### Recent Posts

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##### Quiz 3 / Quiz 3 B
« Last post by ZiqiWang on May 10, 2022, 04:54:19 PM »
The following is my sample answer for quiz 3.
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##### Quiz 2 / Quiz 2 B
« Last post by ZiqiWang on May 10, 2022, 04:51:22 PM »
The following is my sample answer for quiz 2.
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##### Quiz 2 / Quiz 2 D
« Last post by ZiqiWang on May 01, 2022, 03:57:40 PM »
The following is my answer for quiz 2.
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##### Quiz 1 / Quiz1 C
« Last post by ZiqiWang on May 01, 2022, 03:52:18 PM »
This is my answer for quiz 1.
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##### Quiz 6 / Quiz 6
« Last post by ZiqiWang on May 01, 2022, 03:46:28 PM »
The following is my answer for quiz 6.
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##### Quiz 4 / Quiz 4 A
« Last post by ZiqiWang on May 01, 2022, 03:43:46 PM »
The following is the sample for quiz 4.
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##### Quiz 3 / Quiz 3 A
« Last post by ZiqiWang on May 01, 2022, 03:39:23 PM »
The following is the sample of quiz 3.
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##### Quiz 2 / Quiz 2 b
« Last post by ZiqiWang on May 01, 2022, 03:36:42 PM »
The following is the sample of quiz2.
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##### Quiz 1 / Quiz 1 C
« Last post by ZiqiWang on May 01, 2022, 03:33:40 PM »
The following is the sample for quiz 1.
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##### Final Exam / Re: Alternative solution to the optimization in Problem 2 on the practice final
« Last post by Victor Ivrii on April 25, 2022, 07:51:30 AM »
Yes, it can be solved using Lagrange multiplies. However note, if restrictions are $g_1\le 0$, $g_2\le 0$, $g_3\le 0$ you need to consider
• $g_1=0$ (and $g_2\le 0, g_3\le 0)$); there will be only one Lagrange multiplier at $g_1$. Two other cases in the similar way
• $g_1=g_2=0$ (and $g_3\le 0)$); there will be  two Lagrange multipliers. Two other cases in the similar way
It will be, however, more cumbersome. Note that (1) corresponds to two rays and one arc, (2) to two corners.

No, you need not consider quadratic forms after you found all suspicious points. It would serve no purpose.
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