Answer: $(2, -3)$

Compute partial derivatives: $f_x = 2x - 4$ and $f_y = 2y + 6$. Set both equal to zero: $2x - 4 = 0 \Rightarrow x = 2$ and $2y + 6 = 0 \Rightarrow y = -3$. Why the correct answer works: The only critical point is $(2, -3)$, which is Option A. Why distractors fail: Option B has incorrect signs. Option C uses the coefficients $-4$ and $6$ directly without dividing by 2. Option D is the origin, which does not satisfy $f_x = 0$ or $f_y = 0$.

Answer: $\left(\dfrac{12}{7}, \dfrac{18}{7}, \dfrac{6}{7}\right)$

Set up the problem: Minimize $f(x,y,z) = x^2 + y^2 + z^2$ subject to $g(x,y,z) = 2x + 3y + z = 12$. Apply Lagrange conditions: $2x = 2\lambda$, $2y = 3\lambda$, $2z = \lambda$, so $x = \lambda$, $y = \frac{3\lambda}{2}$, $z = \frac{\lambda}{2}$. Solve using the constraint: $2\lambda + 3 \cdot \frac{3\lambda}{2} + \frac{\lambda}{2} = 12$ gives $2\lambda + \frac{9\lambda}{2} + \frac{\lambda}{2} = 12$, so $7\lambda = 12$, thus $\lambda = \frac{12}{7}$. Then $x = \frac{12}{7}$, $y = \frac{18}{7}$, $z = \frac{6}{7}$. Why distractors fail: Option B does not satisfy the constraint: $2(2) + 3(3) + 1 = 14 \neq 12$. Option C gives $2(4) + 0 + 4 = 12$ but $f = 32$, which is much larger than $f = \frac{504}{49} \approx 10.29$ at the correct point. Option D gives $0 + 12 + 0 = 12$ but $f = 16$, also larger.

Answer: $\nabla f = \lambda \nabla g$ and $g(x, y) = c$

The Lagrange multiplier condition: At a constrained extremum, the gradient of $f$ must be proportional to the gradient of $g$. This gives $\nabla f = \lambda \nabla g$ for some scalar $\lambda$, together with the constraint $g(x, y) = c$. Why the correct answer works: Option B correctly states both the gradient proportionality condition and the constraint equation. Why distractors fail: Option A sets the gradient of $f$ to zero, which finds unconstrained critical points. Option C omits the multiplier $\lambda$ and incorrectly constrains $f$ rather than $g$. Option D drops $\lambda$ and incorrectly sets $f = 0$ instead of enforcing $g = c$.

Answer: The student is correct; Lagrange multipliers always identify all extrema, and every point on the constraint is a maximum

Analyze the setup: On the constraint $x^2 + y^2 = 1$, the function $f = x^2 + y^2 = 1$ at every point. So $f$ attains its maximum (and minimum) value of $1$ at every point on the circle. Check the Lagrange conditions: $\nabla f = (2x, 2y)$ and $\nabla g = (2x, 2y)$. Setting $\nabla f = \lambda \nabla g$ gives $\lambda = 1$, and every point on the circle satisfies the system. The method correctly identifies all such points. Why the correct answer works: Option A is correct: the method works properly. Every point on the constraint is both a maximum and a minimum of $f$ on the constraint. Why distractors fail: Option B is wrong — the method does work; it simply returns all points. Option C is wrong — the circle is compact, and the Extreme Value Theorem applies. Option D incorrectly interprets the value of $\lambda$.

Answer: Yes, but the conclusion follows from the sign analysis of $f$ near the origin, not from the Second Derivative Test

Analyze the situation: At $(0,0)$: $f_{xx} = 2y = 0$, $f_{yy} = 0$, $f_{xy} = 2x = 0$. So $D = 0$, and the Second Derivative Test is inconclusive. Check sign behavior: Near the origin, $f(x, y) = x^2 y$ is positive when $y > 0$ and negative when $y < 0$. So $(0,0)$ is indeed a saddle point. Why the correct answer works: Option B correctly notes that the saddle-point conclusion comes from direct sign analysis, not from $D = 0$, which is inconclusive. Why distractors fail: Option A incorrectly claims $D = 0$ always gives a saddle point — it could also give a degenerate extremum. Option C is wrong; $D = 0$ is inconclusive. Option D is factually wrong: $f(1,1) = 1 > 0$ and $f(1,-1) = -1 < 0$.

Answer: Saddle point because $D < 0$

Compute the discriminant: $D = f_{xx} f_{yy} - (f_{xy})^2 = (5)(3) - (4)^2 = 15 - 16 = -1$. Interpret: Since $D = -1 < 0$, the critical point is a saddle point, regardless of the signs of $f_{xx}$ and $f_{yy}$. Why distractors fail: Option A incorrectly assumes $D > 0$ without computing it. Option B misapplies the test — positive $f_{xx}$ and $f_{yy}$ alone do not guarantee a minimum when $D < 0$. Option D is wrong because $D = -1 \neq 0$.

