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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 $0\leq f(x,y)\leq 1, \forall 0\leq x\leq 1, 0\leq y\leq 1$

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3 Answers 3

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$p(x, y) = -x^2 - xy - y^2 + 4x - y$ is an example. The eigenvalues of the Hessian are $-3$ and $-1$.

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Not sure why you want to avoid using an exponential function, but the following function satisfies your constraints:

$f(x, y) = -(y+1)^2 - \frac{1}{x + 1}$

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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 $x,y\in[0,1]$.

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