Solution: Take derivative $ A'(x) = -\frac{4}{x^2} - 2x $. Set $ A'(x) = 0 $: $ -\frac{4}{x^2} - 2x = 0 \Rightarrow -4 - 2x^3 = 0 \Rightarrow x^3 = -2 $. No positive roots. Analyze behavior: as $ x \to 0^+ $, $ A \to \infty $; as $ x \to \infty $, $ A \to -\infty $. Thus, no maximum exists; the function increases without bound near $ x = 0 $. However, if constrained, recheck calculations. Correct approach:

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