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📝 Lagrange Multipliers in calculus (14 MCQs)

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

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What is Lagrange Multipliers in calculus?

Definition:
Method finds extrema of $f(x,y)$ subject to constraint $g(x,y)=k$ by solving $\nabla f = \lambda \nabla g$ and $g=k$ simultaneously.

Example:
Maximize area $A=xy$ subject to perimeter $2x+2y=20$ ; solve $\langle y,x \rangle = \lambda \langle 2,2 \rangle$ ⇒ $x=y=5$ .

Reason:
Transforms constrained optimization into system of equations; geometric insight: optimum occurs where objective and constraint contours are tangent.

Easy

Medium

Hard

📝 All Lagrange Multipliers in calculus MCQs

Q1. A cylindrical can must hold volume $V$ . To minimize surface area $S$ , one sets $\nabla S = \lambda \nabla V$ . If a student solves this but obtains $\lambda < 0$ , what is the most likely conceptual error in their formulation?

A.They minimized volume instead of area.

B.They defined the constraint as

g(x,y,z) = V - \pi r^2 h

and interpreted

\lambda

as cost rather than a rate of change. ✅

C.They correctly solved it;

\lambda

is always negative for minimization problems involving geometry.

D.They forgot to include the top and bottom circles in the surface area function.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The Lagrange multiplier $\lambda$ represents the rate of change of the optimal objective value with respect to the constraint constant. For minimizing surface area subject to fixed volume, increasing volume should increase minimum area, implying $\lambda > 0$ . A negative value suggests a sign error in defining the constraint gradient or misinterpreting the physical meaning of the multiplier within the specific optimization context.

Q2. Consider optimizing $f(x,y)$ subject to $g(x,y)=c$ . At the optimal point, the level curve of $f$ and the constraint curve $g=c$ are tangent. If the curves intersect transversely (non-tangentially) at a feasible point $P$ , which statement best explains why $P$ cannot be an extremum?

A.The gradients are parallel but opposite.

B.The directional derivative of

f

along the constraint curve at

P

is non-zero, allowing movement to increase or decrease

f

. ✅

C.The Hessian matrix of the Lagrangian is indefinite at

P

D.The constraint qualification fails at point

P

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Higher-order thinking requires connecting geometric tangency to calculus. Tangency implies $\nabla f$ is orthogonal to the tangent space of the constraint, making the directional derivative zero. Transverse intersection means $\nabla f$ has a component along the constraint's tangent vector. Therefore, one can move along the constraint curve to find points with both higher and lower function values, precluding a local extremum.

Q3. An economist models utility $U(x,y)$ subject to budget $px+qy=I$ . The optimal bundle yields $\lambda^* = 5$ . If income increases by $\Delta I = 10$ , which prediction relies on the correct interpretation of $\lambda$ as a shadow price, assuming regularity conditions hold?

A.Utility increases by exactly 50 units.

B.Utility increases by approximately 50 units, representing the marginal utility of income. ✅

C.The new optimal bundle shifts such that the marginal rate of substitution becomes 5.

D.The budget line shifts outward, making

\lambda

irrelevant for small changes.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: This tests conceptual understanding of $\lambda$ in applied modeling. $\lambda$ measures the sensitivity of the optimal objective value to the constraint parameter. It provides a linear approximation $\Delta U \approx \lambda \Delta I$ . Students often confuse exact equality with approximation or misunderstand $\lambda$ as a price ratio rather than the rate of change of maximum utility with respect to income.

Q4. When optimizing $f(x,y,z)$ subject to two constraints $g(x,y,z)=0$ and $h(x,y,z)=0$ , the condition is $\nabla f = \lambda \nabla g + \mu \nabla h$ . Why is it insufficient to simply solve $\nabla f \times \nabla g = \vec{0}$ and ignore $h$ ?

A.Cross products only work in two dimensions.

B.This ignores the second constraint entirely; the solution must lie on the intersection curve where

\nabla f

lies in the plane spanned by both constraint normals. ✅

C.The cross product eliminates the Lagrange multipliers needed for verification.

D.It assumes

\nabla g

and

\nabla h

are always parallel.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Multi-step reasoning is required here. With two constraints, the feasible set is a curve. Optimality requires $\nabla f$ to be orthogonal to this curve's tangent, meaning $\nabla f$ must be a linear combination of both normals. Ignoring $h$ finds points where level surfaces of $f$ and $g$ are tangent, but these points generally do not satisfy $h=0$ or fail to optimize on the true intersection manifold.

