This is a problem involving a series circuit with one EMF and, in series with the EMF, an internal resistance Ri and a load resistance, which I will denote simply with R. The current flowing through the entire circuit, and hence also through the load resistance, as a function of R, is i= EMF/(Ri + R), and the power dissipated by R, as a function of R, is P(R) = i*i*R
=(EMF^2*R)/(Ri+R)^2.
In order to find the maximum (extremum) of P(R), we differentiate P(R) with respect to R and set to zero.
This yields an extremum at R=Ri.
In order to show that Ri is a max (and not a min or an inflection point), I just graphed P(x) =x/(3+x)^2 and saw a max at x=3. This suggests that P(x) = x/(n+x)^2 has a max at n, and therefore that P(R) has a max at Ri.
This verifies the statement that was made.