20 50 b 0 for critical points we have

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Unformatted text preview: b) ( )= + + > 0. For critical points, we have , where so = 50 2 and = 20 2 = 50 − = 2 =0 = − 2 4 20 (50/ 2 )2 = 125 20 3 . Now we get =0 = 125, that is, = 5 and = 2. Further, 100 H= 3 1 1 4 5 H (5 2) = 40 1 15 3 and det H = 3 > 0, tr H > 0. Hence (5 2) is a local minimum. (c) ( )= = ln( ln( 2 2 + + 2 2 ). For critical points, we have )+ 2 2 + From = 0, we see that either two cases separately. 2 =0 = 0 or ln( = 2 + ln( 2 )+ 2 + 2 2 )+ 2 2 + 2 =0 2 2+ 2 = 0. Let’s consider these First, if = 0, then in the second equation we have 2 ln | | = 0, so either = 0 or = ±1. But if = = 0, then the value ( ) is not defined, so we must have = ±1 and = 0. Now let’s look at the other equation. Again, we either have = 0 or ln( 2 + 2 ) + 22 = 0 can be done in the same manner as = 0 above. In 2 + 2 = 0. The case fact, the original function is symmetric as we have ( )= ( ) and hence by symmetry we’ll get two critical points = 0, = ±1. Finally, the last possible option is ln( 2 + 2 )+ 22 2+ 2 = ln( 2 2 + 2 )+ 22 2+ 2 =0 which implies 2 = 2 , or = ± . It then implies ln(2 2 ) + 1 = 0, that is, 2 2 = and = ± √...
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