Question

To be answered in full

lc min S. ubject to

Answer 1

(a) i) min (1 + x/2 x) , x greaterthanorequalto 1 1 + x/2 x = 1/2 x + 1/2 = 1/2 + 1/2 x so minimum will attained at x rightarrow infinity lim_x rightarrow infinity (1/2 + 1/2 x) = 1/2 = min (1 + x/2 x) (ii) Weierstrass them is not satisfied: since it requires bounded , closed interval . but we have [1 , infinity) interval unbounded . (iii) function is continuous for x > 1 f(x) is strictly decreasing & bounded below , so it will attain its minimum value (b) min . f(x) = {1/x , x < 2 1 , x = 2 (i) No . minimum is not attained solution does not exist . (ii) Weierstrass them is not valid , since f(x) is not continuous this time (iii) Remark . function is continuous in [1 , 2) open at 2 so , since f(x) is strictly decreasing function . if we assume it attains minimum at let say x elementof [1 , 2)