Example 9 As shown in Figure 6, it is known that a corner of a square steel plate with a side length of 4 is rusted, where $A F=2, B F=1$. To make reasonable use of this steel plate, a rectangular block MPND will be cut from the pentagon $E A B C D$, with point $P$ located on $A B$, and it is required that the area be maximized. What is the maximum utilization rate of the steel plate?
As shown in Figure 6, point $P$ is on line segment $AB$. Draw $PM \perp DE$ and $PN \perp CD$, with the feet of the perpendiculars being $M$ and $N$ respectively. Extend $NP$ to intersect $EF$ at $Q$. Let $PQ = x$, and the area of quadrilateral $MPND$ be $y$. It is easy to see that 
then $\triangle APQ \sim \triangle ABF$.
Therefore, $\frac{PQ}{BF} = \frac{AQ}{AF}$, which means $\frac{x}{1} = \frac{AQ}{2}$.
Thus, $AQ = 2x$.
Hence, $y = PM \cdot PN = EQ(NQ - PQ)$
$$
= (2 + 2x)(4 - x) = -2x^2 + 6x + 8.
$$

Since point $P$ is on line segment $AB$, the range of $x$ is $0 \leqslant x \leqslant 1$.

The x-coordinate of the vertex of the parabola $y = -2x^2 + 6x + 8$ is $x = \frac{3}{2}$, which means this practical problem cannot reach the vertex coordinates. Therefore, when $x = 1$, $y_{\text{max}} = 12$. At this time, the maximum utilization rate of the steel plate is $\frac{12}{15} = 80\%$.

Note: This problem is about finding the maximum area of an inscribed rectangle, which is smoothly solved through the combination of similar triangles and quadratic functions.