Analytically minimize sum of products I am having the following function of 2 variables which I would like to minimize.
$$ f(x,y)={\left(n_{1}\right)\left(1- x\left(1-x\right)^{k_{1}}\left(1-y\right)^{k_{2}}\right)^{T}} + {\left(n_{2}\right)\left(1- y\left(1-x\right)^{k_{1}}\left(1-y\right)^{k_{2}}\right)^{T}}$$


Note that $x \in [0,1]$ and $y \in [0,1]$. In fact, $x$ and $y$ are probability values.

T is a positive integer. (T actually denotes number of tests in my actual problem.) $n_1$ and $n_2$ are also positive integers but much larger than T. ( $n_1$ and $n_2$ represents number of members in  2 classes in my actual problem setup).
$k_1$ and $k_2$ represents number of defective members in the two classes containing $n_1$ and $n_2$ members each. Thus, $T$ can be viewed as the number of tests made to identify the defective members.

The only way I can think of is take the partial derivatives w.r.t $x$ and $y$ and equate them to zero and solve simultaneously.
I tried this, but I don't see any ways of solving the resulting 2 equations as they are not analytically simple atleast for me.
I also notice that, there is some sort of symmetry in $f(x,y)$ with respect to $x$ and $y$. My hunch is that this could be made use of to find a minimum.


Can someone suggest a possible way of solving it?


One more question:


Is there any computer software package or online tools that I can use to minimize this function while keeping the generic letter constants as such?


 A: $f(x,y)=\left(n_{1}\right)\left(1-x\left(1-x\right)^{k_{1}}\left(1-y\right)^{k_{2}}\right)^{T}+\left(n_{2}\right)\left(1-y\left(1-x\right)^{k_{1}}\left(1-y\right)^{k_{2}}\right)^{T}$
define $\alpha=\left(1-x\right)^{k_{1}}\left(1-y\right)^{k_{2}}$
$f(x,y)=\left(n_{1}\right)\left(1-x\alpha(x,y)\right)^{T}+\left(n_{2}\right)\left(1-y\alpha(x,y)\right)^{T}$
minimize with a Lagrange multiplier:
$\left(n_{1}\right)\left(1-x\alpha\right)^{T}+\left(n_{2}\right)\left(1-y\alpha\right)^{T}+\lambda(\left(1-x\right)^{k_{1}}\left(1-y\right)^{k_{2}}-\alpha)$
$\partial_{x}:-\left(T\alpha n_{1}\right)\left(1-x\alpha\right)^{T-1}-\lambda k_{1}(\left(1-x\right)^{k_{1}-1}\left(1-y\right)^{k_{2}})=0$
$\partial_{y}:-\left(T\alpha n_{2}\right)\left(1-y\alpha\right)^{T-1}-\lambda k_{2}(\left(1-x\right)^{k_{1}}\left(1-y\right)^{k_{2}-1})=0$
$\partial_{\alpha}:-\left(xTn_{1}\right)\left(1-x\alpha\right)^{T-1}-\left(yTn_{2}\right)\left(1-y\alpha\right)^{T-1}-\lambda=0$
multiplying the first two equations by $\frac{x}{\alpha};\frac{y}{\alpha}$
  respectively:
$-\left(Txn_{1}\right)\left(1-x\alpha\right)^{T-1}-\lambda xk_{1}(\left(1-x\right)^{k_{1}-1}\left(1-y\right)^{k_{2}})/\alpha=0$
$-\left(Tyn_{2}\right)\left(1-y\alpha\right)^{T-1}-\lambda yk_{2}(\left(1-x\right)^{k_{1}}\left(1-y\right)^{k_{2}-1})/\alpha=0$
subtracting from the $\partial_{\alpha}$
 term:
$-xk_{1}(\left(1-x\right)^{k_{1}-1}\left(1-y\right)^{k_{2}})/\alpha-yk_{2}(\left(1-x\right)^{k_{1}}\left(1-y\right)^{k_{2}-1})/\alpha+1=0$
using the constraint $\alpha=\left(1-x\right)^{k_{1}}\left(1-y\right)^{k_{2}}$
  ( we are allowed to use it now that we have finished differentiating):
$-xk_{1}(\frac{\alpha}{1-x})/\alpha-yk_{2}(\frac{\alpha}{1-y})/\alpha+1=0$
yielding:
$k_{1}(\frac{x}{1-x})+k_{2}(\frac{y}{1-y})=1$
We can now eliminate $y$
  and solve for a one dimensional minimum.
