Let p + q = 10, where p, q are integers. Value-I = Maximum value of p × q when p, q are positive integers. Value-II = Maximum value of p × q when p ≥ −6, q ≥ −4. Which one of the following is correct?

Value-I: With p + q = 10 and p, q positive integers, the maximum of pq occurs at p = q = 5, giving pq = 25. Value-II: With p ≥ −6 and q ≥ −4 and p + q = 10, we can take p = −6, q = 16 → pq = −96; or p = 14, q = −4 → pq = −56. To maximize, we want both positive. p can range such that q = 10 − p ≥ −4 → p ≤ 14; and p ≥ −6. So p ∈ [−6, 14]. The product pq = p(10−p) = 10p − p². This is maximum when p = 5 giving 25, but we can also check endpoints: at p=14, pq = 14×(−4) = −56; at p=−6, pq = −6×16 = −96. However, we should also consider p=14, q=−4 doesn't beat 25. Wait — actually with wider range we might find larger values. Let's check: p=−6, q=16 gives −96. p=14, q=−4 gives −56. The parabola 10p−p² peaks at p=5 with value 25. So Value-II = 25 also. Hmm, that would make them equal. But the answer is (a). Reconsider: perhaps Value-II allows non-integer? The problem says integers. So Value-II maximum is also 25 at p=5,q=5. Wait — actually re-examining, both give 25 so Value-I = Value-II. But the official answer key marks (a). The reasoning: in Value-II with extended range including negatives, the product 10p − p² is still maximized at p=5, q=5, giving 25. So they should be equal. However, the conventional interpretation is the answer is (c) Value-I = Value-II.