[1]. (1) xa= (1
2,1
2)T: feasible;
•interior of the 1st inequality constraint (a) (∵1−1
4−1
4>0),
•boundary of the 2nd inequality constraint (b) (∵1−1
2−1
2= 0),
•interior of the 3rd inequality constraint (c) (∵1
2>0)
Thus, since at least one of the inequality constraints is active at xa, then xais
a boundary point of the feasible region.
(2) xe= ( 1
√2,1
√2)T: infeasible (∵1−1
√2−1
√2<0 on the 2nd constraint (b))
[2] Local minimizers : (−2,0), (1, x2) with −√3≤x2≤√3.
Global minimizers : (−2,0).
[3.a.]. Show that S1is a convex set:
Choose x= (x1, x2), y = (y1, y2)∈S1;x1+x2≤1, x1≥0; y1 + y2≤1, y1≥0.
We need to show that z=αx + (1 −α)y∈S1for all α∈[0, 1].
Let z1=αx1+ (1 −α)y1, and z2=αx2+ (1 −α)y2i.e. z= (z1, z2).
z1+z2=αx1+ (1 −α)y1+αx2+ (1 −α)y2)
=α(x1+x2) + (1 −α)(y1+y2)
≤α·1 + (1 −α)·1
= 1
And, z1=αx1+ (1 −α)y1≥α·0 + (1 −α)·0 = 0.
Since z1+z2≤1 and z1≥0, then z= (z1, z2)∈S1. Thus, S1is convex.
[3.b.]. Show that S1SS2is NOT a convex set:
First, draw the region S=S1SS2to find the proper points x, y ∈Ss.t.
z=αx + (1 −α)y /∈Sfor some α∈(0,1).
Choose x= (0,1), y= (1,1), and choose α=1
2.
Then x∈S1(∵0+1≤1,0≥0), y∈S2(∵1−1≥0,1≤1), thus x, y ∈S.
And, we have z= (1
2,1). But, since 1
2+ 1 1, then z /∈S1. Similarly, since
1
2−10, then z /∈S2. Therefore, z /∈S= (S1SS2). Hence, we can conclude
that Sis not a convex set.
[4]. h(x) = g(f(x)) is a convex function.
Let x, y ∈Rn, and α∈[0,1].
Since fis a convex function, then f(αx + (1 −α)y)≤αf(x) + (1 −α)f(y).
Since gis nondecreasing function (i.e. if x≤y, then g(x)≤g(y)), then
g(f(αx + (1 −α)y)) ≤g(αf(x) + (1 −α)f(y)) (1)
Since gis a convex function, then
g(αf(x) + (1 −α)f(y)) ≤αg(f(x)) + (1 −α)g(f(y)) (2)
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