Maximizing (p+q)(r+s) with distinct digits

To maximize the product of two bracketed sums (p+q) × (r+s), we must select the largest available single-digit positive numbers: \{9, 8, 7, 6\}.

To maximize a multiplied product X × Y when the total combined sum of all elements is fixed (9+8+7+6 = 30), the two independent sums ⟨MATH⟩X⟨/MATH⟩ and ⟨MATH⟩Y⟨/MATH⟩ must be configured to be as close in value to each other as possible.

Let's test distribution permutations:

Wait, let's re-verify if 225 can be achieved. p=9, q=6 \rightarrow 15. r=8, s=7 \rightarrow 15. All digits \{9, 6, 8, 7\} are distinct single digits. Their product is 15 × 15 = 225.

Self-Correction Check on Option boundaries: Let's look closely at the choices. (a) 230, (b) 225, (c) 224, (d) 221. Since 225 is perfectly valid and represents the absolute algebraic maximum for a sum of 30, it is the correct answer.

Answer: (b).