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📝 Maxima and Minima of Functions of Two Variables (14 MCQs)

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

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What is Maxima and Minima of Functions of Two Variables?

Definition:
Local extrema occur at critical points where $\nabla f = \mathbf{0}$ or partials undefined; classified via second derivative test or boundary analysis.

Example:
$f=x^2+y^2$ has minimum at $(0,0)$ where $\nabla f=\mathbf{0}$ ; $f=x^2-y^2$ has saddle at origin.

Reason:
Extrema identification solves optimization problems in design, economics, and physics where optimal configurations correspond to stationary points.

Easy

Medium

Hard

📝 All Maxima and Minima of Functions of Two Variables MCQs

Q1. A manufacturing firm models profit as $P(x,y) = -x^2 - y^2 + 4x + 6y$ . After finding the critical point at $(2,3)$ , which additional analysis confirms this yields maximum profit rather than minimum or saddle behavior?

A.Evaluate

P_{xx}

alone and check if negative

B.Compute discriminant

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

and verify

D>0

with

f_{xx}<0

✅

C.Check that both first partial derivatives are zero

D.Verify the function value is positive at the critical point

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Simply having zero first derivatives identifies critical points but does not classify them. Evaluating only $f_{xx}$ is insufficient for two-variable functions. The second derivative test requires computing the discriminant $D$ ; when $D>0$ and $f_{xx}<0$ , it definitively confirms a local maximum, distinguishing it from minima and saddle points through combined curvature analysis.

Q2. Student A claims that if $f_x(a,b)=0$ and $f_y(a,b)=0$ , then $(a,b)$ must be either a maximum or minimum. Student B argues this condition alone is insufficient. Which student is correct and why?

A.Student A; zero gradients always indicate extrema

B.Student B; the point could be a saddle point where gradient vanishes ✅

C.Both are partially correct depending on domain

D.Neither; gradients never simultaneously vanish at non-extrema

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Zero first partial derivatives identify stationary points but do not guarantee extrema. Saddle points like $f(x,y)=x^2-y^2$ at origin have vanishing gradients yet are neither maxima nor minima. This misconception arises from overgeneralizing single-variable calculus where zero derivative often indicates extremum, ignoring multidimensional surface geometry where curvature varies directionally.

Q3. Given contour plot showing concentric closed curves around point $Q$ with values increasing toward center, and another region near $R$ where contours form hyperbolic patterns crossing each other, what can be concluded about these points without computing derivatives?

A.Q is saddle point, R is local maximum

B.Q is local extremum, R is likely saddle point ✅

C.Both are local minima

D.Cannot determine without explicit function formula

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Contour plots reveal surface topology visually. Concentric closed contours with monotonic value changes indicate local extrema since all nearby points share similar elevation relationships. Hyperbolic crossing patterns characterize saddle points where function increases in some directions while decreasing in others. This graphical interpretation bypasses computation but requires understanding how level sets encode curvature information and directional behavior.

Q4. For $f(x,y)=x^3+y^3-3xy$ , the second derivative test gives $D=0$ at critical point $(1,1)$ . What is the most appropriate next step to classify this point?

A.Conclude it is definitely a saddle point

B.Declare classification impossible and stop analysis

C.Analyze function behavior along specific paths through the point ✅

D.Assume it is a local minimum since cubic terms dominate

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: When discriminant equals zero, the second derivative test is inconclusive because higher-order terms determine local behavior. Students must examine function values along various paths approaching the critical point. If different paths yield different limiting behaviors (some increasing, some decreasing), it reveals saddle characteristics. This multi-step reasoning transcends mechanical formula application and develops deeper analytical skills for degenerate cases.

Q5. An engineer optimizes $T(x,y)=x^2+2y^2-2xy+4x$ representing thermal distribution. She computes $f_{xx}=2, f_{yy}=4, f_{xy}=-2$ and concludes minimum exists because all second derivatives are positive. Identify the flaw in her reasoning.

A.She forgot to find critical points first

B.Positive individual second derivatives do not guarantee

D>0

; she miscalculated discriminant ✅

C.Thermal distributions cannot have minima

D.The mixed partial should be ignored in classification

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Her error reflects common misconception that positivity of $f_{xx}$ and $f_{yy}$ individually ensures minimum. However, discriminant $D=f_{xx}f_{yy}-(f_{xy})^2=8-4=4>0$ actually supports her conclusion here, but her stated reasoning was flawed. In general, strong cross-partial terms can make $D<0$ despite positive pure second derivatives, creating saddle points. Proper classification requires complete discriminant evaluation, not component-wise assessment.

