# simple concave two-variable function

I'm looking for a simple continuous concave two-variable function $$f(x,y)$$, where $$0\leq x,y \leq 1$$, that satisfies all of the following: $$\frac{\partial f(x,y)}{\partial x}>0$$ $$\frac{\partial f(x,y)}{\partial y}<0$$ $$\frac{\partial^2 f(x,y)}{\partial x^2}<0$$ $$\frac{\partial^2 f(x,y)}{\partial y^2}<0$$

I was able to find one: $$1-e^{-(x\cdot(1-y))}$$. However, I would like to find another one without an exponential function. Could anyone give other suggestions or hints please? Thank you!

EDIT: Forgot to mention one another condition. I need $$0\leq f(x,y)\leq 1, \forall 0\leq x\leq 1, 0\leq y\leq 1$$

$$p(x, y) = -x^2 - xy - y^2 + 4x - y$$ is an example. The eigenvalues of the Hessian are $$-3$$ and $$-1$$.
$$f(x, y) = -(y+1)^2 - \frac{1}{x + 1}$$
What about $$f(x,y)=-\frac{(x-1)^2}{2}-\frac{(y+\frac{1}{2})^2}{3}+1,$$ which satisfies $$f_x(x,y)=1-x>0$$, $$f_y(x,y)=-\frac{2}{3}(y+\frac{1}{2})<0$$, $$f_{xx}(x,y)=-1$$ and $$f_{yy}(x,y)=-\frac{2}{3}$$ for all $$y,x\in[0,1]$$ and also $$f(x,y)\in[0,1]$$ for $$x,y\in[0,1]$$.