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📝 Constrained optimization problems (13 MCQs)

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

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What is Constrained optimization problems?

Definition:
Optimization where variables must satisfy equality/inequality constraints $g_i(\mathbf{x})=c_i$ , reducing feasible domain to subset.

Example:
Minimize cost $C(x,y)$ subject to production requirement $xy=100$ ; feasible set is hyperbola branch.

Reason:
Real-world problems inherently involve limitations; constrained methods extend calculus applicability beyond unconstrained idealizations.

Easy

Medium

Hard

📝 All Constrained optimization problems MCQs

Q1. A manufacturer designs a cylindrical can with fixed volume $V$ . If the cost of material for the top and bottom is twice that of the side, which condition must hold at the minimum cost configuration?

A.The height equals the diameter

B.The height equals twice the radius

C.The total cost derivative with respect to radius vanishes only when

h = 4r

✅

D.The Lagrange multiplier equals the marginal cost of volume

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: This problem requires setting up a constrained optimization where the objective function has weighted costs. Students must derive the relationship between dimensions by equating gradients, recognizing that unequal material costs shift the optimal ratio away from the standard geometric result where height equals diameter.

Q2. When maximizing $f(x,y)$ subject to $g(x,y)=c$ , a student finds a point where $\nabla f = \lambda \nabla g$ but $f$ is actually minimized there. What is the most likely conceptual error?

A.Assuming all critical points of the Lagrangian are maxima ✅

B.Forgetting to check the boundary of the constraint curve

C.Miscomputing the partial derivatives of

g

D.Using the wrong sign for

\lambda

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The method of Lagrange multipliers identifies stationary points of the constrained system but does not distinguish between maxima, minima, or saddle points. Students must apply second-order tests or compare function values; assuming every solution yields an extremum of the desired type is a common higher-order misconception.

Q3. Given a contour plot of $f(x,y)$ and a constraint curve $g(x,y)=c$ , at which geometric configuration is a constrained maximum most likely located?

A.Where the constraint curve crosses the highest contour line transversely

B.Where the constraint curve is tangent to a contour line and curves toward lower values ✅

C.Where the gradient of

f

is perpendicular to the constraint

D.Where the constraint passes through the origin

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Graph-based interpretation requires recognizing that tangency indicates a candidate extremum, but determining max versus min depends on how the constraint bends relative to increasing contours. If the constraint curves into regions of lower $f$ after tangency, that point is a local maximum under the constraint.

Q4. In optimizing $f(x,y,z)$ subject to two constraints $g=0$ and $h=0$ , which statement correctly describes the necessary condition at an extremum?

\nabla f

must be parallel to

\nabla g

only

\nabla f

lies in the plane spanned by

\nabla g

and

\nabla h

✅

\nabla f \cdot (\nabla g \times \nabla h) = 1

D.All three gradients must be mutually orthogonal

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: With multiple constraints, the gradient of the objective must be a linear combination of the constraint gradients, meaning it resides in their span. This reflects that no feasible direction exists to increase $f$ ; students often incorrectly assume pairwise parallelism instead of understanding the subspace condition in higher dimensions.

Q5. A student solves $\max xy$ subject to $x^2 + y^2 = 1$ using substitution $y = \sqrt{1-x^2}$ and misses the global maximum. What is the fundamental flaw in this approach?

A.The square root restricts

y

to non-negative values, omitting symmetric solutions ✅

B.The derivative becomes undefined at endpoints

C.Substitution always introduces extraneous solutions

D.The function

xy

is not differentiable on the circle

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Error analysis reveals that algebraic manipulation can implicitly reduce the domain. By choosing the positive root, the student ignores the third quadrant where $xy$ also achieves its maximum. This highlights the advantage of Lagrange multipliers in preserving symmetry and avoiding domain truncation during constrained optimization.

Q6. Which scenario best illustrates when the method of Lagrange multipliers fails to identify a true constrained extremum?

A.When the constraint defines a compact set

B.When

\nabla g = \mathbf{0}

at the optimal point ✅

C.When

f

is linear and

g

is quadratic

D.When there are multiple Lagrange multipliers

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The Lagrange condition requires $\nabla g \neq \mathbf{0}$ to define a smooth constraint manifold. If the gradient vanishes at the optimum, the constraint may have a singularity (e.g., cusp), and the standard multiplier equations become invalid. Recognizing regularity conditions is essential for correct application beyond mechanical computation.

Q7. Compare solving $\min x^2 + y^2$ subject to $x + y = 1$ via Lagrange multipliers versus direct substitution. Which insight is uniquely gained from the multiplier method?

