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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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 Spring '13

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