= Solution
Put $\xi=x/\delta_0$ and $n=2p-q$. The <eddy viscosity> is $\nu_T=C_\mu\sqrt{k_0}\delta_0\xi^n$, since $\delta_0=k_0^{3/2}/\epsilon_0$. In the energy equation, advection scales as $\xi^{p-1}$, turbulent diffusion as $\xi^{n+p-2}$, and destruction as $\xi^q$. A leading advection–diffusion <K-epsilon turbulence front> requires $n=1$, hence
$$
\boxed{q=2p-1.}
$$
Comparing the leading coefficients then gives $U_0=C_\mu p\sqrt{k_0}/\sigma_k$. Similarly, the leading coefficients of the dissipation equation give $U_0=C_\mu q\sqrt{k_0}/\sigma_\epsilon$. Their equality implies $q=\sigma p$, where $\sigma=\sigma_\epsilon/\sigma_k$. Combining this with $q=2p-1$ yields
$$
\boxed{p=\frac1{2-\sigma},\qquad q=\frac{\sigma}{2-\sigma},
\qquad U_0=\frac{C_\mu(2p-1)\sqrt{k_0}}{\sigma_\epsilon}.}
$$
The assumed positive exponents require $0<\sigma<2$. Under these values, the omitted destruction terms are indeed higher order, as checked in part (d), so the leading balance is not circular.
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