<p>Answer a is not possible since it reduces to </p>

<p>$m = \frac{1}{2} 2m +  \frac{3}{2}\frac{2m}{3} = 2m$</p>

<p>which has no sense. Idem for answer c, $m\neq\frac{1}{4}m$.</p>

<p>Remains answers b or d.</p>

<p>For answer b, we have </p>

<p>$u(2m,0)=8m$</p>

<p>and for answer d,</p>

<p>$u(0,\frac{2m}{3}) = \frac{28}{3}m = (9+\frac{1}{3})m$</p>

<p>As you can see, $ (9+\frac{1}{3}) &gt; 8$.</p>

<p>In the general case, to find the maximum of your utility function given a monetary constrain, you can formalize and maximize the following <a href="https://en.wikipedia.org/wiki/Lagrange_multiplier" rel="nofollow noreferrer">Lagrangian</a> function</p>

<p>$L(x_1,x_2,\lambda)=u(x_1,x_2)+\lambda(m-p_1 x_1 - p_2 x_2)$</p>

<p>where $p_1$ and $p_2$ are prices. In your case, those are $\frac{1}{2}$ and $\frac{3}{2}$ respectively.</p>

<h3> Suite </h3>

<p>As usual, <strong><em>the story behind equations is of first importance</em></strong>.</p>

<p>Remember that $x_1$ and $x_2$ are two perfect substitutes. This means that the individual will spend all her income either in $x_1$ or in $x_2$, and will chose the good which provides her with the highest utility. And this is actually the story that the use of the Lagrangian function tells you. </p>

<p>As you mentioned in your comment below, you get <em>first order conditions which seem to be contradictory. But they are not</em>. Indeed, the two goods will never be bought simultaneously. Which means that you will get either $\lambda = 8$ or $\lambda = \frac{28}{3}$ in a <em>mutually exclusive</em> manner.</p>

<p>Recall what $\lambda$ is : it is a shadow price. In other words, it expresses how much your objective function, $u(x_1,x_2)$, will increase, if your constrain, $m$, increases by $1$.</p>

<p>Thus, if $m$ increases by $1$, and the individual spends all her income in $x_1$, $u$ will increase by $8$.
If she spends all her income in $x_2$, $u$ will increase by $\frac{28}{3}$.</p>

<p>Which good will the individual chose ?</p>

CC BY-SA 3.0
2017-03-08T17:31:24.877Z
Thank you ! But I think that the Lagrangian function won'´t work here, because we would have $L=4x_1+14x_2+\lambda(m-\frac{x_1}{2}-\frac{3x_2}{2})$ and the first order condition would imply $\lambda=8$ and $\lambda = \frac{28}{3}$
pinkpanther5
CC BY-SA 3.0
2017-03-09T14:05:23.363Z
Actually this is not $\lambda=8$ and $\lambda=\frac{28}{3}$ but $\lambda=8$ **xor** $\lambda=\frac{28}{3}$. See update.
keepAlive
CC BY-SA 3.0
2017-03-08T16:59:10.837Z
2017-11-17T22:16:10.487Z
2017-11-17T22:16:10.487Z
keepAlive
<p>Utility maximization problem is:</p>

<p>$\max\limits_{(x_1, x_2)\in\mathbb{R}^2_+} 4x_1 + 14x_2 \\ \text{s.t.} \ \ m = \frac{1}{2}x_1 + \frac{3}{2}x_2$</p>

<p>This problem can be converted into an optimization problem involving just one variable $x_1$:</p>

<p>$\max\limits_{x_1} 4x_1 + 14\left[\frac{2}{3}\left(m - \frac{1}{2}x_1\right)\right] \\ \text{s.t.} \ \ 0 \leq x_1 \leq 2m$</p>

<p>Differentiating the objective $4x_1 + 14\left[\frac{2}{3}\left(m - \frac{1}{2}x_1\right)\right]$ with respect to $x_1$, we get $\left(4- \frac{14}{3}\right) = \frac{-2}{3} &lt; 0$. So we observe that the objective is decreasing in $x_1$. Therefore optimal choice must be the smallest $x_1$ we can have i.e. $x_1 = 0$ and hence $x_2 = \frac{2m}{3}$.</p>