A.The exact minimum value

B.That the solution lies on the line

y=x

C.The sensitivity of the minimum value to changes in the constraint constant ✅

D.That the constraint is linear

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: While both methods yield the same numerical answer, the Lagrange multiplier $\lambda$ quantifies the rate of change of the optimal value with respect to the constraint level. This shadow price interpretation connects optimization to economics and physics, offering conceptual depth that pure substitution cannot provide without additional differentiation.

Q8. An engineer models heat dissipation $T(x,y)$ on a plate with boundary $x^4 + y^4 = 1$ . Why might numerical methods be preferred over analytical Lagrange multipliers here?

A.The constraint is not differentiable anywhere

B.The resulting system of equations is highly nonlinear and lacks closed-form solutions ✅

C.Heat functions are never optimizable analytically

D.The boundary is unbounded

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Although $x^4 + y^4 = 1$ is smooth, the high-degree terms lead to polynomial systems resistant to symbolic solution. This scenario emphasizes practical modeling trade-offs: theoretical existence of extrema doesn’t guarantee tractable computation, pushing practitioners toward numerical approaches despite valid analytical foundations.

Q9. If $f(x,y)$ has a constrained maximum at $P$ under $g(x,y)=c$ , and $c$ increases slightly, what determines whether the new maximum value increases or decreases?

A.The sign of

\lambda

P

✅

B.The curvature of

f

P

C.Whether

g

is convex

D.The magnitude of

\nabla f

P

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The Lagrange multiplier $\lambda$ represents $df^*/dc$ , the sensitivity of the optimal value to the constraint parameter. A positive $\lambda$ means relaxing the constraint (increasing $c$ ) improves the objective. This links abstract calculus to real-world resource allocation decisions, requiring interpretation beyond solving equations.

Q10. A student claims that if $\nabla f$ and $\nabla g$ are parallel at a point on $g=c$ , then $f$ must have a local extremum there. Which counterexample disproves this?

f=x^2+y^2, g=x+y

f=x^3, g=x

x=0

✅

f=xy, g=x^2+y^2-1

(1,0)

D.No counterexample exists; the claim is always true

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: At $x=0$ for $f=x^3$ constrained to $x=0$ , gradients are trivially parallel, yet $f$ has an inflection, not an extremum, along the constraint. This tests deep understanding: parallel gradients are necessary but insufficient; second-order behavior or feasibility directions must also be considered to confirm extrema.

Q11. In a production model, output $Q(K,L)$ is maximized subject to budget $pK + wL = B$ . If wages $w$ rise, how does the optimal capital-labor ratio typically adjust, assuming diminishing marginal returns?

A.Capital decreases and labor increases to maintain output

B.Both inputs decrease proportionally

C.Capital increases relative to labor as firms substitute away from expensive labor ✅

D.The ratio remains unchanged due to fixed technology

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: This application integrates constrained optimization with economic intuition. Rising $w$ steepens the budget line, shifting tangency to a point with higher $K/L$ . Students must connect the mathematical condition $MP_L/w = MP_K/p$ to behavioral responses, demonstrating synthesis of calculus and microeconomic theory beyond rote calculation.

Q12. Consider maximizing $f(x,y) = x$ subject to $y^2 = x^3$ . At $(0,0)$ , $\nabla g = \mathbf{0}$ . Despite this, $f$ attains a minimum there. What does this imply about Lagrange multipliers?

A.They are unnecessary for polynomial constraints

B.The method gives sufficient but not necessary conditions

C.Regularity conditions are prerequisites for necessity; extrema can occur at singular points ✅

D.The constraint is not binding at the origin

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: This challenging case shows that constrained extrema may exist where $\nabla g = 0$ , violating the regularity assumption. Lagrange multipliers provide necessary conditions only under constraint qualifications. Advanced students must recognize that failure of the multiplier method doesn’t preclude extrema—it signals need for alternative analysis like parametrization or direct inspection.

Q13. A student uses Lagrange multipliers to minimize distance from origin to $y = x^3$ and obtains $\lambda = 0$ at $(0,0)$ . They conclude it’s not an extremum because $\lambda = 0$ . What is the error?

A.Confusing

\lambda = 0

with

\nabla g = 0

✅

B.Assuming

\lambda \neq 0

is required for minima

C.Miscalculating the gradient of the distance function

D.Believing distance functions cannot have zero multipliers

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Here $\lambda = 0$ arises because $\nabla f = \mathbf{0}$ at origin (for squared distance), not because the constraint is degenerate. The point is indeed the global minimum. The error lies in misattributing the cause of $\lambda = 0$ ; students must distinguish between objective stationarity and constraint singularity in interpreting multiplier values.

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