7. Given $0<k<4$, line $l_{1}: k x-2 y-2 k+$ $8=0$ and line $l_{2}: 2 x+k^{2} y-4 k^{2}-4=0$ form a quadrilateral with the two coordinate axes. The value of $k$ that minimizes the area of this quadrilateral is ( ).
(A)2
(B) $\frac{1}{2}$
(C) $\frac{1}{4}$
(D) $\frac{1}{8}$
7.D.

It is known that the lines $l_{1}: y-4=\frac{k}{2}(x-2)$ and $l_{2}$: $y-4=-\frac{2}{k^{2}}(x-2)$ both pass through the fixed point $P(2,4)$.
In $l_{1}$, let $x=0$, we get $y_{A}=4-k$;
In $l_{2}$, let $y=0$, we get $x_{B}=2 k^{2}+2$. As shown in Figure 6, connect $O P$.
Then $S_{\text {quadrilateral PAOB }}$
$$
\begin{aligned}
= & S_{\triangle P O A}+S_{\triangle P O B} \\
= & \frac{1}{2}|O A| \cdot x_{p}+ \\
& \frac{1}{2}|O B| \cdot y_{p} \\
= & \frac{1}{2}(4-k) \times 2+\frac{1}{2}\left(2 k^{2}+2\right) \times 4 \\
= & 4\left(k-\frac{1}{8}\right)^{2}+\frac{127}{16} .
\end{aligned}
$$

Therefore, when $k=\frac{1}{8}$, the area of quadrilateral $P A O B$ is minimized.