Answer: Local maximum

Compute the discriminant: $D = f_{xx} f_{yy} - (f_{xy})^2 = (-12)(-12) - (4)^2 = 144 - 16 = 128$. Apply the test: Since $D = 128 > 0$ and $f_{xx} = -12 < 0$, the critical point $(1, 1)$ is a local maximum. Why distractors fail: Option A would require $f_{xx} > 0$ with $D > 0$. Option C requires $D < 0$. Option D requires $D = 0$.

Answer: $(0, 0)$ and $(1, 1)$

Compute partial derivatives: $f_x = 3x^2 - 3y = 0$ gives $y = x^2$. $f_y = -3x + 3y^2 = 0$ gives $x = y^2$. Solve the system: Substituting $y = x^2$ into $x = y^2$: $x = x^4$, so $x^4 - x = 0$, giving $x(x^3 - 1) = 0$. Thus $x = 0$ or $x = 1$, yielding critical points $(0, 0)$ and $(1, 1)$. Why the correct answer works: Option C lists both valid critical points obtained from solving the system. Why distractors fail: Options A and B each miss one critical point. Option D includes $(-1, -1)$, which does not satisfy $f_x = 0$: $3(1) - 3(-1) = 6 \neq 0$.

Answer: The candidates are incomplete; solving for $\lambda$ in (1) and (2) reveals an additional relationship $x^2 - 6xy - 4y^2 = 0$, which yields further critical points.

Identifying the Hidden Points: To find all candidates, eliminate $\lambda$. From (1), $\lambda = \frac{2x+y}{2x}$ (for $x \neq 0$). From (2), $\lambda = \frac{x+2y}{8y}$ (for $y \neq 0$). Equating them: $\frac{2x+y}{2x} = \frac{x+2y}{8y} \implies 8y(2x+y) = 2x(x+2y) \implies 16xy + 8y^2 = 2x^2 + 4xy$. Simplifying gives $x^2 - 6xy - 4y^2 = 0$. Why Option B is the Correct Critique: The relationship $x^2 - 6xy - 4y^2 = 0$ is a hyperbola that intersects the ellipse $x^2 + 4y^2 = 4$ at four additional points. These points must be checked to find the true absolute maximum and minimum. Visualizing the Error: The points $(\pm 2, 0)$ and $(0, \pm 1)$ are 'easy' solutions where one variable is zero, but the temperature function $T(x,y)$ has an $xy$ term, which tilts the level curves. This tilt means extrema usually occur at points where both $x$ and $y$ are non-zero.

Answer: $4$

Set up the Lagrange equations: $\nabla f = \lambda \nabla g$ gives $y = 2\lambda x$ and $x = 2\lambda y$, with constraint $x^2 + y^2 = 8$. Solve the system: From the first two equations: $y/(2x) = \lambda = x/(2y)$, so $y^2 = x^2$, meaning $y = \pm x$. Substituting $y = x$ into the constraint: $2x^2 = 8$, so $x^2 = 4$, giving $x = \pm 2$. Evaluate the function: At $(2, 2)$: $f = 4$. At $(-2, -2)$: $f = 4$. At $(2, -2)$ and $(-2, 2)$: $f = -4$. The maximum is $4$. Why distractors fail: Option A ($2$) results from an algebra error. Option C ($8$) is the constraint value, not $f$. Option D ($2\sqrt{2}$) confuses coordinate values with function values.

Answer: $6$

Check for interior critical points: $f_x = 1$ and $f_y = 2$ are never zero, so there are no interior critical points. Evaluate on the boundary: At the vertices: $f(0,0) = 0$, $f(4,0) = 4$, $f(0,3) = 6$. Along each edge, since $f$ is linear, extrema occur at endpoints. Identify the maximum: The largest value is $f(0, 3) = 6$. Why distractors fail: Option A is $f(4, 0)$. Option C ($7$) does not correspond to any vertex or boundary evaluation. Option D ($5$) is also not attained on the region.

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

Recall the discriminant formula: The Second Derivative Test uses the discriminant $D = f_{xx} f_{yy} - (f_{xy})^2$, evaluated at the critical point. Why the correct answer works: Option C matches the standard formula exactly: $D = f_{xx} f_{yy} - (f_{xy})^2$. Why distractors fail: Option A gives the trace of the Hessian, not the determinant. Option B uses a plus sign instead of minus, which is incorrect. Option D rearranges terms incorrectly and does not represent the Hessian determinant.

