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📝 Extrema of two variable functions (13 MCQs)

📖 From Calculus • 14. Partial Derivatives Calculus • 13 questions available

←Previous Topic:Maxima and Minima of Functions of Two Variables📝 Extrema of two variable functionsBounded sets in multivariable calculus:Next Topic→

What is Extrema of two variable functions?

Definition:
Point $(a,b)$ is local max if $f(x,y) \leq f(a,b)$ nearby; local min if $\geq$ ; absolute extrema compare all local and boundary values.

Example:
$f(x,y) = -(x-1)^2-(y-2)^2+5$ has absolute max 5 at $(1,2)$ ; no minima on unbounded domain.

Reason:
Distinguishing local vs absolute extrema prevents suboptimal solutions; boundary consideration ensures global optimality in constrained domains.

Easy

Medium

Hard

📝 All Extrema of two variable functions MCQs

Q1. A function $f(x,y)$ has a critical point at $(0,0)$ where $f_{xx} = 0$ , $f_{yy} = 0$ , and $f_{xy} = 1$ . A student concludes it is a saddle point using the Second Derivative Test. Which statement best evaluates this reasoning?

A.The conclusion is correct because

D = -1 < 0

definitively indicates a saddle point regardless of zero second partials. ✅

B.The conclusion is incorrect because the test is inconclusive when

f_{xx} = 0

, requiring higher-order analysis.

C.The conclusion is correct but for the wrong reason; the sign of

f_{xy}

alone determines saddle behavior when pure partials vanish.

D.The conclusion is incorrect because

D = f_{xx}f_{yy} - (f_{xy})^2 = -1

actually indicates a local minimum when mixed partials dominate.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The discriminant $D = f_{xx}f_{yy} - (f_{xy})^2 = (0)(0) - (1)^2 = -1$ . Since $D < 0$ , the Second Derivative Test conclusively identifies a saddle point. The student’s conclusion is valid; the zero values of pure second partials do not invalidate the test when the discriminant is strictly negative, making option A the accurate evaluation of the reasoning presented.

Q2. An engineer models temperature on a metal plate as $T(x,y) = x^3 - 3xy^2$ . At the critical point $(0,0)$ , the Hessian determinant is zero. To classify this extremum in an applied context, which approach is most appropriate and why?

A.Compute third-order partial derivatives to apply the higher-order derivative test for degenerate critical points.

B.Restrict the function to lines

y = mx

and analyze sign changes of

T(x,mx)

near the origin to detect directional extrema behavior. ✅

C.Assume it is a saddle point since cubic functions in two variables always have saddles at degenerate critical points.

D.Use numerical optimization algorithms to sample nearby points and statistically infer the nature of the critical point.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: When the Hessian determinant vanishes, the Second Derivative Test fails. For $T(x,y) = x^3 - 3xy^2$ , restricting to $y = mx$ yields $T(x,mx) = x^3(1 - 3m^2)$ , which changes sign depending on $m$ , confirming a saddle. This directional analysis is more reliable than blind higher-order tests or numerical sampling for classification in modeling scenarios.

Q3. Given $f(x,y) = x^4 + y^4 - 4x^2y^2$ , a student claims $(0,0)$ is a local minimum because all fourth-degree terms are positive. Identify the fundamental flaw in this conceptual understanding.

A.Positive even-degree terms guarantee global minima only when cross terms are absent or dominated by pure terms.

B.The function can take negative values along certain paths despite positive pure quartic terms, violating the minimum condition.

C.Fourth-degree polynomials cannot have local minima at the origin unless they are sums of squares with positive coefficients.

D.The student confused necessary conditions for extrema with sufficient conditions, ignoring that positivity of individual terms does not imply positivity of the entire expression. ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: While $x^4$ and $y^4$ are nonnegative, the cross term $-4x^2y^2$ can dominate along paths like $y = x$ , yielding $f(x,x) = 2x^4 - 4x^4 = -2x^4 < 0$ for $x \neq 0$ . Thus, $(0,0)$ is not a minimum. The error lies in assuming term-wise positivity ensures function positivity, neglecting interaction effects between variables in multivariable extrema analysis.

Q4. A contour plot of $f(x,y)$ shows closed elliptical level curves near point $P$ , but at point $Q$ , level curves form hyperbolic patterns crossing each other. Without computing derivatives, what can be definitively concluded about extrema at these points?

P

is a local extremum and

Q

is a saddle point based solely on contour geometry. ✅

B.Both

P

and

Q

could be saddle points if the ellipses represent inflection-like behavior in 3D projection.

P

must be a global extremum within the plotted region, while

Q

indicates a degenerate critical point.

D.Contour shapes alone cannot distinguish extrema from non-critical points without gradient magnitude information.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Closed elliptical contours indicate the function increases or decreases monotonically toward $P$ , characteristic of local maxima or minima. Hyperbolic crossing contours at $Q$ signify directions of both increase and decrease, definitive of saddle points. This graph-based interpretation relies on topological properties of level sets and does not require derivative computation, making it a reliable visual classification method for extrema in applied settings.

