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📝 Extreme value theorem multivariable (14 MCQs)

📖 From Calculus • 14. Partial Derivatives Calculus • 14 questions available

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What is Extreme value theorem multivariable?

Definition:
Continuous $f$ on closed bounded (compact) set $D$ attains absolute maximum and minimum values at some points in $D$ .

Example:
$f=x^2+y^2$ on disk $x^2+y^2 \leq 1$ achieves min 0 at origin and max 1 on boundary circle.

Reason:
Guarantees optimization solvability for continuous functions on nice domains; motivates checking interior critical points AND boundary separately.

Easy

Medium

Hard

📝 All Extreme value theorem multivariable MCQs

Q1. A continuous function $f(x,y)$ is defined on the set $S = \{(x,y) \mid x^2 + y^2 < 1\}$ . Despite being smooth and bounded, $f$ fails to attain a global maximum on $S$ . Which theoretical gap explains this failure?

A.The partial derivatives do not exist at the origin.

B.The function is not differentiable at the boundary.

C.The domain

S

is open and therefore not compact. ✅

D.The Hessian matrix is indefinite everywhere.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The Extreme-Value Theorem strictly requires the domain to be both closed and bounded (compact) in addition to the function being continuous. Since $S$ is an open disk, it lacks its boundary points, allowing the function to approach a supremum asymptotically without ever actually achieving that maximum value within the set.

Q2. Consider $f(x,y) = x^2 - y^2$ restricted to the closed unit disk. A student claims the global extrema must occur where $\nabla f = \mathbf{0}$ . Why is this reasoning fundamentally flawed for finding global extrema on this domain?

A.The gradient never equals zero on this domain.

B.Global extrema on closed bounded sets can occur on the boundary where

\nabla f \neq \mathbf{0}

. ✅

C.The function is not continuous on the unit disk.

D.Critical points only identify local minima, never maxima.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: While critical points where $\nabla f = \mathbf{0}$ are candidates, the Extreme-Value Theorem guarantees existence on compact sets but does not restrict location to interior stationary points. For this saddle function, the global maximum and minimum actually occur on the boundary circle, necessitating separate boundary analysis via parameterization or Lagrange multipliers.

Q3. Given a contour plot of a continuous function on a closed rectangular region where level curves become infinitely dense near the top-right corner but no critical point markers exist there, what can be definitively concluded about the global maximum?

A.It occurs at an interior critical point not shown.

B.The function has no global maximum despite continuity.

C.The global maximum occurs at the top-right corner boundary. ✅

D.The contour plot indicates a removable discontinuity.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Densely packed contour lines indicate rapid change and typically extreme values. Since the EVT guarantees a maximum exists on this closed bounded region and no interior critical points are indicated, the extremum must lie on the boundary. The density pattern specifically suggests the maximum is located at that corner rather than along flat boundary segments.

Q4. A model for temperature distribution $T(x,y)$ on a metal plate includes a point heat source creating a singularity at $(0,0)$ . If the plate is the closed unit square including the origin, why does the EVT fail to guarantee a maximum temperature?

A.The square is not a bounded region.

B.Temperature functions are inherently non-differentiable.

C.Continuity is violated at the singularity, breaking EVT hypotheses. ✅

D.Partial derivatives are undefined at all boundary points.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The Extreme-Value Theorem has two necessary conditions: continuity of the function and compactness of the domain. Even though the closed unit square is compact, the presence of a singularity means $T$ is not continuous at $(0,0)$ . Without continuity, the theorem provides no guarantee that any finite maximum temperature actually exists or is attained.

Q5. When optimizing $f(x,y) = xy$ on the closed triangular region with vertices $(0,0), (1,0), (0,1)$ , which multi-step procedure correctly applies EVT principles to ensure no candidate is missed?

A.Set

\nabla f = \mathbf{0}

and evaluate only interior solutions.

B.Evaluate

f

at vertices and the single interior critical point.

