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Let $\alpha\gt 0$, $\gamma\gt 0$, and $\beta\gt 0$ be real numbers. Let $$M=\{x\in\mathbb{R}^2_+ \mid \alpha x_1+\gamma x_2\leq \beta\}$$ Prove $M$ is a convex set. Prove that $M$ is bounded. What does this set resemble (in economics)?

Attempt: If $(x_1,x_2),(y_1,y_2)\in M$ we get $$\begin{align*} \alpha x_1 + \gamma x_2&\leq \beta\\ \alpha y_1 + \gamma y_2 &\leq \beta \end{align*}$$

We want to prove $$\alpha(ax_1 + (1-a)y_1) + \gamma(ax_2 + (1-a)y_2)\leq \beta.$$

The question is how do I prove this inequality?

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    I think the best proof is just to say: the set is a triangle! (I don't what if anything triangles "resemble in economics")2012-04-25
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    nope. not convincing enough.2012-04-25

2 Answers 2

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Algebra! (pronounced like Jon Lovitz's Master Thespian character)

$$\begin{align*} \alpha(ax_1 + (1-a)y_1) + \gamma(ax_2+(1-a)y_2) &= \alpha ax_1 + \gamma ax_2 + \alpha(1-a)y_1 + \gamma(1-a)y_2\\ &= a(\alpha x_1+\gamma x_2) + (1-a)(\alpha y_1 + \gamma y_2)\\ &\leq a\beta + (1-a)\beta. \end{align*}$$

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    ok you expanded it and rearranged terms.I did the same thing, but the last expression a(αx1+γx2)+(1−a)(αy1+γy2) should be leq than β. I do not understand why you are saying that a(αx1+γx2)+(1−a)(αy1+γy2)≤aβ+(1−a)β2012-04-21
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    oh wait so if I expand aβ+(1−a)β it will β, right?2012-04-21
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    @Dostre: $\alpha x_1+\gamma x_2\leq \beta$ by assumption; multiplying through by $a$ we get $a(\alpha x_1+\gamma x_2)\leq a\beta$. Similarly, form $\alpha y_1+\gamma y_2\leq \beta$, multiplying through by $(1-a)$ we get $(1-a)(\alpha y_1+\gamma y_2)\leq (1-a)\beta$. Add both inequalities to get the one I have; finally, $a\beta + (1-a)\beta = (a+(1-a))\beta = \beta$.2012-04-21
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    @Dostre: That's the last step, yes; but you said you didn't understand the last step I *did* do; I explained it in the comment just above this one.2012-04-21
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    I see now thank you very much. This problem occupied me for a long time. Thanks.2012-04-21
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    @Dostre I approved this edit firstly, just so that this edit information gets saved. (I was not sure, if I rejected this edit, your edit might go away.) I have rolled it back to the original version. I wanted your edit to stay, because, I wanted to tell you that you can write your own answer below. (So, you may want to copy the edit information by going through the edit info available. ) If Arturo agrees that this edit is fine, I leave it to him to bring back to your version. Hope I have not been a source of chaos. Regards,2012-04-23
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    @KannappanSampath: I will post it as my separate answer. Since Arthuro's answer is accepted I wanted to make it complete, but never mind. So remove my edits to Arthuro's answer.2012-04-23
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    @Dostre You can always toggle the accepted answer to a new one. I agree with your contention that it is better to keep the answer complete (I don't know if Arturo's answer is complete or it is not.). So, you may write your answer and toggle the accepted answer to that if you wish. I am sure Arturo will not be offended with this. But, this is an important thing to keep in mind: It is always good to keep the question open without accepting an answer for a sufficiently long time. Regards,2012-04-23
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    @Dostre: Generally speaking, it is not a good idea to add in a lot of information to someone else's answer. In essence, it's like putting (lots and lots of) words into their mouths. Write your own answer, by all means!2012-04-23
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    @ArturoMagidin: ok my bad.2012-04-23
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    @ArturoMagidin:sorry man I did not mean to misspell your name.2012-04-23
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    @Dostre: No worries; you aren't the first one to misspell my name, you won't be the last one (and I've mispelled my share of other people's name in the past).2012-04-23
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Same thing Arturo posted in more detail:

We know that the below two inequalities on the far left are true. So lets use them to prove the one we need to prove$[α(ax_1+(1−a)y_1)+γ(ax_2+(1−a)y_2)≤β]$:

$αx_1+γx_2≤β\;\;|*a\Rightarrow a(\alpha x_1+\gamma x_2)\leq a\beta$

$αy_1+γy_2≤β\;\;|*(1-a)\Rightarrow (1-a)(αy_1+γy_2)\leq (1-a)\beta$

Now add the inequalities on the far right side and we get:

$$a(\alpha x_1+\gamma x_2) + (1-a)(\alpha y_1 + \gamma y_2)\leq a\beta+(1-a)\beta$$

After expanding the expressions in parenthesis on the LHS and rearranging the terms we get:

$$α(ax_1+(1−a)y_1)+γ(ax_2+(1−a)y_2)≤a\beta+(1-a)\beta$$

Which almost looks exactly like the one we need to prove. The RHS after expanding:

$$a\beta+(1-a)\beta=a\beta +\beta -a\beta =\beta \Rightarrow$$

$$\Rightarrow a(\alpha x_1+\gamma x_2) + (1-a)(\alpha y_1 + \gamma y_2)\leq\beta$$

Which is what we needed to show.

2&3 questions:

This set M={$x∈ℝ^2_+∣αx_1+γx_2≤β$} looks like a budget constraint and is bounded by:

if $x_1=0;\;$ $\gamma x_2\leq \beta$;$\;\;x_2\leq \frac{\beta}{\gamma}$

if $x_2=0;\;$ $\alpha x_1\leq \beta$;$\;\;x_1\leq \frac{\beta}{\alpha}$

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