<p>As mentioned in the other answer, the Lagrange multiplier is the marginal effect on the value (optimized) function, when the constrained is "relaxed" marginally. In your case then it should be interpreted "how much studying time changes as the required average expected grade..." increases?</p>

<p>Well, this is a good example to show that how we set up the Lagrangian function matters in order to interpret what we get meaningfully. Your is a minimization case so we have 
$$\min T = t_1+ t_2 \\
s.t\;\;\; \displaystyle m-\frac{[{g}_{1}({t}_{1})+{g}_{2}({t}_{2})]}{2}=0$$</p>

<p>How do we write the Lagrangian? Do we write</p>

<p>$$\Lambda = t_1+ t_2 + \lambda\left[m-\frac{[{g}_{1}({t}_{1})+{g}_{2}({t}_{2})]}{2}\right]$$</p>

<p>or</p>

<p>$$\Lambda = t_1+ t_2 - \lambda\left[m-\frac{[{g}_{1}({t}_{1})+{g}_{2}({t}_{2})]}{2}\right]$$</p>

<p>Since the constraint is an equality constraint, you may hear that "it doesn't matter". Mathematically, it indeed doesn't, but it does matter when the time comes to <em>interpret</em> the value of the multiplier. And, maybe in Utility maximization problems we don't much care because utility is ordinal. But in a problem like the one the OP is solving, the objective function is measured in a very real and measurable and quantifiable unit, time.</p>

<p>Now how we should understand "what happens when the constraint is <em>relaxed</em>"? In the specific minimization problem, it intuitively means "what happens when the required $m$ is <em>reduced</em>". (Since "relax" has a "desirable" connotation, and so we would desire to be asked for less effort, a lower target). </p>

<p>I guess it is now clear how the obtained value for $\lambda$ can be interpreted. And do ponder what variant of the Lagrangian is the appropriate one for your case.</p>

CC BY-SA 3.0
2016-05-30T13:48:42.757Z
2016-05-30T20:52:57.577Z
2016-05-30T20:52:57.577Z
Alecos Papadopoulos
<p>At the optimal values, by duality, this Lagrangian problem is equivalent to maximizing mean score given a time budget, where $t_1 + t_2 = 46$</p>

<p>$$m = \max_{t_1, t_2} \frac{[{g}_{1}({t}_{1})+{g}_{2}({t}_{2})]}{2} - \lambda(t_1 + t_2 - 46) $$</p>

<p>Plug in the optimal values of $t_1, t_2$ into your original Lagrangian:</p>

<p>$$V(m) = 46 + \lambda(m - 70)$$</p>

<p>Note that $$\frac{\partial V}{\partial m} = \lambda$$</p>

<p>I think the interpretation here is that the Lagrange multiplier is the rate of change of the optimal score as you gain more time.</p>

CC BY-SA 3.0
2016-05-26T05:44:27.383Z
2016-05-26T05:44:27.383Z
Kitsune Cavalry
<p>A student wishes to minimize the time required to gain a given expected average
grade, , in her end-of-semester examinations. Let $\displaystyle {t}_{i}$ be the time spent studying
subject i ∈ {1,2}. </p>

<p>Suppose that the expected grade functions are $\displaystyle {g}_{1} ({t}_{1})=40+8\sqrt{{t}_{1}}$ and  $\displaystyle {g}_{2} ({t}_{2}) = 2{t}_{2}$.</p>

<p>Thus the individual's optimization problems is to choose ${t}_{1}$ and ${t}_{2}$ to minimize total studying time $ = {t}_{1} + {t}_{2}$ subject to obtaining a mean grade of  where</p>

<p>$\displaystyle m-\frac{[{g}_{1}({t}_{1})+{g}_{2}({t}_{2})]}{2}=0$</p>

<p>I need to solve for the optimal choices of ${t}_{1} , {t}_{2}$ and $$ in the case where the student wishes to obtain an expected mean grade of 70.</p>

<p>Working through the problem I end with </p>

<p>${t}_{1}=4$ </p>

<p>${t}_{2}=42$</p>

<p>$T=46$</p>

<p>$\displaystyle\lambda=-1$</p>

<p>How would I interpret the lagrange multiplier in this case?</p>

CC BY-SA 3.0
2016-05-26T05:42:48.940Z
Hope this link helps: https://math.stackexchange.com/questions/780097/the-meaning-of-lambda-in-lagrange-multipliers at the bottom you'll find a response to a problem similar to yours.
Ulises Genis
<optimization>
Interpretation of lagrange multiplier
Franzie
# Interpretation of lagrange multiplier

(Asked by: username_0 on 2016-05-26)

A student wishes to minimize the time required to gain a given expected average
grade, , in her end-of-semester examinations. Let $\displaystyle {t}\_{i}$ be the time spent studying
subject i ∈ {1,2}.

