# Is the utility function $u(x)=x_1 x_2 + \gamma x_2$ concave or quasi-concave?

I want to prove, given $$\gamma>0$$ and $$x\in \mathbb{R}^2_+$$, if the utility function:

$$u(x)=x_1 x_2 + \gamma x_2$$

is concave, strictly concave, quasi-concave or strictly quasi-concave.

I have tried with Hessian but I do not reach anywhere.

Then I tried with the definition of concavity and quasi-concavity and I got that it is strictly quasi-concave if (supossing without loss of generality that $$u(y)>u(x)$$ ): $$\alpha(y_1-x_2)(y_2-x_2) < (y_1 y_2 + \gamma y_2)(x_1 x_2 + \gamma x_2)$$ Which I think is sattisfied.

Can someone give me a hint on this?

Consider $$A_\alpha = \{(x_1, x_2) \in \mathbb{R}_+^2: (x_1 + \gamma)x_2 \ge \alpha\}$$.

If $$\alpha \le 0$$, then $$A_\alpha = \mathbb{R}_+^2$$ which is convex but not strictly convex.

If $$\alpha > 0$$, then $$A_{\alpha} = \{(x_1, x_2)\in \mathbb{R}_+^2:x_2 \ge \frac{\alpha}{x_1+\gamma}\}$$ which is convex.

Hence $$u$$ is quasi-concave.

It is easay to check that it is not concave by examining the Hessian.

The hessian matrix is $$\begin{equation} H= \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix} \end{equation}$$

Therefore $$\begin{equation}x^THx=\begin{bmatrix} x_1\\ x_2 \end{bmatrix}^T\begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix} \begin{bmatrix} x_1\\ x_2 \end{bmatrix}=2x_1x_2\geq 0 \end{equation}$$

• What if $x_1>0$ and $x_2<0$? May 24, 2021 at 1:40