Maximizing power dissipated in a variable resistor

[xAly]

Homework Statement

A simple circuit consists of a supply with a voltage of $60V$ as well as a variable resistor (which can be varied between $1$ and $25 \Omega$) and resistor ($5\Omega$) connected in series. Determine the resistance of the variable resistor such that the power dissipated by it is maximised.

Relevant Equations

$P = I^2 R, R = \frac{V}{I}$

Let $R$ denote resistance of standard resistor and $R_v$ the resistance of the variable resistor. I know that $I = \frac{V}{(R_v + R)}$. Now I also know that $P = I^2 R_v$represents the power dissipated by the variable resistor and that I need to maximise $P$. The problem I am having is that both and $I$ and $R_v$ are dependent on each other, a decrease in $R_v$ leads to an increase in $I$ and vice versa. Therefore I don't see how to maximise $P$, usually I would differentiate to find the critical points but I am not sure which one is the independent variable here.

 

[TSny]

$I = \frac{V}{(R_v + R)}\,\,\,\,\,$ $P = I^2 R_v$

Try to use your equation for $I$ to express $P$ in terms of a single variable.