CC BY-SA 3.0
2017-03-09T00:14:35.440Z
2017-03-09T00:14:35.440Z
Amit
<p>In this question we can see that ..
Px=1/2 &amp; Py=3/2
And MRS= 4/14 
Px/Py=1/3>4/14(MRS)
So, the optimal point will be on y axis and it will be(0,m/Py)
i.e. (0,2m/3).</p>

CC BY-SA 4.0
2019-10-24T12:31:14.487Z
2019-10-24T12:31:14.487Z
user24317
<blockquote>
<p>Given <span class="math-container">$u(x_1,x_2)=4x_1+14x_2$</span> and <span class="math-container">$m=\frac{1}{2}x_1+\frac{3}{2}x_2$</span>, I shall choose the optimal decision among:</p>
<p><span class="math-container">$a)(2m,\frac{2m}{3})$</span></p>
<p><span class="math-container">$b)(2m,0)$</span></p>
<p><span class="math-container">$c)(\frac{m}{2},0)$</span></p>
<p><span class="math-container">$d)(0,\frac{2m}{3})$</span></p>
<p>The correct answer is <span class="math-container">$(d)$</span></p>
</blockquote>
<p>But not sure how to find this, what I did:
<span class="math-container">$$m=\frac{1}{2}x_1+\frac{3}{2}x_2 \Leftrightarrow x_2=\frac{2m}{3}-\frac{1}{3}x_1 $$</span>
And the marginal rate of substitution is <span class="math-container">$-\frac{2}{7}$</span>,so we have :</p>
<p><a href="https://i.sstatic.net/n8o0G.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/n8o0G.png" alt="enter image description here" /></a></p>
<p>Obviously <span class="math-container">$a)$</span> isn't the right one and <span class="math-container">$c)$</span> doesn't seem to be the best one... but really don't know 'how to think', some explaination would be great.
Thanks in advance</p>

CC BY-SA 3.0
2017-03-09T09:30:23.753Z
I'm voting to close this question because, although you have shown some effort, it contains many specific figures which make it less useful to future visitors - please see our policy on homework questions (http://meta.economics.stackexchange.com/questions/1465/economics-se-policy-on-homework-questions).
Adam Bailey
<microeconomics><consumer-theory>
Optimal decision for perfect substitutes utility function ?
pinkpanther5
# Optimal decision for perfect substitutes utility function ?

(Asked by: username_0 on 2017-03-08)

> Given $u(x\_1,x\_2)=4x\_1+14x\_2$ and $m=\frac{1}{2}x\_1+\frac{3}{2}x\_2$, I shall choose the optimal decision among:
>
> $a)(2m,\frac{2m}{3})$
>
> $b)(2m,0)$
>
> $c)(\frac{m}{2},0)$
>
> $d)(0,\frac{2m}{3})$
>
> The correct answer is $(d)$

But not sure how to find this, what I did:
$$m=\frac{1}{2}x\_1+\frac{3}{2}x\_2 \Leftrightarrow x\_2=\frac{2m}{3}-\frac{1}{3}x\_1 $$
And the marginal rate of substitution is $-\frac{2}{7}$,so we have :

[![enter image description here](https://i.sstatic.net/n8o0G.png)](https://i.sstatic.net/n8o0G.png)

Obviously $a)$ isn't the right one and $c)$ doesn't seem to be the best one... but really don't know 'how to think', some explaination would be great.
Thanks in advance

**Tags:** microeconomics, consumer-theory
> **Comments:**
> **username_1**: I'm voting to close this question because, although you have shown some effort, it contains many specific figures which make it less useful to future visitors - please see our policy on homework questions (http://meta.economics.stackexchange.com/questions/1465/economics-se-policy-on-homework-questions).  

