<p>Solving</p>
<p><span class="math-container">$Min(-F[x])$</span> s.t. <span class="math-container">$G[x]\leq 0$</span></p>
<p>is same as solving</p>
<p><span class="math-container">$Max(F[x])$</span> s.t. <span class="math-container">$G[x]\leq0 $</span></p>
<p>So, the Lagrangian for the minimizing problem will be:</p>
<p><span class="math-container">$L = -F[x] - \lambda (0-G[x])$</span></p>
<p>For the maximization problem the Lagrangian will be:</p>
<p><span class="math-container">$L = F[x] + \lambda (0-G[x])$</span></p>
<p>In both cases, <span class="math-container">$\lambda \geq 0$</span></p>

CC BY-SA 4.0
2022-01-27T06:29:59.887Z
I would have thought that $\lambda$ for the minimisation problem is non-positive, $\lambda \le 0$. The first Lagrangian is the negative of the second one, so the first mulitplier should be the negative of the second one. Also the inequalities in the maximisation and minmisation problem should be weak ($\le$) instead of strict.
tdm
CC BY-SA 4.0
2022-01-27T19:01:16.810Z
Correct and correct! Thanks for pointing it out.
erik
CC BY-SA 4.0
2022-01-26T16:52:46.467Z
2022-01-27T19:02:15.007Z
2022-01-27T19:02:15.007Z
erik
<p>if I have a problem min(-f) s.t. g&lt;0, I can rewrite it as -max(f) s.t. g&lt;0.
In this case, if I take Lagrangians, would my lagrangian be
L=f- lambda(g-0)
or would I have to have a negative in front of the f?</p>

<mathematical-economics><lagrangian>
minimisation problem as a maximisation problem for lagrangians?
theshadowers
# minimisation problem as a maximisation problem for lagrangians?

(Asked by: username_0 on 2022-01-25)

if I have a problem min(-f) s.t. g<0, I can rewrite it as -max(f) s.t. g<0.
In this case, if I take Lagrangians, would my lagrangian be
L=f- lambda(g-0)
or would I have to have a negative in front of the f?

**Tags:** mathematical-economics, lagrangian


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## 1 answers

### Answer by username_1 on 2022-01-26. Score: 1

Solving

$Min(-F[x])$ s.t. $G[x]\leq 0$

is same as solving

$Max(F[x])$ s.t. $G[x]\leq0 $

So, the Lagrangian for the minimizing problem will be:

$L = -F[x] - \lambda (0-G[x])$

For the maximization problem the Lagrangian will be:

$L = F[x] + \lambda (0-G[x])$

In both cases, $\lambda \geq 0$

> **Comments:**
> **username_2**: I would have thought that $\lambda$ for the minimisation problem is non-positive, $\lambda \le 0$. The first Lagrangian is the negative of the second one, so the first mulitplier should be the negative of the second one. Also the inequalities in the maximisation and minmisation problem should be weak ($\le$) instead of strict.  
> **username_1**: Correct and correct! Thanks for pointing it out.  

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