I am trying to prove that the Carnot efficiency for a heat engine with maximum power is $1-\sqrt{\frac{\tau_l}{\tau_h}}$, where $\tau_h$ is the high temperature, $\tau_2$ the lower temperature, and we have the working temperature $\tau_{hw},\tau_{lw}$ such that $\tau_h>\tau_{hw}>\tau_{lw}>\tau_l$. During the hot isothermal stage, the rate of heat flow from hot reservoir to working fluid is proportional to the temperature difference $x=\tau_h-\tau_{hw}$, and for the cold it is proportional to $y = \tau_{lw}-\tau_l$. If we assume that both isothermal stages take time $t$, and that both isentropic stages take negligible time, then we can write $Q_h=Ktx$ for the hot isothermal stage, $Q_l=Kty$ for the cold stage.
I am aware of the derivation by Curzon and Ahlborn but I do not understand their derivation that well. So, I want to approach it a different way. I thought about expressing the power output as a function of $K,x $ and $y$ and then finding the optimal value for $x,y$. However, I am having trouble with this first step. Any help would be appreciated on finding this expression for the power output and also on how to write $y$ in terms of $x,\tau_h,\tau_l$. (We are also assuming the working fluid undergoes an ideal Carnot cycle between temperatures $\tau_{hw}$ and $\tau_{lw}$)