📝 Bounded sets in multivariable calculus (13 MCQs)

📖 Explanation: The error lies in conflating analytic properties with topological ones. Partial derivatives depend on local behavior near each point and do not impose global constraints like boundedness. For example, $f(x,y)=x+y$ has partials everywhere on the unbounded plane. Boundedness is a separate geometric condition not implied by differentiability, making option B the correct identification of the misconception.

📖 Explanation: The extreme value theorem requires the domain to be compact, meaning both closed and bounded in $\mathbb{R}^n$. Here, $S$ excludes its circular boundary (not closed) and extends infinitely rightward (unbounded). Even with smooth partials, a function could approach a supremum asymptotically without attaining it. This tests understanding that compactness, not just differentiability, ensures extremum existence.

📖 Explanation: Contour spacing reflects the magnitude of the gradient $\|\nabla f\|$, not the extent of the domain. A function like $f(x,y) = e^{-(x^2+y^2)} \sin(x^2+y^2)$ has increasingly dense contours at infinity yet is defined on all $\mathbb{R}^2$. The misconception confuses rate of change with spatial extent. Proper analysis requires distinguishing local derivative information from global domain topology.

📖 Explanation: Only the annulus is both closed and bounded (compact) in $\mathbb{R}^2$, satisfying the extreme value theorem’s hypotheses. The open disk lacks boundary points where extrema might occur; the half-plane and punctured plane are unbounded. Although $f$ has singularities at the origin, the annulus excludes it, ensuring continuity throughout. This applies topological criteria rather than computing critical points directly.

📖 Explanation: A set is bounded if there exists $M > 0$ such that $\|(x,y)\| \leq M$ for all points in the set. The vertical line $x=2, y \in \mathbb{R}$ contains points with arbitrarily large $y$, so no such $M$ exists. Student A incorrectly assumes partial boundedness suffices. This reinforces that boundedness is a global property requiring uniform containment, not component-wise assessment.

📖 Explanation: Boundedness plus a gradient bound implies Lipschitz continuity regardless of closure, as $|f(p)-f(q)| \leq K \|p-q\|$ holds for all $p,q \in D$. However, without closedness, $D$ isn’t compact, so extrema aren’t guaranteed (e.g., $f(x)=x$ on $(0,1)$). Uniform continuity follows from Lipschitz but isn’t the strongest conclusion. Option C captures both the valid implication and the limitation regarding extrema.

📖 Explanation: Engineers confuse empirical data scope with mathematical domain properties. Smoothness and finite readings within a sampled area don’t imply the physical domain is bounded. Temperature could grow linearly or exponentially outside sensor coverage while maintaining continuous partials locally. Relying on boundedness without verification risks missing extreme values in unmodeled regions. This highlights the danger of equating observational limits with topological constraints in applied modeling.

📖 Explanation: Set $A$ is closed and bounded, hence compact, ensuring $f$ attains max/min by EVT. Set $B$ is unbounded in $x$; although $\sin(y) \in [-1,1]$, $f(x,y) = x \sin(y)$ achieves arbitrarily large positive and negative values as $x \to \infty$ when $\sin(y) \neq 0$. Thus, no global extrema exist on $B$. This contrasts compact vs. non-compact domains under identical functional form.

📖 Explanation: The student conflates function range boundedness with domain boundedness. While $\arctan(x)+\arctan(y)$ maps $\mathbb{R}^2$ to $(-\pi,\pi)$, the domain itself contains points with arbitrarily large norms. Metric boundedness depends solely on point distances, not function values. Bounded partials ensure Lipschitz continuity but say nothing about domain extent. This Olympiad-style question exposes deep confusion between codomain and domain properties in multivariable analysis.

📖 Explanation: For unconstrained interior optima to be global maxima, concavity ensures any critical point is a global maximizer. Combined with a bounded (hence compact if closed) feasible region, this avoids boundary checks. Open regions lack guaranteed extrema even with concavity. Positive Hessian indicates local minima, not maxima. Vanishing partials alone don’t ensure global optimality without convexity/concavity structure. This integrates boundedness, calculus, and optimization theory.

📖 Explanation: When $x=0$, $0 \cdot y^2 = 0 \leq 1$ for all $y$, so the entire y-axis belongs to $S$; similarly for the x-axis. Thus $S$ contains unbounded lines. Lagrange multipliers find critical points but don’t guarantee global extrema without compactness. Solutions may exist but not be global optima. This tests recognition that algebraic constraints can hide unbounded components, undermining standard optimization assumptions.

📖 Explanation: Boundedness depends only on whether the set fits in a ball, independent of function behavior. A bounded region like $[0,1]^2$ can host functions with singular partials at boundaries (e.g., $f(x,y)=\sqrt{1-x}$). The extreme value theorem requires continuity on a compact set; discontinuous partials imply possible discontinuity, so extrema aren’t guaranteed despite boundedness. This separates topological domain properties from analytical function regularity.

📖 Explanation: Although $D$ is compact, $f$ is discontinuous at $(0,0)$: approaching along $y=x$ gives limit $1/2 \neq f(0,0)$. The extreme value theorem requires continuity on the entire compact set. Without it, $f$ approaches $1/2$ but never attains it near the origin, so no maximum exists. This subtle case shows compactness is necessary but not sufficient without continuity, testing deep understanding of EVT conditions beyond mere boundedness.