How to find a function with certain properties? I'm trying to find a function $f(x,y):\mathbb{R^2_{+}} \rightarrow \mathbb{R_{+}}$. $\mathbb{R}_+$ denotes the set of non-negative real numbers. The function $f$ is supposed to satisfies following properties:


*

*$f(x,0)=0,\forall x$; $f(0,y)=0, \forall y$.

*$f$ is twice continuously differentiable, increasing and strictly concave in $x$ and $y$, i.e., $\frac{\partial f}{\partial x} \geq 0,\frac{\partial f}{\partial y} \geq 0,\frac{\partial^2f}{{\partial x}^2}<0,\frac{\partial^2f}{\partial y^2}<0$.

*$\frac{\partial^2 f}{\partial x \partial y}>0, \frac{\partial^2f}{\partial y \partial x} >0$.


I tried to form some quadratic function but failed to find one that satisfies all the properties. What is the way of thinking when we are asked to find a function with certain properties. 
 A: You just need twice continuously differentiable functions $g_i:[0,\infty)\to[0,\infty)$ such that $g_i(0)=0$, $g_i(x)>0$, $g_i'(x)>0$ and $g_i''(x)<0$ for $x>0$ and $i=1,2$. Then for $f(x,y)=g_1(x)g_2(y)$ we get


*

*$f(x,0)=g_1(x)g_2(0)=0$ and $f(0,y)=g_1(0)g_2(y)=0$ for all $x,y\geq 0$.

*$f$ is twice continuously differentiable, increasing and strictly concave in $x$ and $y$ on the interior of $\mathbb R_+^2$.
$$\frac{\partial f}{\partial x}=g_1'(x)g_2(y)\geq 0\\
\frac{\partial f}{\partial y}=g_1(x)g_2'(y)\geq 0\\
\frac{\partial^2 f}{\partial x^2}=g_1''(x)g_2(y)<0\\
\frac{\partial^2 f}{\partial y^2}=g_1(x)g_2''(y)<0
$$

*On the interior of $\mathbb R_+^2$ we have
$$
\frac{\partial^2 f}{\partial x\partial y}=
\frac{\partial^2 f}{\partial y\partial x}=g_1'(x)g_2'(y)>0
$$


Obviously you can choose $g_1=g_2$. Models for $g_i$ could be


*

*$g(x)=1-e^{-\alpha x}$ for $\alpha>0$ since $$g(0)=1-e^0=0,~ g(x)>0,~ g'(x)=\alpha e^{-\alpha x}>0\text{ and }g''(x)=-\alpha^2e^{-\alpha x}<0\text{ for }x>0.$$

*$g(x)=1-\frac1{(x+1)^\alpha}$ for $\alpha>0$ since $$g(0)=1-\frac1{(0+1)^\alpha}=0,~g(x)>0, ~g'(x)=\frac{\alpha}{(x+1)^{\alpha+1}}>0\\\text{ and }g''(x)=-\frac{\alpha(\alpha+1)}{(x+1)^{\alpha+2}}<0\text{ for }x>0.$$

*$g(x)=x^\alpha$ for $\alpha\in(0,1)$ (i.e. $g(x)=\sqrt{x}$) since $$g(0)=0^\alpha=0, ~g(x)>0, ~g'(x)=\alpha x^{\alpha-1}>0\text{ and }g''(x)=-\alpha(1-\alpha)x^{\alpha-2}<0\text{ for }x>0.$$
