How to prove that a utility function U(x,y)=min(x,2y) is quasiconcave? I have a question that asks:
&quot;Let there be two goods 1 and 2.Let $x$ and $y$ denote their respective quantities.$(x,y)$ represents a bundle. Suppose a consumer’s preferences over bundles in $R^2_+$  can be represented by the utility function $U(x,y)=min(x,2y)$. Prove that the consumer’s preferences over bundles in $R^2_+$ are convex by proving that the utility function is quasiconcave in $R^2_+$&quot;
I was taught that if a utility function is quasiconcave, then $\begin{vmatrix} 0 &amp; U_x &amp; U_y \\ U_x &amp; U_{xx} &amp; U_{xy} \\ U_y &amp; U_{yx} &amp; U_{yy} \\\end{vmatrix} &gt; 0 \forall (x, y) \in R^2_+$
However, when I start doing the partial derivatives, I notice that I get:
$\frac{\partial U}{\partial x} = \begin{cases}1, &amp;&amp; x \le 2y \\ 0, &amp;&amp; x &gt; 2y\end{cases},$
$\frac{\partial U}{\partial y} = \begin{cases}0, &amp;&amp; x \le 2y \\ 2, &amp;&amp; x &gt; 2y\end{cases},$
$\frac{\partial U}{\partial x^2} = \begin{cases}0, &amp;&amp; x \le 2y \\ 0, &amp;&amp; x &gt; 2y\end{cases},$
$\frac{\partial U}{\partial y^2} = \begin{cases}0, &amp;&amp; x \le 2y \\ 0, &amp;&amp; x &gt; 2y\end{cases},$
$\frac{\partial U}{\partial xy} = \begin{cases}0, &amp;&amp; x \le 2y \\ 0, &amp;&amp; x &gt; 2y\end{cases},$
$\frac{\partial U}{\partial yx} = \begin{cases}0, &amp;&amp; x \le 2y \\ 0, &amp;&amp; x &gt; 2y\end{cases}$
I then got $\begin{vmatrix} 0 &amp; U_x &amp; U_y \\ U_x &amp; U_{xx} &amp; U_{xy} \\ U_y &amp; U_{yx} &amp; U_{yy} \\\end{vmatrix} = \begin{cases}0, &amp;&amp; x \le 2y \\ 0, &amp;&amp; x &gt; 2y\end{cases}$
This is where I am stuck. I am unsure if:
a) I made a mistake somewhere
b) my method is wrong
c) Everything I've done so far is correct, but I need to do another step
Please let me know if I need to provide more information, thanks
A function $f:D\rightarrow \mathbb{R}$ is said to be quasiconcave if the following set is a convex set for every value of $a\in\mathbb{R}$:
$P_a = \{x\in D: f(x) \geq a\}$
To show that $f(x,y) =\min(x, 2y)$ is quasiconcave, we just need to show that $P_a = \{(x,y)\in \mathbb{R}^2: \min(x, 2y) \geq a\}$ is a convex set. For that we consider arbitrary $(x', y')$ and $(x'', y'')$ from the set $P_a$ and arbitrary $\lambda\in [0,1]$ and show that
$\lambda (x', y')+(1-\lambda)(x'', y'')$ is in $P_a$. Observe that $x'\geq \min(x', 2y')\geq a$ and $x''\geq \min(x'', 2y'')\geq a$, so $\lambda x'+(1-\lambda)x''\geq a$. Likewise, $\lambda 2y'+(1-\lambda)2y''\geq a$. Therefore, it follows that $\min(\lambda x'+(1-\lambda)x'',2(\lambda y'+(1-\lambda)y'')) \geq a$ and consequently, $\lambda (x',y')+(1-\lambda)(x'', y'')$ is in $P_a$.