Prove the set $M=\{x\in\mathbb{R}^2_+ \mid \alpha x_1+\gamma x_2\leq \beta\}$ is convex

Question

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?

Answer 1

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

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

Answer 2

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