Q5. A student attempts to maximize $f(x,y)=x+y$ subject to $x^2+y^2=0$ . They set up $\nabla f = \lambda \nabla g$ and find no solution because $\nabla g = (0,0)$ at the only feasible point. What does this reveal about the method's limitations?

A.The function has no maximum on this constraint.

B.The constraint qualification fails;

\nabla g = \vec{0}

at the feasible point, so the Lagrange multiplier theorem does not apply. ✅

C.The student made an algebraic error in computing partial derivatives.

D.Lagrange multipliers only work for inequality constraints.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Error analysis question. The single point $(0,0)$ is trivially the max/min, but the standard Lagrange condition requires $\nabla g \neq \vec{0}$ at the optimum. This highlights that the method identifies candidates under regularity assumptions. When the gradient vanishes, the constraint surface may have a singularity, and extrema can exist without satisfying $\nabla f = \lambda \nabla g$ . Students must recognize when theoretical prerequisites break down.

Q6. Given contour plots of $f(x,y)$ and constraint curve $g(x,y)=c$ , you observe three intersection points: A (tangent, $f$ contours increasing inward), B (transverse crossing), and C (tangent, $f$ contours decreasing inward). Without calculation, how would you classify these points?

A.A is local max, B is saddle/not extremum, C is local min. ✅

B.A is local min, B is not extremum, C is local max.

C.All three are potential extrema requiring second derivative test.

D.B is global max, A and C are minima.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Graph-based interpretation skill. Tangency indicates candidate extrema. Direction of increasing $f$ relative to constraint determines type: if moving along constraint away from tangency decreases $f$ , it is a local max. Transverse intersections cannot be extrema since $f$ changes monotonically through them. Visual reasoning complements analytical methods and helps verify computational results in constrained optimization scenarios.

Q7. To find the shortest distance from origin to surface $z=x^2+y^2-1$ , one minimizes $D^2=x^2+y^2+z^2$ subject to $g=x^2+y^2-z-1=0$ . A student gets four critical points but cannot determine which gives minimum distance. Which mixed-concept approach resolves this ambiguity efficiently?

A.Compute the bordered Hessian determinant at each point.

B.Substitute constraint into objective to reduce variables, then use unconstrained second derivative test.

C.Evaluate

D^2

at all critical points and compare numerically since the feasible set is closed and bounded below. ✅

D.Use geometric intuition: the closest point must have positive

z

-coordinate.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Combines Lagrange multipliers with practical evaluation strategy. While bordered Hessians provide theoretical classification, direct comparison is often more efficient when candidates are finite. This tests judgment in selecting appropriate tools. Students sometimes overcomplicate by insisting on second-order tests when simple evaluation suffices, or they forget that existence of minimum is guaranteed here, making comparison valid.

Q8. In maximizing profit $\Pi(K,L)$ subject to cost $C(K,L)=B$ , suppose $\frac{\partial \Pi}{\partial K} / \frac{\partial C}{\partial K} \neq \frac{\partial \Pi}{\partial L} / \frac{\partial C}{\partial L}$ at current input levels. What does this inequality imply about resource allocation efficiency?

A.The firm is at optimal allocation; ratios need not be equal.

B.Resources are misallocated; reallocating spending toward the input with higher marginal profit per dollar increases total profit. ✅

C.The constraint is binding too tightly, requiring budget increase.

D.Diminishing returns have set in for both inputs.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Application and conceptual synthesis. Equal marginal productivity per dollar is the economic interpretation of $\nabla \Pi = \lambda \nabla C$ . Inequality signals inefficiency. This connects abstract mathematics to real-world decision-making. Distractors reflect common misconceptions about optimality conditions or confuse necessary conditions with sufficient ones. Students must translate mathematical disequilibrium into actionable managerial insight.

Q9. A physics problem requires minimizing energy $E(x,y,z)$ on intersection of sphere $x^2+y^2+z^2=R^2$ and plane $ax+by+cz=d$ . Setting up two constraints yields system with symmetry. If $d=0$ , how does symmetry simplify finding extrema compared to general case?

A.No simplification occurs; full system must be solved.

B.The origin satisfies both constraints, making it automatically the minimum.

C.Extrema occur at antipodal points; solving reduces to eigenvalue problem for quadratic form restricted to plane. ✅

D.Only one Lagrange multiplier is needed since plane passes through origin.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Olympiad-style reasoning combining geometry and algebra. When plane passes through origin, the problem becomes finding principal axes of quadratic form on subspace. Symmetry guarantees paired solutions. Recognizing structural properties avoids brute-force computation. This tests ability to leverage problem-specific features rather than mechanically applying algorithms, distinguishing deep understanding from procedural knowledge.

