7. As shown in Figure 3, the graph of the linear function $y=k x+b$ passes through the point $P(1,4)$, and intersects the positive x-axis and y-axis at points $A$ and $B$, respectively. $O$ is the origin. When the area of $\triangle A O B$ is minimized, the values of $k$ and $b$ are ( ).
(A) $-4,8$
(B) $-4,4$
(C) $-2,4$
(D) $-2,2$
7.A.

Since point $P(1,4)$ lies on the graph of the linear function $y=k x+b$, the linear function can be expressed as $y=k x+(4-k)$.
Let $x=0$, we get $B(0,4-k)$.
Let $y=0$, we get $A\left(\frac{k-4}{k}, 0\right)$.
Connecting $P O$. Then
$$
\begin{array}{l}
S_{\triangle B O A}=S_{\triangle B O P}+S_{\triangle P O H} \\
=\frac{1}{2}[1 \times(4-k)]+\frac{1}{2} \times 4 \times \frac{k-4}{k} \\
=4-\frac{1}{2}\left(k+\frac{16}{k}\right) .
\end{array}
$$

Obviously, $k0, v>0$, and $u v=16$.
Thus $S_{\triangle B O}=4+\frac{1}{2}(u+v)$
$$
\begin{array}{l}
=4+\frac{1}{2}\left[(\sqrt{u}-\sqrt{v})^{2}+2 \sqrt{u w}\right] \\
\geqslant 4+\sqrt{u w}=8(u>0, v>0) .
\end{array}
$$

When and only when $\sqrt{u}=\sqrt{v}$, i.e., $k=-u=-4$, the above expression takes the equal sign, and $S_{\triangle s o n}$ takes the minimum value 8. At this time, $b=4-(-4)=8$.