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?

Correct Option: Value-I = Value-II

Let us analyze the problem step-by-step to find the maximum values for both scenarios.

We are given that $p+q=10$, where $p$ and $q$ are integers.

Step 1: Calculate Value-I
Value-I is the maximum value of $p\times q$ when $p$ and $q$ are positive integers.
Since $p+q=10$ and both are positive integers, the possible pairs for $(p,q)$ are:
• $(1,9)\Rightarrow p\times q=9$
• $(2,8)\Rightarrow p\times q=16$
• $(3,7)\Rightarrow p\times q=21$
• $(4,6)\Rightarrow p\times q=24$
• $(5,5)\Rightarrow p\times q=25$
Thus, the maximum value of $p\times q$ for positive integers occurs when $p=5$ and $q=5$.
Value-I = 25.

Step 2: Calculate Value-II
Value-II is the maximum value of $p\times q$ when $p\ge <!--\; \ge \; -->-6$ and $q\ge <!--\; \ge \; -->-4$.
Let us analyze the range of values $p$ and $q$ can take under these constraints.
Since $p+q=10$:
• If $q=-4$, then $p=10-(-4)=14$. Here, $p\times q=14\times (-4)=-56$.
• If $p=-6$, then $q=10-(-6)=16$. Here, $p\times q=(-6)\times 16=-96$.
• For positive values of $p$ and $q$ satisfying $p\ge <!--\; \ge \; -->-6$ and $q\ge <!--\; \ge \; -->-4$ (for instance, $p=5,q=5$), $p\times q=25$.
Mathematically, for any two numbers with a constant sum $S=10$, their product $p\times q=p(10-p)=10p-p^2$ is maximized when $p=q=5$.
Since $p=5\ge <!--\; \ge \; -->-6$ and $q=5\ge <!--\; \ge \; -->-4$, the pair $(5,5)$ is valid in this range.
Value-II = 25.

Step 3: Comparison
Comparing Value-I and Value-II:
Value-I = 25
Value-II = 25
Therefore, Value-I = Value-II.