C.Check interior critical points, then parameterize and optimize

f

on each of three boundary segments, comparing all candidates. ✅

D.Use Lagrange multipliers on the entire triangular region simultaneously.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Proper application requires exhaustive search: first finding interior critical points where $\nabla f = \mathbf{0}$ , then treating each boundary edge as a separate constrained optimization problem (often reducing to single-variable calculus), and finally evaluating vertices. Simply checking vertices and one critical point misses potential extrema occurring at non-vertex boundary locations where the directional derivative along the edge vanishes.

Q6. A student argues that since $f(x,y) = e^{-(x^2+y^2)}$ is continuous everywhere, it must have a global minimum by the EVT. What is the precise error in applying the theorem to this unbounded domain?

A.The exponential function is not continuous at infinity.

B.EVT requires compact domains;

\mathbb{R}^2

is closed but unbounded. ✅

C.The function has no critical points in

\mathbb{R}^2

D.Global minima cannot exist for positive functions.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The Extreme-Value Theorem demands both closed AND bounded (compact) domains. While $\mathbb{R}^2$ is closed, it is unbounded, so EVT does not apply. In this case, $f$ approaches 0 asymptotically but never attains it, confirming no global minimum exists. Students often confuse 'closed' with 'compact' or assume continuity alone suffices for extremum existence.

Q7. Two surfaces $f$ and $g$ are continuous on the same closed disk. Graphs show $f$ has higher peak but $g$ has deeper valley. If $h = f + g$ , which statement about $h$ 's global extrema follows from EVT and superposition principles?

h

's maximum equals sum of individual maxima.

h

's extrema locations are averages of

f

and

g

extrema locations.

C.EVT guarantees

h

has global max and min, but their values/locations require analyzing

f+g

directly. ✅

h

inherits the convexity properties of both functions.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: While EVT guarantees that the sum of continuous functions on a compact set attains global extrema, the location and value cannot be deduced simply from individual extrema. Superposition can create new critical points or shift extrema to boundaries. Students must recognize that existence is preserved under addition, but specific extremal properties are not linearly combinable without full re-analysis.

Q8. In modeling population density $P(x,y)$ over a closed habitat, data shows $P$ is continuous but measurements suggest no stable equilibrium. If EVT guarantees a maximum density exists, what practical implication does this have for resource allocation planning?

A.Maximum density always occurs at geographic center.

B.Planning can ignore boundary regions since maxima are interior.

C.Resource capacity must accommodate the guaranteed maximum, even if its exact location requires further computation. ✅

D.Continuous models cannot represent real populations accurately.

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: EVT assures planners that some finite maximum population density exists somewhere in the habitat, providing a theoretical upper bound for infrastructure design. However, the theorem is existential, not constructive—it doesn't reveal where or what that maximum is. Effective planning thus requires additional optimization techniques to locate the maximum, while trusting that a worst-case scenario is well-defined and finite.

Q9. A function $f$ on a closed annulus has $\nabla f \neq \mathbf{0}$ everywhere in the interior. A peer concludes no global extrema exist. Refute this using EVT and geometric reasoning about multiply-connected domains.

A.Annuli are not compact, so EVT doesn't apply.

B.Global extrema must occur on either inner or outer circular boundary when no interior critical points exist. ✅

C.Multiply-connected domains cannot support continuous functions.

D.The gradient condition implies discontinuity at the hole.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Compactness depends on closed and bounded, not simple connectivity—an annulus is compact. With no interior critical points, EVT still guarantees extrema exist, forcing them onto the boundary. For an annulus, this means checking both the inner and outer circles. The misconception confuses topological complexity with compactness; EVT applies regardless of holes as long as the set is closed and bounded.

Q10. During numerical optimization of $f(x,y)$ on a closed square, an algorithm converges to $(0.5, 0.5)$ with $\nabla f \approx \mathbf{0}$ . Verification shows $f(0,0) > f(0.5,0.5)$ . What does this reveal about algorithmic limitations versus EVT guarantees?

A.EVT was incorrectly applied to the square.

