Principles of Mathematics in Operations Research

Solutions 233

Let us check the boundary denned by xj\

/(0, X 2 ) = l-x\ - X 2 + 19 =» ^^^ = a* - 1 = 0 => Z2 = 1.

Since —£.? = 1 > 0, x 2 = 1 > 0 is the local minimizer outside the feasible

region. As the first derivative is negative for — 4 < x 2 < 0, we will check x 2 = 0

for minimizer and x 2 = —4 for maximizer (see Figure S.5).

Let us check the boundary denned by xn'-

Since —{> f^2 ' = 1 > 0, x 2 = — 5 < —4 is the local minimizer outside the

feasible region. As the first derivative is positive for — 4 < x 2 < 0, we will

check x 2 = —4 for minimizer and £2 = 0 for maximizer (see Figure S.5).

Let us check the boundary defined by xju:

f(Xl,0) = ±x* + \x\ + 19 =* ^-ft =x^21 +x 1 =0=>x 1 =0, -1.

Since d2f£i'0) = 2xx + 1, a;i = 0 is the local minimizer (d2^°s'0) = 1 > 0)

on the boundary, and x\ = —1 is the local maximizer (—^~ 2 ' ' = — 1 < 0)

outside the feasible region. As the first derivative is positive for 0 < x 2 < 3,

we will check x 2 — 3 for maximizer (see Figure S.5).

Let us check the boundary denned by x/y:

/(an, -4) = \x\ + \x\ - 8Xl + 31 => rf/(2'~4) =^ + ^1-8 = 0

^ -l±v/l + 32

Xl = _ _.

Since c'2/(jcxV~4) = 2xx + 1 again, the positive root xi = =i±sM = 2.3723
is the local minimizer (— dxi— > 0), and the negative root is the local
maximizer but it is outside the feasible region. As the first derivative is positive
for 0 < x 2 < 3, we will check x 2 = 3 for maximizer again (see Figure S.5).
To sum up, we have to consider (2,-3), (0,0) and (2.3723,-4) for the
minimizer; (3,0) and (0, —4) for the maximizer:

/(2,-3) = 19.16667, /(0,0) = 19, /(2.3723,-4) = 19.28529

=> (0,0) is the minimizer!

/(3,0) = 32.5, /(0, -4) = 31 => (3,0) is the maximizer!