I have got the following function:
$$L(x,y) = \frac{(16.2x +0.9y+5.2)^{24} e^{-(16.2x +0.9y+5.2)}}{24!} \cdot \frac{(2.1x +4.2y+0.9)^8 e^{-(2.1x +4.2y+0.9)}}{8!} \tag{1}$$
and I am attempting at finding the values for x and y that give me the global maximum.
I have attempted doing this using MATLAB by setting an equal range for the x and y coordinates, and using islocalmax(L),but I simply get a row of 0s and 1s.
Can this be done any other way?
Done by hand.
For the coordinates of the maximum, forget the factorial and take the logarithm of the function and compute the partial derivatives
$$\frac{\partial\log[L(x,y)]}{\partial x}=\frac{56}{7 x+14 y+3}+\frac{3888}{162 x+9 y+52}-\frac{183}{10}=0$$ $$\frac{\partial\log[L(x,y)]}{\partial y}=\frac{112}{7 x+14 y+3}+\frac{216}{162 x+9 y+52}-\frac{51}{10}=0$$ which is more than simple if you let $$A=\frac{1}{7 x+14 y+3} \qquad \text{and} \qquad B=\frac{1}{162 x+9 y+52}$$ Two linear equations in $(A,B)$ which, back to $(x,y)$ give $$x=\frac{2419}{2205}\qquad \text{and} \qquad y=\frac{2518}{2205}$$