Q10. Student claims: 'If $\nabla f = \lambda \nabla g$ at point $P$ on constraint $g=c$ , then $P$ must be a local extremum.' Which counterexample best refutes this claim while illustrating nuanced understanding?

f=x^2+y^2

g=x^2+y^2-1

: every point satisfies condition but none are extrema.

f=x^3

g=x-y=0

: condition holds at origin but it is inflection-like along constraint. ✅

f=x+y

g=x^2+y^2-1

: condition holds at two points, one max one min.

f=e^x

g=x=0

: condition never holds despite extremum existing.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Error analysis targeting overgeneralization. Necessary condition is not sufficient. Option B shows point satisfying Lagrange equations that is neither max nor min along constraint curve. Option A is degenerate (entire curve critical). Option C actually supports the claim. Option D illustrates constraint qualification failure. Selecting B demonstrates understanding that stationarity doesn't guarantee extremality, requiring second-order or comparative analysis.

Q11. When optimizing $f(x,y)$ subject to $g(x,y)=c$ , suppose numerical solver returns $\lambda \approx 0$ . Beyond computational error, what legitimate scenario produces this result?

A.The constraint is inactive at optimum.

B.The unconstrained optimum of

f

happens to lie exactly on the constraint surface. ✅

C.The constraint curve is asymptotic to level curves of

f

D.The objective function is constant everywhere.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Conceptual depth question. $\lambda=0$ means constraint doesn't affect optimal value locally; the free optimum satisfies constraint coincidentally. This differs from inactive constraints in inequality problems. Students often assume $\lambda \neq 0$ always or misinterpret near-zero values as errors. Understanding this edge case reveals grasp of multiplier as measure of constraint tightness and distinguishes equality from inequality constraint behavior.

Q12. Designing a rectangular box with fixed volume $V$ and minimal surface area, you derive $x=y=z$ via Lagrange multipliers. If material costs differ per face (top/bottom cost $2c$ , sides cost $c$ ), how does the optimal shape qualitatively change compared to uniform cost case?

A.Remains cube; cost scaling doesn't affect ratios.

B.Height decreases relative to base dimensions to reduce expensive top/bottom area. ✅

C.Height increases to compensate for cheaper side materials.

D.Shape becomes spherical to distribute cost evenly.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Scenario-based modeling extending standard problem. Modified objective $S' = 4xy + 2xz + 2yz$ changes optimality conditions. Higher top/bottom cost incentivizes reducing those areas, flattening the box. Tests transfer of method to realistic variations. Distractors include intuitive but incorrect compensatory reasoning or assumption of geometric invariance. Requires adapting mathematical framework to altered economic parameters while preserving core technique.

Q13. Comparing substitution method versus Lagrange multipliers for optimizing $f(x,y)$ on $g(x,y)=c$ : in which situation does Lagrange offer decisive advantage despite added variables?

A.When

g

can be easily solved for

y=h(x)

explicitly.

B.When dealing with multiple constraints or implicit relationships where elimination creates messy expressions. ✅

C.When the domain is unbounded and global extrema are sought.

D.When second derivative test is simpler in reduced dimension.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Mixed concepts evaluating method selection. Substitution works well for simple explicit constraints but fails for complex or multiple constraints. Lagrange handles implicit relations and higher dimensions systematically. This metacognitive question assesses strategic thinking beyond mechanical application. Students must weigh trade-offs between variable count and algebraic complexity, recognizing contexts where each method excels based on problem structure rather than preference.

Q14. Maximizing $f(x,y)=xy$ subject to $x^2+y^2=1$ yields four critical points. A student argues all give same $|f|$ due to symmetry, so classification is unnecessary. What flaw exists in this reasoning regarding optimization theory?

A.Symmetry doesn't guarantee equal objective values at critical points.

B.While magnitudes match, signs differ; maxima and minima are distinct optima requiring identification. ✅

C.The student ignored boundary points of the constraint.

D.Critical points from Lagrange include saddle points even on compact sets.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Challenging conceptual check. Symmetry ensures magnitude equality but not sign; $+\frac{1}{2}$ and $-\frac{1}{2}$ represent fundamentally different optima. Dismissing classification overlooks that optimization seeks specific extreme values, not just magnitudes. Tests precision in interpreting results. Distractor A is false here; C misunderstands compactness; D mischaracterizes constrained critical points. Correct answer emphasizes rigorous distinction between max/min despite symmetric structure.

🔗 Related Topics (MCQs)

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