Suppose that the expected grade functions are $\displaystyle {g}\_{1} ({t}\_{1})=40+8\sqrt{{t}\_{1}}$ and $\displaystyle {g}\_{2} ({t}\_{2}) = 2{t}\_{2}$.

Thus the individual's optimization problems is to choose ${t}\_{1}$ and ${t}\_{2}$ to minimize total studying time $ = {t}\_{1} + {t}\_{2}$ subject to obtaining a mean grade of where

$\displaystyle m-\frac{[{g}\_{1}({t}\_{1})+{g}\_{2}({t}\_{2})]}{2}=0$

I need to solve for the optimal choices of ${t}\_{1} , {t}\_{2}$ and $$ in the case where the student wishes to obtain an expected mean grade of 70.

Working through the problem I end with

${t}\_{1}=4$

${t}\_{2}=42$

$T=46$

$\displaystyle\lambda=-1$

How would I interpret the lagrange multiplier in this case?

**Tags:** optimization
> **Comments:**
> **username_1**: Hope this link helps: https://math.stackexchange.com/questions/780097/the-meaning-of-lambda-in-lagrange-multipliers at the bottom you'll find a response to a problem similar to yours.  

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## 2 answers

### Answer by username_2 on 2016-05-30. Score: 5

As mentioned in the other answer, the Lagrange multiplier is the marginal effect on the value (optimized) function, when the constrained is "relaxed" marginally. In your case then it should be interpreted "how much studying time changes as the required average expected grade..." increases?

Well, this is a good example to show that how we set up the Lagrangian function matters in order to interpret what we get meaningfully. Your is a minimization case so we have
$$\min T = t\_1+ t\_2 \\
s.t\;\;\; \displaystyle m-\frac{[{g}\_{1}({t}\_{1})+{g}\_{2}({t}\_{2})]}{2}=0$$

How do we write the Lagrangian? Do we write

$$\Lambda = t\_1+ t\_2 + \lambda\left[m-\frac{[{g}\_{1}({t}\_{1})+{g}\_{2}({t}\_{2})]}{2}\right]$$

or

$$\Lambda = t\_1+ t\_2 - \lambda\left[m-\frac{[{g}\_{1}({t}\_{1})+{g}\_{2}({t}\_{2})]}{2}\right]$$

Since the constraint is an equality constraint, you may hear that "it doesn't matter". Mathematically, it indeed doesn't, but it does matter when the time comes to *interpret* the value of the multiplier. And, maybe in Utility maximization problems we don't much care because utility is ordinal. But in a problem like the one the OP is solving, the objective function is measured in a very real and measurable and quantifiable unit, time.

Now how we should understand "what happens when the constraint is *relaxed*"? In the specific minimization problem, it intuitively means "what happens when the required $m$ is *reduced*". (Since "relax" has a "desirable" connotation, and so we would desire to be asked for less effort, a lower target).

I guess it is now clear how the obtained value for $\lambda$ can be interpreted. And do ponder what variant of the Lagrangian is the appropriate one for your case.


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### Answer by username_3 on 2016-05-26. Score: 3

At the optimal values, by duality, this Lagrangian problem is equivalent to maximizing mean score given a time budget, where $t\_1 + t\_2 = 46$

$$m = \max\_{t\_1, t\_2} \frac{[{g}\_{1}({t}\_{1})+{g}\_{2}({t}\_{2})]}{2} - \lambda(t\_1 + t\_2 - 46) $$

Plug in the optimal values of $t\_1, t\_2$ into your original Lagrangian:

$$V(m) = 46 + \lambda(m - 70)$$

Note that $$\frac{\partial V}{\partial m} = \lambda$$

I think the interpretation here is that the Lagrange multiplier is the rate of change of the optimal score as you gain more time.


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