Question

Which of the following set is convex.
A. The set $\{(x_1, x_2) \mid \frac{x_1^2}{4} + \frac{x_2^2}{9} \le 1, x_1 \ge 0, x_2 \ge 0\}$ B. The set $\{(x_1, x_2) \mid x_1 x_2 \ge 1, x_1 \ge 0, x_2 \ge 0\}$ C. The set $\{(x_1, x_2) \mid x_1 \ge 1 \text{ or } x_2 \ge 1\}$ D. The set $\{(x_1, x_2) \mid x_1 + x_2 \le 1, x_1 \ge 0, x_2 \ge 0\}$ Choose the correct answer from the options given below:

• A. A, B, C Only
• B. A, B, D Only
• C. B, C, D Only
• D. A, D Only

Hint:

Sets defined with "OR" conditions (like $x_1 \ge 1 \text{ or } x_2 \ge 1$) correspond to unions of sets and are generally NOT convex!

Answer 1

The Correct Option is B

Step 1 : Concept:
This question tests convexity of sets in Linear Programming Problem (LPP). A set $S \subseteq \mathbb{R}^n$ is convex if for any two points $\mathbf{x}, \mathbf{y} \in S$, the line segment connecting them lies entirely within $S$, i.e., $\lambda \mathbf{x} + (1-\lambda)\mathbf{y} \in S$ for all $\lambda \in [0, 1]$.

Step 2 : Key Formulas and Approach:
1. Sublevel sets of convex functions are convex sets. 2. Intersections of convex sets are convex. 3. Half-spaces defined by linear inequalities are convex.

Step 3 : Step-by-step Explanation:

• Set A: The region $\frac{x_1^2}{4} + \frac{x_2^2}{9} \le 1$ represents a solid ellipse, which is a convex set. The conditions $x_1 \ge 0$ and $x_2 \ge 0$ define convex half-spaces. The intersection of these convex sets is convex. Hence, Set A is convex.

• Set B: The inequality $x_1 x_2 \ge 1$ with $x_1, x_2 \ge 0$ can be rewritten as $x_2 \ge \frac{1}{x_1}$. The function $f(x) = \frac{1}{x}$ is convex for $x > 0$ because $f''(x) = \frac{2}{x^3} > 0$. The epigraph of a convex function is a convex set. Hence, Set B is convex.

• Set C: Consider points $\mathbf{x} = (2, -10)$ and $\mathbf{y} = (-10, 2)$. For $\mathbf{x}$, $x_1 = 2 \ge 1$, so $\mathbf{x} \in C$. For $\mathbf{y}$, $y_2 = 2 \ge 1$, so $\mathbf{y} \in C$. Take the midpoint ($\lambda = 0.5$): \[ \mathbf{z} = \frac{1}{2}(2, -10) + \frac{1}{2}(-10, 2) = (-4, -4) \] For $\mathbf{z}$, $z_1 = -4 < 1$ and $z_2 = -4 < 1$, so $\mathbf{z} \notin C$. Thus, Set C is not convex.

• Set D: The region bounded by linear inequalities $x_1 + x_2 \le 1$, $x_1 \ge 0$, and $x_2 \ge 0$ forms a triangular region (polyhedron), which is convex. Hence, Set D is convex.

Step 4 : Final Answer:
Sets A, B, and D are convex. Therefore, option (B) is the correct answer.