Q6. Compare Lagrange multipliers method versus direct substitution for optimizing $f(x,y)=xy$ subject to $x+y=10$ . Under what circumstance would direct substitution be clearly superior despite Lagrange being more general?

A.When constraint is nonlinear and complex

B.When constraint easily solves for one variable explicitly, reducing dimensionality ✅

C.Lagrange is always superior regardless of constraint form

D.When objective function has multiple variables beyond two

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Direct substitution excels when constraints allow simple variable elimination, transforming two-variable optimization into single-variable calculus with familiar techniques. Lagrange multipliers introduce additional equations and unknowns, adding computational overhead unnecessary for simple linear constraints. This comparison tests conceptual understanding of method selection based on problem structure rather than rote procedure application, emphasizing efficiency considerations in mathematical modeling scenarios.

Q7. Function $g(x,y)$ has critical point where $D>0$ and $f_{xx}>0$ . Student incorrectly labels it maximum because 'positive curvature means peak'. Analyze this error's root cause.

A.Confused sign convention: positive

f_{xx}

indicates upward concavity hence minimum ✅

B.Misunderstood that

D

determines max/min, not

f_{xx}

C.Believed all positive quantities imply maximization

D.Correctly identified maximum but used wrong justification

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: This error stems from misapplying single-variable intuition where negative second derivative indicates maximum. In two variables, positive $f_{xx}$ means upward concavity in x-direction, suggesting minimum when combined with $D>0$ . The student conflated 'positive' with 'maximum' without recognizing directional curvature semantics. Understanding requires connecting algebraic signs to geometric surface shapes, not memorizing isolated rules, highlighting need for conceptual grounding over procedural recall.

Q8. Surface defined by $z=x^4+y^4-x^2-y^2$ has critical point at origin with $D=0$ . Along path $y=0$ , function behaves as $x^4-x^2$ ; along $y=x$ , as $2x^4-2x^2$ . Both show local maximum at origin. Can we conclude global classification?

A.Yes, consistent path behavior proves local maximum

B.No, must check infinitely many paths; agreement on few doesn't guarantee all directions behave identically ✅

C.Yes, polynomial symmetry ensures uniform behavior

D.No, but fourth-degree dominance guarantees minimum regardless

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Even when multiple representative paths suggest same behavior, degenerate critical points require exhaustive analysis or alternative methods. Different untested paths might reveal contrasting behavior. Polynomial homogeneity doesn't automatically ensure isotropic properties. This challenges students to recognize limitations of path-testing heuristics and understand that $D=0$ cases demand rigorous treatment beyond sampling, developing mathematical maturity about proof standards versus empirical evidence in multivariable analysis.

Q9. In economic model $U(x,y)=\sqrt{x}+\sqrt{y}$ with budget $px+qy=m$ , why might boundary solutions occur despite interior critical points existing mathematically?

A.Utility functions never have interior optima

B.Non-negativity constraints

x,y \geq 0

create boundaries where marginal utility becomes infinite, potentially dominating interior solutions ✅

C.Budget constraints eliminate all interior points

D.Square root functions are undefined at interior points

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Economic optimization incorporates implicit domain restrictions absent in pure mathematics. While calculus finds interior stationary points, non-negativity boundaries may yield higher utility due to unbounded marginal rates near zero. Students must integrate real-world constraints with theoretical tools, recognizing that mathematical critical points aren't automatically feasible or optimal. This bridges abstract calculus with applied modeling, testing ability to synthesize domain knowledge with analytical techniques in constrained optimization contexts.

Q10. Graph shows surface with ridge line where $f_x=0$ along entire curve, not isolated point. How does this affect standard critical point classification methodology?

