<p>In the attached inequality, constrained, optimisation, problem. Looking at the specific case where <span class="math-container">$\lambda_1 = 0, \lambda_2 &gt; 0$</span> that I am trying to solve, you can see that I have managed to find half of the critical points, however, the mark scheme identifies additional critical points from this case (-2,0) and (2,0) these can be seen fairly easily from inspection of the constraint. However, I am concerned that there should be a systematic way of finding them apart from just guessing values in the constraint, and I think I would have missed them.</p>
<ol>
<li>Is my approach correct so far?</li>
<li>How would I have found these other critical points systematically?</li>
<li>Is there a better or faster way of solving this case for the 4 critical points?</li>
</ol>
<p>Apologies for uploading hand written solutions, i have RSI from typing, and so excessive typing is problematic atm. Thanks for your understanding!</p>
<p><a href="https://i.sstatic.net/FBbyt.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/FBbyt.png" alt="enter image description here" /></a></p>
<p><strong>Update</strong></p>
<p>Thanks to an interesting point raised by @tdm in the comments, i am also uploading the mark scheme! All thoughts welcome.</p>
<p><a href="https://i.sstatic.net/Q0ndp.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Q0ndp.png" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2024-04-22T09:05:07.223Z
How does $(x,y) = (2,0)$ satisfy the constraint $y + \lambda_1 5 + \lambda_2 2x = 0$ if $\lambda_1 = 0$ and $\lambda_2 > 0$?
tdm
CC BY-SA 4.0
2024-04-22T14:31:56.600Z
yeah i see your point, and technically $(-2, 0)$ also seems to fail 3, and $(2,0)$ fails 4.  I think $(-2,0)$ works for 4 i believe because technically if $y = 0$ the constraint becomes $x + \lambda_1 10 + \lambda_2 8y ≤ 0$ I'm uploading the mark scheme additionally, is it an error or am i missing something. Sadly there are no professors i can contact!
CormJack
<optimization><lagrangian>
Missing Solutions in KKT Optimisation Problem
CormJack
# Missing Solutions in KKT Optimisation Problem

(Asked by: username_0 on 2024-04-21)

In the attached inequality, constrained, optimisation, problem. Looking at the specific case where $\lambda\_1 = 0, \lambda\_2 > 0$ that I am trying to solve, you can see that I have managed to find half of the critical points, however, the mark scheme identifies additional critical points from this case (-2,0) and (2,0) these can be seen fairly easily from inspection of the constraint. However, I am concerned that there should be a systematic way of finding them apart from just guessing values in the constraint, and I think I would have missed them.

1. Is my approach correct so far?
2. How would I have found these other critical points systematically?
3. Is there a better or faster way of solving this case for the 4 critical points?

Apologies for uploading hand written solutions, i have RSI from typing, and so excessive typing is problematic atm. Thanks for your understanding!

[![enter image description here](https://i.sstatic.net/FBbyt.png)](https://i.sstatic.net/FBbyt.png)

**Update**

Thanks to an interesting point raised by @username_1 in the comments, i am also uploading the mark scheme! All thoughts welcome.

[![enter image description here](https://i.sstatic.net/Q0ndp.png)](https://i.sstatic.net/Q0ndp.png)

**Tags:** optimization, lagrangian

> **Comments:**
> **username_1**: How does $(x,y) = (2,0)$ satisfy the constraint $y + \lambda_1 5 + \lambda_2 2x = 0$ if $\lambda_1 = 0$ and $\lambda_2 > 0$?  
> **username_0**: yeah i see your point, and technically $(-2, 0)$ also seems to fail 3, and $(2,0)$ fails 4.  I think $(-2,0)$ works for 4 i believe because technically if $y = 0$ the constraint becomes $x + \lambda_1 10 + \lambda_2 8y ≤ 0$ I'm uploading the mark scheme additionally, is it an error or am i missing something. Sadly there are no professors i can contact!  

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## 0 answers

