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?

  • $\begingroup$ I think the best proof is just to say: the set is a triangle! (I don't what if anything triangles "resemble in economics") $\endgroup$ – Omar Antolín-Camarena Apr 25 '12 at 3:16
  • $\begingroup$ nope. not convincing enough. $\endgroup$ – Koba Apr 25 '12 at 14:06

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

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

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

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:


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


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.