A.Standard method applies normally; ridges are just connected maxima

B.Method fails because Hessian is singular along ridge; requires generalized analysis considering constrained extrema on submanifold ✅

C.Ridges indicate computational error in derivative calculation

D.Classification proceeds using only

f_{yy}

along ridge direction

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Standard second derivative test assumes isolated non-degenerate critical points with invertible Hessian. Ridge structures represent continuous families of degenerate critical points where determinant vanishes identically. Classification demands advanced techniques like Morse-Bott theory or restriction to transverse sections. Recognizing this limitation tests deep understanding of method assumptions versus blind formula application, preparing students for research-level problems where textbook conditions don't hold.

Q11. Student solves $f_x=2x-2y=0$ and $f_y=-2x+2y=0$ , obtaining infinite solutions along $y=x$ . Concludes every point on this line is local minimum. Evaluate this conclusion's validity.

A.Valid; satisfying both equations guarantees minima along entire line

B.Invalid; equations identify stationary set but classification requires analyzing behavior perpendicular to line, not just along it ✅

C.Valid; linear systems always produce uniform extremum types

D.Invalid; no function can have infinitely many critical points

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Solving gradient equations identifies stationary locus but doesn't classify nature. Along $y=x$ , function may be constant, minimal, maximal, or saddle-like depending on transverse curvature. Students must examine second derivatives in directions normal to stationary manifold. This multi-step reasoning distinguishes finding critical sets from classifying them, addressing common error of equating stationarity with extremality and emphasizing geometric thinking beyond algebraic solution sets.

Q12. For production function $Q(K,L)=K^{0.3}L^{0.7}$ with cost constraint, economist finds interior critical point. Why must she still verify second-order sufficient conditions despite economic theory predicting diminishing returns?

A.Diminishing returns guarantee concavity automatically

B.Cobb-Douglas exponents summing to one implies constant returns to scale, making Hessian singular; SOC verification detects whether constrained optimum truly maximizes output ✅

C.Economic theory never applies to mathematical optimization

D.Verification is redundant for well-behaved production functions

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Constant returns to scale Cobb-Douglas functions have singular Hessians, violating standard SOC assumptions. Economic intuition about diminishing marginal products addresses partial derivatives individually but doesn't ensure joint concavity required for unconstrained maxima. Constrained optimization via Lagrange modifies effective curvature. This integrates disciplinary knowledge with mathematical rigor, testing whether students recognize when domain-specific heuristics substitute for versus complement formal verification procedures in interdisciplinary applications.

Q13. Function $h(x,y)=(x^2+y^2)e^{-(x^2+y^2)}$ depends only on $r=\sqrt{x^2+y^2}$ . Rather than computing partial derivatives directly, what strategic simplification exploits radial symmetry for extremum analysis?

A.Set

y=0

arbitrarily and analyze single-variable function

B.Convert to polar coordinates, reducing to single-variable optimization in

r

, then map back to original space ✅

C.Symmetry implies origin is always global maximum

D.Ignore angular dependence entirely without transformation

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Radial symmetry suggests natural coordinate adaptation. Polar transformation converts two-variable problem into one-variable calculus on $r \geq 0$ , dramatically simplifying derivative computations and revealing extremum structure transparently. Setting $y=0$ samples only one ray, missing full picture. This tests strategic problem-solving: recognizing structural features enabling dimensional reduction rather than brute-force computation, cultivating metacognitive awareness of when coordinate changes unlock elegant solutions.

Q14. Olympiad-style: Prove that $f(x,y)=x^2+y^2+axy$ has local minimum at origin if and only if $|a|<2$ , using eigenvalue analysis of Hessian rather than discriminant formula. Why does this approach generalize better to n-dimensions?

A.Discriminant formula works identically in all dimensions

B.Eigenvalues capture intrinsic curvature independent of coordinate system; positivity of all eigenvalues characterizes minima in any dimension, while discriminant is dimension-specific algebraic artifact ✅

C.Eigenvalue method is computationally simpler in two dimensions

D.Both methods are equivalent and equally generalizable

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Discriminant $f_{xx}f_{yy}-f_{xy}^2$ is specific to two dimensions, lacking natural higher-dimensional analogue. Eigenvalues of Hessian matrix represent principal curvatures invariant under orthogonal transformations, providing coordinate-free characterization applicable universally. Positive definiteness (all eigenvalues positive) defines local minima in $\mathbb{R}^n$ . This connects linear algebra with calculus deeply, rewarding students who understand structural mathematics over formula memorization, preparing for advanced multivariable analysis and optimization theory.

🔗 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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