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Find x and y that optimise

\begin{align} f(x,y) &= (-a-y)(\Psi(y)-\Psi(x+y)) + (b-x)(\Psi(x)-\Psi(x+y)) \\ &-\log \Gamma(x+y) + \log\Gamma(x) + \log\Gamma(y) \end{align}

where a, b are strictly positive constant.

Taking the derivative of $f(x,y)$ w.r.t. $x$, we get \begin{align*} \frac{\partial}{\partial x}f(x,y) = (b-x)\Psi'(x) - (b-x-a-y)\Psi'(x+y) \end{align*}

Taking the derivative of $f(x,y)$ w.r.t. $y$, we get \begin{align*} \frac{\partial}{\partial x}f(x,y) = (-a-y)\Psi'(y)-(b-x-a-y)\Psi'(x+y) \end{align*}

It gives $x = b$ and $y = -a$ which means $y<0$ since $a>0$. Here $\Psi$ is digamma and $\Psi'$ is trigamma.

How can I proceed from here to optimise $f(x,y)$ such that $x > 0$ and $y > 0$.?

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    $\begingroup$ What are the functions $\Psi$ and $\Gamma$? $\endgroup$ – David M. May 11 at 13:56
  • $\begingroup$ $\Psi$ is digamma function and $\Gamma$ is gamma function. $\endgroup$ – bemma May 11 at 13:56
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    $\begingroup$ In general, nonlinear optimization models and algorithms are not capable of handling strict inequality constraints. The typical way to handle this is to instead require $x \ge \epsilon$ for small $\epsilon$. $\endgroup$ – LarrySnyder610 May 11 at 14:01
  • $\begingroup$ @LarrySnyder610 If we introduce $x \ge \epsilon$ then is $y = \epsilon$ the solution? $\endgroup$ – bemma May 11 at 14:02
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    $\begingroup$ @ThePheromoneKid I want to maximise the function. $\endgroup$ – bemma May 13 at 12:02

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