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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

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

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    $\begingroup$ 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 (meta.economics.stackexchange.com/questions/1465/…). $\endgroup$ – Adam Bailey Mar 9 '17 at 9:30
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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 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 ?

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  • $\begingroup$ 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 $\lambda = \frac{28}{3}$ $\endgroup$ – pinkpanther5 Mar 8 '17 at 17:31
  • $\begingroup$ Actually this is not $\lambda=8$ and $\lambda=\frac{28}{3}$ but $\lambda=8$ xor $\lambda=\frac{28}{3}$. See update. $\endgroup$ – keepAlive Mar 9 '17 at 14:05
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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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