HW4_Sol - .5 =-15<0 Thus f(x 1 x 2 x 3 is concave on R 3...

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E160 Operations Research I Spring 2009 Homework #4 Solution Chapter 12 Section 12.3 1. f''(x) = 6x 0, (for x 0) so f(x) is convex on S. 3. f''(x) = 2x -3 >0 (for x>0). Thus f(x) is convex on S. 5. f''(x) = -x -2 <0, so f(x) is a concave function on S. 7. f(x 1 , x 2 ) is the sum of convex functions and is therefore a convex function. 9. 4 0 0 0 2 5 . 0 5 . 2 - - - = H 1st Order PM's are -2, -2, -4 which are all <0. 2nd Order PM's are all >0 75 . 3 2 5 . 5 . 2 det = - - 8 4 0 0 2 det = - - 8 4 0 0 2 det = - - Expanding by Row 3 we find 3rd Order PM = -4((-2)(-2)-(.5)
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Unformatted text preview: (.5)) = -15<0. Thus f(x 1 , x 2 , x 3 ) is concave on R 3 . 12. We know that (1) f(cx + (1 - c)y) ≤ cf(x) + (1 - c)f(y) (2) g(cx + (1 - c)y) ≤ cg(x) + (1 - c)g(y) Adding (1) and (2) yields h(cx + (1 - c)y) ≤ ch(x) + (1 - c)h(y) which shows that h is also a convex function. 13. Since f is convex we know that for 0 ≤ k ≤ 1 (1) f(kx + (1 - k)y) ≤ kf(x) + (1 - k)f(y). Multiplying both sides of (1) by c ≥ 0 shows that g is also a convex function....
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HW4_Sol - .5 =-15<0 Thus f(x 1 x 2 x 3 is concave on R 3...

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