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## 3 answers

### Answer by username_2 on 2017-03-08. Score: 0 | Accepted answer

Answer a is not possible since it reduces to

$m = \frac{1}{2} 2m + \frac{3}{2}\frac{2m}{3} = 2m$

which has no sense. Idem for answer c, $m\neq\frac{1}{4}m$.

Remains answers b or d.

For answer b, we have

$u(2m,0)=8m$

and for answer d,

$u(0,\frac{2m}{3}) = \frac{28}{3}m = (9+\frac{1}{3})m$

As you can see, $ (9+\frac{1}{3}) > 8$.

In the general case, to find the maximum of your utility function given a monetary constrain, you can formalize and maximize the following [Lagrangian](https://en.wikipedia.org/wiki/Lagrange_multiplier) function

$L(x\_1,x\_2,\lambda)=u(x\_1,x\_2)+\lambda(m-p\_1 x\_1 - p\_2 x\_2)$

where $p\_1$ and $p\_2$ are prices. In your case, those are $\frac{1}{2}$ and $\frac{3}{2}$ respectively.

### Suite

As usual, ***the story behind equations is of first importance***.

Remember that $x\_1$ and $x\_2$ are two perfect substitutes. This means that the individual will spend all her income either in $x\_1$ or in $x\_2$, and will chose the good which provides her with the highest utility. And this is actually the story that the use of the Lagrangian function tells you.

As you mentioned in your comment below, you get *first order conditions which seem to be contradictory. But they are not*. Indeed, the two goods will never be bought simultaneously. Which means that you will get either $\lambda = 8$ or $\lambda = \frac{28}{3}$ in a *mutually exclusive* manner.

Recall what $\lambda$ is : it is a shadow price. In other words, it expresses how much your objective function, $u(x\_1,x\_2)$, will increase, if your constrain, $m$, increases by $1$.

Thus, if $m$ increases by $1$, and the individual spends all her income in $x\_1$, $u$ will increase by $8$.
If she spends all her income in $x\_2$, $u$ will increase by $\frac{28}{3}$.

Which good will the individual chose ?

> **Comments:**
> **username_0**: Thank you ! But I think that the Lagrangian function won'´t work here, because we would have $L=4x_1+14x_2+\lambda(m-\frac{x_1}{2}-\frac{3x_2}{2})$ and the first order condition would imply $\lambda=8$ and $\lambda = \frac{28}{3}$  
> **username_2**: Actually this is not $\lambda=8$ and $\lambda=\frac{28}{3}$ but $\lambda=8$ **xor** $\lambda=\frac{28}{3}$. See update.  

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### Answer by username_3 on 2017-03-09. Score: 2

Utility maximization problem is:

$\max\limits\_{(x\_1, x\_2)\in\mathbb{R}^2\_+} 4x\_1 + 14x\_2 \\ \text{s.t.} \ \ m = \frac{1}{2}x\_1 + \frac{3}{2}x\_2$

This problem can be converted into an optimization problem involving just one variable $x\_1$:

$\max\limits\_{x\_1} 4x\_1 + 14\left[\frac{2}{3}\left(m - \frac{1}{2}x\_1\right)\right] \\ \text{s.t.} \ \ 0 \leq x\_1 \leq 2m$

Differentiating the objective $4x\_1 + 14\left[\frac{2}{3}\left(m - \frac{1}{2}x\_1\right)\right]$ with respect to $x\_1$, we get $\left(4- \frac{14}{3}\right) = \frac{-2}{3} < 0$. So we observe that the objective is decreasing in $x\_1$. Therefore optimal choice must be the smallest $x\_1$ we can have i.e. $x\_1 = 0$ and hence $x\_2 = \frac{2m}{3}$.


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### Answer by username_4 on 2019-10-24. Score: 0

In this question we can see that ..
Px=1/2 & Py=3/2
And MRS= 4/14
Px/Py=1/3>4/14(MRS)
So, the optimal point will be on y axis and it will be(0,m/Py)
i.e. (0,2m/3).


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