^2
]
= 0 (E 6 )
Equations(E 5 ) nda (E 6 ) an be simplified to obtainc
1 − 2 x 12 −x 22 = 0
1 −x^21 − 2 x 22 = 0
from which it follows thatx∗ 1 =x∗ 2 = 1 /
√
3 and hencex 3 ∗= 1 /
√
- This solution gives
the maximum volume of the box as
fmax=
8
3
√
3
To find whether the solution found corresponds to a maximum or a minimum,
we apply the sufficiency conditions tof (x 1 , x 2 ) f Eq.o (E 4 ) The second-order partial.
derivatives off at(x 1 ∗, x∗ 2 ) re given bya
∂^2 f
∂x 12
= −
32
√
3
at(x∗ 1 , x 2 ∗)
∂^2 f
∂x 22
= −
32
√
3
at(x∗ 1 , x 2 ∗)
∂^2 f
∂x 1 ∂x 2
= −
16
√
3
at(x∗ 1 , x 2 ∗)
Since
∂^2 f
∂x 12
< 0 and
∂^2 f
∂x 12
∂^2 f
∂x^22
−
(
∂^2 f
∂x 1 ∂x 2
) 2
> 0
the Hessian matrix off is negative definite at(x∗ 1 , x 2 ∗) Hence the point. (x 1 ∗, x 2 ∗)
corresponds to the maximum off.
2.4.2 Solution by the Method of Constrained Variation
The basic idea used in the method of constrained variation is to find a closed-form
expression for the first-order differential off (df)at all points at which the constraints
gj( X)= 0 ,j= 1 , 2 ,... , m, are satisfied. The desired optimum points are then obtained
by setting the differentialdfequal to zero. Before presenting the general method,
