图中显示了两条宽度分别为 9 m 和 4 m 且呈直角的走廊的交叉口。P 和 Q 是动点，且 PBQ 为一条直线。

1. Identify similar triangles formed by the horizontal and vertical lines

Let $D$ be the point on $AQ$ such that $BD$ is horizontal and perpendicular to $AQ$. Then $AD = 4$ m and $QD = y - 4$ m. Also, $DB = 9$ m.
$\triangle QDB$ is similar to $\triangle BCP$ because $\angle QDB = \angle BCP = 90^\circ$ and $\angle DQB = \angle CBP$ (corresponding angles, since $DB$ is parallel to $AP$).

2. Set up the ratio of corresponding sides

$\frac{QD}{BC} = \frac{DB}{CP}$
$\frac{y - 4}{4} = \frac{9}{x}$

3. Rearrange to show the required equation

$y - 4 = \frac{36}{x}$
$y = 4 + \frac{36}{x}$ (shown)

1. Formulate an expression for the area of triangle APQ in terms of $x$.

The base of triangle APQ is $AP = 9 + x$ and the height is $AQ = y$.
Area $A = \frac{1}{2}(9 + x)y$
Substitute $y = 4 + \frac{36}{x}$ from part (a):
$A = \frac{1}{2}(x + 9)\left(4 + \frac{36}{x}\right)$
$A = \frac{1}{2}\left(4x + 36 + 36 + \frac{324}{x}\right)$
$A = 2x + 36 + \frac{162}{x}$

2. Differentiate the area expression with respect to $x$.

$\frac{dA}{dx} = 2 - \frac{162}{x^2}$

3. Find the value of $x$ that gives the stationary point.

For minimum area, set $\frac{dA}{dx} = 0$:
$2 - \frac{162}{x^2} = 0$
$2x^2 = 162$
$x^2 = 81$
$x = 9$ (since $x > 0$ as it is a length)

4. Calculate the least possible area and verify it is a minimum.

$\frac{d^2A}{dx^2} = \frac{324}{x^3}$
When $x = 9$, $\frac{d^2A}{dx^2} = \frac{324}{9^3} > 0$, confirming it is a minimum.
Substitute $x = 9$ into the area expression:
$A = 2(9) + 36 + \frac{162}{9}$
$A = 18 + 36 + 18 = 72$