Question

Let
\(𝑆 = x\left\{(𝑥, 𝑦) ∈ ℝ^2 : 𝑥 > 0, 𝑦 > 0\right\} ,\)
and f: S → ℝ be given by
\(f(x,y)=2x^2+3y^2-\log x-\frac{1}{6}\log y.\)
Then, which of the following statements is/are TRUE ?

• A. There is a unique point in S at which f(x, y) attains a local maximum
• B. There is a unique point in S at which f(x, y) attains a local minimum
• C. For each point (x₀, y₀) ∈ S, the set {(x, y) ∈ S : f(x, y) = f(x₀, y₀) } is bounded
• D. For each point (x₀, y₀) ∈ S, the set {(x, y) ∈ S : f(x, y) = f(x₀, y₀) } is unbounded

Answer 1

The Correct Option is B, C

- (A) The function \( f(x, y) = 2x^2 + 3y^2 - \log{x} - \frac{1}{6} \log{y} \) is convex in both \( x \) and \( y \) since both the quadratic terms are convex and the logarithmic terms are concave. The combination of convex and concave functions does not yield a local maximum, thus no local maximum exists. - (B) The function \( f(x, y) \) is convex, so it attains a unique global minimum. This minimum is found by setting the first partial derivatives equal to zero and solving for \( x \) and \( y \). The global minimum is unique. - (C) For each point \( (x_0, y_0) \in S \), the set of points \( \{ (x, y) \in S : f(x, y) = f(x_0, y_0) \} \) forms a level set. Since \( f(x, y) \) is continuous and differentiable, the level sets for continuous functions over bounded domains are bounded. - (D) The level set of \( f(x, y) \) for any given point \( (x_0, y_0) \) does not become unbounded. The quadratic and logarithmic terms imply that the level sets are bounded and do not stretch to infinity. Thus, the correct answers are (B) and (C).