B.Numerical methods may find local, not global, extrema; EVT ensures global exists but doesn't guide algorithms to it. ✅

C.The function is discontinuous at corners.

D.Gradient-based methods fail on squares specifically.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: EVT guarantees a global maximum exists on the compact square, but provides no mechanism to find it. Gradient descent naturally converges to nearby critical points, potentially missing higher boundary values. This discrepancy highlights that theoretical existence (EVT) and computational discovery are distinct challenges. Robust optimization requires combining EVT-aware boundary checks with interior search, not relying solely on gradient convergence.

Q11. If $f(x,y)$ is continuous on a closed bounded region $D$ and $f(a,b) = M$ is the global maximum at an interior point $(a,b)$ , which additional condition MUST hold that distinguishes this from a boundary maximum?

\nabla f(a,b) = \mathbf{0}

and the Hessian is negative semi-definite.

f

must be differentiable at

(a,b)

with

\nabla f(a,b) = \mathbf{0}

. ✅

C.The value

M

must exceed all boundary values.

D.All second partial derivatives must be negative at

(a,b)

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: For an interior global maximum where $f$ is differentiable, Fermat's theorem for multivariable functions requires $\nabla f = \mathbf{0}$ . This is a necessary condition distinguishing interior extrema from boundary ones, where gradients need not vanish. Note that negative semi-definiteness of Hessian is sufficient but not necessary for maximality, and exceeding boundary values is a consequence, not a prerequisite condition derived from EVT itself.

Q12. A physics model defines energy $E(x,y,z)$ on a closed ball. Symmetry suggests extrema at center, but calculations show $\nabla E \neq \mathbf{0}$ at origin. Using EVT and symmetry-breaking concepts, where should one next search for global extrema?

A.Symmetry guarantees center is extremum; recalculate gradient.

B.Search entire spherical surface boundary and any asymmetric interior critical points. ✅

C.Energy functions cannot have extrema in balls.

D.Only check points where

x=y=z

due to symmetry.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: When expected symmetric critical points fail to satisfy $\nabla E = \mathbf{0}$ , symmetry is broken and extrema may occur at unexpected locations. EVT still guarantees existence on the compact ball, so one must systematically check: (1) actual interior critical points solving $\nabla E = \mathbf{0}$ without assuming symmetry, and (2) the entire boundary sphere. Relying on intuitive symmetry without verification is a common error in physical modeling.

Q13. Comparing optimization of $f(x,y) = x^2 + y^2$ on a closed disk versus an open disk: which pair of statements correctly contrasts EVT applicability and extremum attainment?

A.Both domains yield global minimum at origin; EVT applies equally.

B.Closed disk: EVT guarantees min at origin. Open disk: min exists at origin but EVT doesn't guarantee it. ✅

C.Open disk has no minimum; closed disk minimum is on boundary.

D.EVT applies to open disk but not closed disk due to boundary issues.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: On the closed disk, EVT guarantees both max and min exist; min occurs at origin. On the open disk, the same minimum value is attained at origin (which is included), but EVT cannot be invoked because the domain isn't compact. This subtle distinction shows that extremum attainment can occur without EVT, but only EVT provides the guarantee. Students often conflate 'attainment' with 'guaranteed attainment.'

Q14. A continuous function on a closed rectangle has exactly one critical point in the interior, which is a saddle point. Boundary analysis reveals constant function values along all four edges. What can be concluded about global extrema locations?

A.Global max and min both occur at the saddle point.

B.No global extrema exist due to saddle point.

C.Global max and min occur on boundary edges, but EVT alone cannot distinguish which edge without further analysis. ✅

D.Constant boundary values imply function is constant everywhere.

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Since the only interior critical point is a saddle (not an extremum), global extrema must lie on the boundary by EVT. Constant boundary values mean every boundary point is simultaneously a candidate for both max and min. However, without knowing interior behavior relative to boundary constant, we cannot determine if boundary represents max, min, or both. EVT confirms existence on boundary but additional evaluation is needed to classify.

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