The answers above give the usual methods. Here's a method I've come up with taking up your clue of a link with $e$.
You know that $x+y+z+t = 10 \:\:\;\:\;\;\;\;\; x,y,z,t > 0$ and wish to maximise $P=xyzt$. Take the natural logarithm to get $\log P = \log x + \log y + \log z + \log t.$ Now $P$ is maximised exactly when $\log P$ is maximised. So we may rephrase the problem as follows:
Find when the maximum of $f(x,y,z,t) = x + y + z + t$ occurs given the constraint $e^x + e^y + e^z + e^t = 10$.
Following the technique of Lagrange multipliers, we define the function $g(x,y,z,t,\lambda) = x + y + z + t + \lambda(e^x + e^y + e^z + e^t - 10).$ Then $\partial_x g = 1 + \lambda e^x$ $\partial_y g = 1 + \lambda e^y$ $\partial_z g = 1 + \lambda e^z$ $\partial_t g = 1 + \lambda e^t$ and $\partial_{\lambda} g = e^x + e^y + e^z + e^t - 10.$
To find when the maximum occurs we set the partial derivatives to be zero. From the first four partial derivatives we have $e^x=e^y=e^z=e^t = \frac{-1}{\lambda}$. Substituting this into the fifth partial derivative gives $-\frac{4}{\lambda}-10=0$, which gives $\lambda = -\frac{2}{5}$. This then gives $e^x=e^y=e^z=e^t = \frac{5}{2} = 2.5$, returning the result that the product is maximised when all four numbers are $2.5$.