Math 164/2 Comments:
- Reminder: Final Exam on Monday, March 20, time 3-6pm, room MS 5137.
- Sample problems for the final with solutions are posted on the class webpage. - Office
Hours with the instructor before the final: Thursday 5pm, Friday 2-3pm (and at 5pm),
Sunday 12-2pm (please note that I will not be able to hold regular office hour on Wednesday),
and probably Monday before the final.
- Last homework due on Friday or on Monday.
- When you apply the Complementary Slackness Thm. 6.3, page 154, Chapter 6, make
sure that the primal problem is in standard form.
- Problems 6 and 8 from the sample final problems are examples from Section 14.5.
Hints for the last homework #9:
Problem [2] I want you to use the general method (Lemmas 14.1 and 14.2), using a
basis matrix Zof the null-space of A(and not the more “naive” method where we could
express x3= 3 −(2x1+ 5x2+x3) to obtain an unconstrained problem; this method works
in this case too, but I want you to learn the general method introduced in Section 14.2).
Recall that xis a stationary point for problem [2] if Ax =band if ZT∇f(x) = ~
0 (or
equivalently, if Ax =band if ∇f(x) = ATλfor some vector λ). Here A= [2 5 1] ∈ M(1×3),
AT∈ M(3 ×1) and λis a scalar (or vector with just one component).
Problem [3] You do not need to change the maximization problem into a minimization
problem. You have to find local maximizers. Apply Lemma 14.2, page 431, with “negative
definite” instead of “positive definite”. First you need to find all stationary points, then see
which ones are local maximizers, if any.
Here A= [ 1
a1
1
a2
1
a3]∈ M(1 ×3) and b= 1. To find the stationary points first, find Z
as a 3 ×2 matrix, then find all x= (x1, x2, x3)Tsuch that
(x1
a1+x2
a2+x3
a3= 1
ZT∇f(x) = ~
0,(1)
System (1) has 3 equations and 3 unknows x1, x2, x3. You will show that all possible so-
lutions of the system are x= ( a1
3,a2
3,a3
3)T,x= (0,0, a3)T,x= (0, a2,0)Tand x= (a1,0,0)T.
For each of the 4 points found, compute ZT∇2f(x)Z(a 2 ×2 matrix) and see if it is
negative definite. Since the problem has some symmetry, it is sufficient to test for the first
point and for one of the last 3 points only.
You will obtain that the first point (a1
3,a2
3,a3
3)Tis a local maximizer (the 2 ×2 matrix
ZT∇2f(x)Zcan be transformed into an upper triangular matrix with all entries on the
diagonal negative, so ZT∇2f(x)Zwill be negative definite). The other 3 points are not
local maximizers (the corresponding 2 ×2 matrices ZT∇2f(x)Zhave one eigenvalue strictly
positive and one eigenvalue strictly negative, therefore ZT∇2f(x)Zis not negative semi-
definite, so the 2nd order necessary condition for local maximizer is not satisfied).
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