Problem Sheet 9 with Solutions GRA 6035 Mathematics

4⎧⎪⎨ g 1 (x 1 ,...,x n ) ≤ b 1max f (x 1 ,...,x n ) subject to..⎪⎩g m (x 1 ,...,x n ) ≤ b mby applying the following steps:a) L = f − λ 1 g 1 − ··· − λ m g mb) L 1 ′ = 0,L 2 ′ = 0,...,L n ′ = 0 (FOC’s)c) λ j ≥ 0 and λ j = 0 if g j (x 1 ,...,x n ) < b jd) Require g j (x 1 ,...,x n ) ≤ b jTo transform the problem into this setting, we definef (x,y) = −(4ln(x 2 + 2) + y 2 )since minimizing 4ln(x 2 + 2) + y 2 is the same as maximizing −(4ln(x 2 + 2) + y 2 ).We also rewrite the constraints asWe define the Lagrange function:g 1 (x,y) = −x 2 − y ≤ −2g 2 (x,y) = −x ≤ −1L = −(4ln(x 2 + 2) + y 2 ) − λ 1 (−x 2 − y) − λ 2 (−x)The first order conditions are the= −4ln(x 2 + 2) − y 2 + λ 1 (x 2 + y) + λ 2 xL 1 ′ 1= −4x 2 + 2 · 2x + 2λ 1x + λ 2 = −8xx 2 + 2 + 2λ 1x + λ 2 = 0L 2 ′ = −2y + λ 1 = 0Since there are two constraints, there are four cases to consider:The case −x 2 − y = −2 and −x = −1 :Since x = 1, we deduce from L ′1 = 0 that−8 · 11 2 + 2 + 2λ 1 · 1 + λ 2 = 0 ⇐⇒ 2λ 1 + λ 2 − 8 3 = 0From x 2 + y = 2 and x = 1 we obtain that y = 1. From L 2 ′ = −2y + λ 1 = 0 we theobtain that λ 1 = 2. Substituting this into 2λ 1 + λ 2 − 8 3= 0 we get2 · 2 + λ 2 − 8 3 = 0 ⇐⇒ λ 2 = − 4 3 < 0