1
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ I think the best proof is just to say: the set is a triangle! (I don't what if anything triangles "resemble in economics") $\endgroup$ Apr 25, 2012 at 3:16
  • $\begingroup$ nope. not convincing enough. $\endgroup$
    – Koba
    Apr 25, 2012 at 14:06

2 Answers 2

3
$\begingroup$

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

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

$\endgroup$
12
  • $\begingroup$ 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)β $\endgroup$
    – Koba
    Apr 21, 2012 at 23:48
  • $\begingroup$ oh wait so if I expand aβ+(1−a)β it will β, right? $\endgroup$
    – Koba
    Apr 21, 2012 at 23:50
  • $\begingroup$ @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, $a\beta + (1-a)\beta = (a+(1-a))\beta = \beta$. $\endgroup$ Apr 21, 2012 at 23:50
  • $\begingroup$ @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. $\endgroup$ Apr 21, 2012 at 23:51
  • $\begingroup$ I see now thank you very much. This problem occupied me for a long time. Thanks. $\endgroup$
    – Koba
    Apr 21, 2012 at 23:53
1
$\begingroup$

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

enter image description here

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.