Q5. Consider $f(x,y) = (x^2 + y^2)e^{-(x^2 + y^2)}$ . A student finds critical points at $(0,0)$ and on the circle $x^2 + y^2 = 1$ , then applies the Second Derivative Test at $(0,0)$ and gets $D = 0$ . What is the most efficient next step to classify $(0,0)$ ?

A.Switch to polar coordinates to reduce the problem to single-variable analysis of

g(r) = r^2 e^{-r^2}

near

r = 0

. ✅

B.Compute third-order partial derivatives to resolve the degenerate case via Taylor expansion.

C.Sample values numerically on a grid around the origin to empirically determine extremum type.

D.Conclude the test is inconclusive and report

(0,0)

as unclassifiable without further constraints.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: In polar form, $f = r^2 e^{-r^2}$ , and $g(r)$ has a clear minimum at $r = 0$ since $g'(r) = 2re^{-r^2}(1 - r^2) > 0$ for small $r > 0$ . This exploits radial symmetry to bypass messy Cartesian higher-order tests. Recognizing structural simplifications through coordinate transformation is a key application skill for efficiently resolving degenerate extrema in symmetric functions.

Q6. Two students analyze $f(x,y) = x^2y + xy^2$ at $(0,0)$ . Student A says it’s a saddle because $f(x,0) = 0$ and $f(x,x) = 2x^3$ changes sign. Student B argues it’s not a critical point since $\nabla f(0,0) \neq \vec{0}$ . Who is correct and why?

A.Student A is correct; sign change along paths confirms saddle despite zero gradient components.

B.Student B is correct;

f_x(0,0) = 0

and

f_y(0,0) = 0

, so it is critical, but their reasoning about non-zero gradient is flawed.

C.Neither is fully correct;

(0,0)

is critical and a saddle, but Student A’s path choice is insufficient without verifying gradient first. ✅

D.Student B is correct because

f_x = 2xy + y^2

and

f_y = x^2 + 2xy

, both zero at origin, contradicting their own claim.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: First, verify $\nabla f(0,0) = (0,0)$ , so it is critical—Student B’s premise is wrong. Student A correctly identifies sign change via $f(x,x) = 2x^3$ , but proper analysis requires confirming criticality before path testing. Option C captures that both students have flaws: B misstates gradient, and A skips verification. Multi-step reasoning demands checking prerequisites before applying classification methods to avoid erroneous conclusions.

Q7. For $f(x,y) = x^3 + y^3 - 3xy$ , the critical points are $(0,0)$ and $(1,1)$ . At $(1,1)$ , $D = 9 > 0$ and $f_{xx} = 6 > 0$ , indicating a local minimum. However, $f(2,-1) = 8 - 1 + 6 = 13 > f(1,1) = -1$ . Does this contradict the local minimum classification?

A.Yes, because a true local minimum must be less than all nearby values, and

(2,-1)

is sufficiently close.

B.No, local minima only require comparison within some neighborhood;

(2,-1)

may lie outside the basin of attraction. ✅

C.Yes, the Second Derivative Test gives false positives for cubic functions due to unboundedness.

D.No, but it implies

(1,1)

is actually a global minimum since no lower values exist.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Local extrema are defined relative to an open neighborhood, not globally. While $f(2,-1) > f(1,1)$ , this doesn’t violate locality; one must check if $(2,-1)$ is within the specific neighborhood where $f \geq f(1,1)$ . Cubic functions can have local minima without being global. The misconception confuses local and global properties, highlighting the need for precise definitions in extremum classification beyond computational results.

Q8. A constrained optimization problem uses Lagrange multipliers to find extrema of $f(x,y)$ subject to $g(x,y) = c$ . At a candidate point, $\nabla f = \lambda \nabla g$ holds, but the bordered Hessian is singular. What does this imply about the extremum classification?

A.The point cannot be an extremum; singularity indicates constraint qualification failure.

B.Classification requires analyzing higher-order bordered minors or reducing dimensionality via constraint substitution. ✅

C.The extremum is guaranteed to be degenerate and thus neither max nor min.

D.Lagrange multiplier method is invalid here; switch to penalty function methods immediately.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Singular bordered Hessian means the standard second-order sufficient conditions fail, but the point may still be an extremum. One must either examine higher-order terms in the Lagrangian expansion or substitute the constraint to reduce variables and apply unconstrained tests. This reflects advanced application where algorithmic outputs require deeper analytical intervention, distinguishing procedural compliance from genuine mathematical resolution in constrained extrema problems.

Q9. Compare classifying extrema of $f(x,y) = x^2 - y^2$ using the Second Derivative Test versus eigenvalue analysis of the Hessian. In what scenario does eigenvalue analysis provide decisive insight that the discriminant $D$ alone cannot?

A.When

D > 0

but eigenvalues have opposite signs, indicating misclassification risk.

B.Never;

D

and eigenvalue signs are mathematically equivalent for

2 \times 2

Hessians.

