Given two independent random variables $X$and $Y$ with distribution $X\sim Poisson(\phi)$ and $Y\sim Poisson(2\phi)$, and the observations $X=2$ and $Y=4$ of these. Also given: $\phi \in [0,10]$.
The expression for log-likelihood function is given by: $$L(\phi)=[4\ln(2)-\ln(2!)-\ln(4!)]+6\ln(\phi)-3 \phi$$
Show that the probability maximum estimator is $$\hat{\phi}= \frac{6}{3}$$ and calculate it for the observed values of $X$ and $Y$.
Now, I found the maximum, by brute force, to be: $$L(2)\approx -2.9$$
But, once I have found the maximum, how to I use it for this proof?
For $\phi \in [0,10]$ the log likelihood is: $$\log\left[ \frac{\exp(-\phi) \phi^2}{2!} \frac{\exp(-2\phi) (2\phi)^4}{4!} \right]=-3\phi+6 \log\phi +\log16-\log(2!)-\log(4!)$$ Setting the derivative to $0$ gives $-3+6/\phi=0$ or $\phi=2$.