Area constraint: $[\triangle APB] = \tfrac{1}{2}\cdot AB \cdot y = \tfrac{1}{2}\cdot 10\cdot y = 5y = 20 \Rightarrow y = 4.$

So $P$ lies on the horizontal line $y=4$, with $P=(x,4)$ for some real $x$.

Minimising $PA+PB$ on the line $y=4$: Reflect $A=(0,0)$ across this line to $A'=(0,8)$. For any $P$ on the line, $PA = PA'$, so

$

PA + PB = PA' + PB \ge A'B.

$

Equality iff $A', P, B$ are collinear. $A'B = \sqrt{(10-0)^{2} + (0-8)^{2}} = \sqrt{164} = 2\sqrt{41}.$

Line $A'B$: from $(0,8)$ to $(10,0)$, slope $-0.8$, equation $y = 8 - 0.8x.$ At $y=4$: $x = 5.$ So $P^{*} = (5,4).$

\textbf{Computing $PA\cdot PB$ at $P^{*}$:} By symmetry of $(5,4)$ relative to $A=(0,0)$ and $B=(10,0)$:

$

PA = \sqrt{25+16} = \sqrt{41}, \qquad PB = \sqrt{25+16} = \sqrt{41}.

$

Hence $PA\cdot PB = 41.$

Why this is CAT-level: The reflection trick is the canonical fast route, but the question disguises it -- students may try Lagrange multipliers, derivatives, or even ellipse properties (since $PA+PB$ constant defines an ellipse with foci $A,B$, and we want the smallest ellipse touching $y=4$). All routes converge, but the reflection method is by far the cleanest under time pressure. Wrong option (D) $50$ tempts those who compute $PA \cdot PB$ at $P=(0,4)$ (a boundary candidate, $\sqrt{16}\cdot\sqrt{116}\approx 4\cdot 10.77 = 43$, close to but not $50$); wrong option (A) $36$ corresponds to $P=(5,4)$ but with an area constraint of $25$ (a misreading).

Answer: (C) $41$.

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