# How to prove that a utility function U(x,y)=min(x,2y) is quasiconcave?

I have a question that asks:

"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 $$R^2_+$$"

I was taught that if a utility function is quasiconcave, then $$\begin{vmatrix} 0 & U_x & U_y \\ U_x & U_{xx} & U_{xy} \\ U_y & U_{yx} & U_{yy} \\\end{vmatrix} > 0 \forall (x, y) \in R^2_+$$

However, when I start doing the partial derivatives, I notice that I get:

$$\frac{\partial U}{\partial x} = \begin{cases}1, && x \le 2y \\ 0, && x > 2y\end{cases},$$ $$\frac{\partial U}{\partial y} = \begin{cases}0, && x \le 2y \\ 2, && x > 2y\end{cases},$$

$$\frac{\partial U}{\partial x^2} = \begin{cases}0, && x \le 2y \\ 0, && x > 2y\end{cases},$$ $$\frac{\partial U}{\partial y^2} = \begin{cases}0, && x \le 2y \\ 0, && x > 2y\end{cases},$$

$$\frac{\partial U}{\partial xy} = \begin{cases}0, && x \le 2y \\ 0, && x > 2y\end{cases},$$ $$\frac{\partial U}{\partial yx} = \begin{cases}0, && x \le 2y \\ 0, && x > 2y\end{cases}$$

I then got $$\begin{vmatrix} 0 & U_x & U_y \\ U_x & U_{xx} & U_{xy} \\ U_y & U_{yx} & U_{yy} \\\end{vmatrix} = \begin{cases}0, && x \le 2y \\ 0, && x > 2y\end{cases}$$

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

• min(x, 2y) is a concave function and every concave function is quasiconcave. You can watch the following playlist for the proof that min(x, 2y) is concave : youtube.com/… – Amit Apr 4 at 6:41
• Or you can just use another definition of quasiconcavity, one that does not rely on differentiability. – Giskard Apr 4 at 8:56
• That $U$ is quasi-concave means that for any $s,t$ in the convex domain (here $\mathbb{R}_+^2$) and any $\lambda$ in $[0,1]$ (you could, equivalently, use an open interval), $U\big(\lambda s+(1-\lambda)t\big)\geq\min\big\{U(s),U(t)\big\}.$ It is wonderful that there exit calculus characterizations for differentiable functions, but this function is very much non-differentable, and all the action is at a kink. Start directly with the definition. – Michael Greinecker Apr 4 at 8:58
• Thanks @MichaelGreinecker that makes sense – DoubleRainbowZ Apr 4 at 12:21

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 $$P_a$$.