C.When extending to three or more variables, where

D

generalizes poorly but eigenvalues directly reveal definiteness. ✅

D.Only when the Hessian is diagonal, making eigenvalues trivial to compute versus calculating

D

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: In two variables, $D = \lambda_1 \lambda_2$ , so sign equivalence holds. But in $n > 2$ , the Hessian’s definiteness depends on all eigenvalue signs, not a single determinant. Relying on generalized discriminants becomes ambiguous, while eigenvalues unambiguously indicate positive/negative definiteness for extrema classification. This mixed-concept question highlights dimensional limitations of scalar tests and advocates spectral methods for robust multivariable analysis beyond textbook cases.

Q10. A student computes $f_{xx}f_{yy} - (f_{xy})^2 = 4 > 0$ and $f_{xx} = -2 < 0$ at a critical point, concluding a local maximum. Later, they discover $f_{xy}$ was miscalculated as 1 instead of 3. How does this error affect the original conclusion?

A.The conclusion remains valid since

D = (-2)f_{yy} - 9

could still be positive with adjusted

f_{yy}

B.The conclusion is definitely invalid because recalculating with

f_{xy} = 3

makes

D = (-2)f_{yy} - 9

, which is negative if

f_{yy}

unchanged.

C.The error might reverse the conclusion depending on

f_{yy}

; without its value, impact is indeterminate. ✅

D.Miscalculating

f_{xy}

never affects maxima/minima classification when

f_{xx} < 0

, only saddle detection.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Original $D = 4$ assumed $f_{xy} = 1$ . With $f_{xy} = 3$ , new $D = f_{xx}f_{yy} - 9$ . If original $f_{yy}$ satisfied $(-2)f_{yy} - 1 = 4 \Rightarrow f_{yy} = -2.5$ , then new $D = (-2)(-2.5) - 9 = 5 - 9 = -4 < 0$ , changing max to saddle. But if $f_{yy}$ were different, outcome varies. Error analysis requires propagating uncertainty through dependencies, not assuming fixed outcomes.

Q11. Which function has a critical point at $(0,0)$ where the Second Derivative Test is conclusive for a local minimum, yet the function takes smaller values arbitrarily close to the origin along a non-smooth path?

A.No such function exists; conclusive Second Derivative Test guarantees local minimality in all directions. ✅

f(x,y) = x^2 + y^4

, where parabolic paths show decrease despite positive definite Hessian.

f(x,y) = (x - y^2)(x - 2y^2)

, which has positive definite Hessian but negative values between parabolas.

D.Such functions violate continuity assumptions underlying the Second Derivative Test.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The Second Derivative Test’s sufficiency relies on $C^2$ smoothness and positive definiteness implying strict local minimum in a full neighborhood. If $D > 0$ and $f_{xx} > 0$ conclusively, no path—smooth or not—can yield smaller values nearby. Options B and C describe cases where the test is inconclusive or misapplied. This Olympiad-style question probes deep understanding of theorem hypotheses, exposing misconceptions about ‘conclusive’ meaning absolute guarantee under stated regularity conditions.

Q12. In modeling population dynamics, $P(x,y) = x(a - bx - cy)$ represents species growth. At coexistence equilibrium $(x^*, y^*)$ , the Hessian has negative trace and positive determinant. What does this imply biologically about stability and extremum type?

A.Unstable saddle point indicating competitive exclusion is inevitable.

B.Locally stable node corresponding to a local maximum of per-capita growth rate.

C.Asymptotically stable equilibrium representing a local maximum of total population fitness landscape. ✅

D.Indeterminate stability requiring center manifold analysis due to ecological nonlinearities.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Negative trace and positive determinant imply both eigenvalues have negative real parts, indicating asymptotic stability. In ecological models, equilibria often correspond to extrema of Lyapunov functions or fitness landscapes. Here, the stable node aligns with a local maximum in the effective potential governing dynamics. This integrates calculus extrema with dynamical systems interpretation, demanding translation between mathematical classification and biological meaning beyond rote computation.

Q13. A researcher claims that if $f(x,y)$ has a local minimum at $(a,b)$ , then every cross-section $f(x,b)$ and $f(a,y)$ must have a local minimum at that point. Is this claim valid, and what does its validity imply about verifying extrema?

A.Valid; it provides a necessary condition useful for quick elimination of non-extrema candidates. ✅

B.Invalid; cross-sectional minima are necessary but not sufficient, as demonstrated by

f(x,y) = x^2 + y^2 - 4x^2y^2

near origin.

C.Valid only for quadratic functions; higher-order interactions can create cross-sectional minima without true 2D minima.

D.Invalid; the converse is true—if cross-sections have minima, then

(a,b)

is a 2D minimum.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: If $f$ has a local min at $(a,b)$ , then restricting to $x$ - or $y$ -lines must yield 1D minima—this is necessary. However, it’s not sufficient, as shown by functions with positive cross-sections but negative diagonal values. The claim’s validity as a necessary condition makes it a practical screening tool in error analysis and preliminary verification, emphasizing hierarchical reasoning in multivariable extrema assessment without overclaiming sufficiency.

🔗 Related Topics (MCQs)

📝 Functions of Two or More Variables MCQs 📝 Multivariable function notation and terminology MCQs 📝 Functions of two variables from tables MCQs 📝 Level curves and contour plots MCQs 📝 Graphing functions of two variables MCQs

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