In the circuit shown, the resistance r is a variable resistance. If for $$r = fR$$, the heat generation in r is maximum then the value of f is

$$R_{total} = R + \frac{Rr}{R+r} = \frac{R^2 + Rr + Rr}{R + r} = \frac{R^2 + 2Rr}{R + r}$$

$$I = \frac{V}{R_{total}} = \frac{V(R + r)}{R^2 + 2Rr}$$

$$I_r = I \times \left( \frac{R}{R + r} \right)$$

$$I_r = \left( \frac{V(R + r)}{R^2 + 2Rr} \right) \times \frac{R}{R + r} = \frac{VR}{R^2 + 2Rr}$$

$$P = I_r^2 \cdot r$$: $$P = \left( \frac{VR}{R^2 + 2Rr} \right)^2 \cdot r$$

$$P = \frac{V^2 R^2 r}{(R^2 + 2Rr)^2}$$

$$P = \frac{V^2 R^2}{\frac{R^4}{r} + 4R^2r + 4R^3}$$

For $$P$$ to be maximum, the denominator $$D = \left( \frac{R^4}{r} + 4R^2r + 4R^3 \right)$$ must be minimum.

Using the AM-GM inequality, a sum of the form $$\frac{a}{x} + bx$$ is minimum when the two terms are equal:

$$\frac{R^4}{r} = 4R^2r$$

$$R^2 = 4r^2$$

$$R = 2r \implies r = \frac{R}{2}$$

$$f = \frac{1}{2}$$