4. If the surface area and volume of a circular cone are divided into upper and lower parts by a plane parallel to the base in the ratio $k$, then the minimum value of $c$ that makes $k c>1$ is $\qquad$

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4.7.

As shown in Figure 3, let $C O_{1}=r_{1}, A O=r_{2}$, the surface area of the smaller cone is $S_{1}$. The lateral surface area of the frustum is $S_{2}$, then
$$
\frac{S_{2}+S_{\text {base }}}{S_{1}}=\frac{1}{k},
$$

which means
$$
\frac{S_{1}}{S_{\text {lateral }}+S_{\text {base }}}=\frac{k}{k+1} \text {. }
$$

Given $S_{1}=\frac{\pi r_{1}^{2}}{\sin \theta}, S_{\text {lateral }}=\frac{\pi r_{2}^{2}}{\sin \theta}, S_{\text {base }}=\pi r_{2}^{2}$, we have
$$
\frac{k}{k+1}=\frac{r_{1}^{2}}{r_{2}^{2}} \cdot \frac{1}{1+\sin \theta} \text {. }
$$

Also, $\frac{V_{\text {frustum }}}{V_{\text {small cone }}}=\frac{1}{k}$, then
$$
\frac{V_{\text {cone }}}{V_{\text {small cone }}}=\frac{1+k}{k} \text {, }
$$

which means $\frac{r_{1}^{3}}{r_{2}^{3}}=\frac{k}{k+1}$.
Thus,
$$
\begin{array}{l}
\left(\frac{k}{k+1}(1+\sin \theta)\right)^{\frac{1}{2}} \\
=\frac{r_{1}}{r_{2}}=\left(\frac{\dot{k}}{k+\dot{\mathrm{i}}}\right)^{\frac{1}{3}} .
\end{array}
$$

Therefore, $\frac{1}{k}=(1+\sin \theta)^{3}-1\frac{1}{k}=(1+\sin \theta)^{3}-1$.
Hence, the minimum value of $c$ is 7.