GeoGebra link: https://www.geogebra.org/m/hscshcj8
This simulation demonstrates the power dissipated in a variable resistor given that the battery has an internal resistance (made variable in this app as well).
Since the power dissipated by the resistor is given by
[latex]P=I^2R[/latex]
and the current is given by
[latex]I=E(R+r)[/latex],
[latex]P=E^2\times\dfrac{R}{(R+r)^2}=\dfrac{E^2}{\dfrac{r^2}{R}+R+2r}[/latex]
This power will be a maximum if the expression for the denominator [latex]\dfrac{r^2}{R}+R+2r[/latex] is a minimum.
Differentiating the expression with respect to R, we get
[latex]\dfrac{d(\dfrac{r^2}{R}+R+2r)}{dR}=-\dfrac{r^2}{R^2}+1[/latex]
When the denominator is a minimum,
[latex]-\dfrac{r^2}{R^2}+1=0[/latex]
Therefore,
[latex]r=R[/latex] when the power dissipated by the resistor is highest.