Answer: Because the level curve of $f$ must be tangent to the constraint curve at the extremum, so their normal vectors are parallel

Geometric interpretation of Lagrange multipliers: At a constrained extremum, the level curve $f(x,y) = k$ is tangent to the constraint curve $g(x,y) = c$. Since gradients are normal to level curves, tangency implies the gradients are parallel. Why the correct answer works: Option B correctly identifies the tangency condition between the level curve and the constraint, and the resulting parallelism of gradients. Why distractors fail: Option A confuses function equality with gradient conditions. Option C states perpendicularity, which is the opposite of the requirement. Option D incorrectly assumes both gradients are zero; $\nabla g \neq 0$ is a standard assumption.

Answer: The claim is partially correct; $D = 0$ makes the test inconclusive, but the origin is in fact a local (and global) minimum since $f \geq 0$ everywhere with $f(0,0) = 0$

Verify the critical point: $f_x = 4x(x^2 + y^2)$ and $f_y = 4y(x^2 + y^2)$. Both are zero at $(0, 0)$, confirming it is a critical point. Apply the Second Derivative Test: At $(0,0)$: $f_{xx} = 4(3x^2 + y^2)\big|_{(0,0)} = 0$, $f_{yy} = 0$, $f_{xy} = 0$. So $D = 0$, and the test is inconclusive. Use direct analysis: $f(x,y) = (x^2+y^2)^2 \geq 0$ for all $(x,y)$, and $f(0,0) = 0$. Therefore the origin is a global minimum. Why distractors fail: Option A is wrong — the origin is a minimum. Option C is wrong — $D = 0$ does not indicate a saddle point. Option D is wrong — we verified $(0,0)$ is a critical point.

Answer: $x = y = 2z$

Set up the Lagrange system: Maximize $V = xyz$ subject to $g = xy + 2xz + 2yz = S$. The conditions $\nabla V = \lambda \nabla g$ give: $yz = \lambda(y + 2z)$, $xz = \lambda(x + 2z)$, $xy = \lambda(2x + 2y)$. Derive the relationship: From the first two equations: $yz/(y + 2z) = xz/(x + 2z)$, which simplifies (for $z \neq 0$) to $y(x + 2z) = x(y + 2z)$, giving $2yz = 2xz$, so $x = y$. From equations 1 and 3 with $x = y$: $xz/(x + 2z) = x^2/(4x)$. This gives $4xz = x(x + 2z)$, so $4z = x + 2z$, yielding $x = 2z$. Conclusion: The optimal dimensions satisfy $x = y = 2z$. Why distractors fail: Option A ($x = y = z$) would apply to a closed box. Option C ($z = 0$) gives zero volume. Option D does not satisfy the Lagrange conditions.

Answer: $P$ is a local min, $Q$ is a local max, $R$ is a saddle point

Apply the Second Derivative Test to each point: At $P$: $D > 0$ and $f_{xx} > 0$ → local minimum. At $Q$: $D > 0$ and $f_{xx} < 0$ → local maximum. At $R$: $D < 0$ → saddle point. Why the correct answer works: Option B correctly classifies all three points according to the standard criteria. Why distractors fail: Option A swaps the max and min classifications. Option C misclassifies $Q$ as a saddle point and $R$ as a max. Option D is wrong because the test is conclusive at all three points.

Answer: $D > 0$ indicates a local extremum, but it is a minimum only if $f_{xx} > 0$

Complete conditions for the Second Derivative Test: When $D > 0$, we know the critical point is a local extremum (not a saddle point). To determine whether it is a max or min, we check $f_{xx}$: if $f_{xx} > 0$ it is a local minimum; if $f_{xx} < 0$ it is a local maximum. Why the correct answer works: Option B correctly identifies that $D > 0$ alone is insufficient — one must also check the sign of $f_{xx}$. Why distractors fail: Option A is wrong because $D > 0$ indicates an extremum, not a saddle point ($D < 0$ gives a saddle). Option C is wrong because $D > 0$ is conclusive (the inconclusive case is $D = 0$). Option D is nonsensical — a single point cannot be both a max and a min.

Answer: $f(x, y) = x^2 - y^2$

Analyze each option: For $f = x^2 - y^2$: $f_x = 2x = 0$, $f_y = -2y = 0$ gives $(0,0)$. $D = (2)(-2) - 0 = -4 < 0$, so it's a saddle point. Check the other functions: $x^2 + y^2$: one critical point at origin, $D = 4 > 0$, $f_{xx} = 2 > 0$ → local min. $(x-1)^2 + (y+2)^2$: critical point at $(1, -2)$, same analysis → local min. $x^2 + 4y^2 + 2x$: critical at $(-1, 0)$, $D = 8 > 0$, $f_{xx} = 2 > 0$ → local min. Why the correct answer works: Only $f(x,y) = x^2 - y^2$ has its single critical point as a saddle point.

Answer: Find all critical points in the interior and evaluate $f$ along the boundary, then compare all values

Extreme Value Theorem approach: A continuous function on a closed, bounded region is guaranteed to attain absolute extrema. These can occur at interior critical points or on the boundary. Why the correct answer works: Option C correctly identifies the two-part procedure: check interior critical points and boundary values, then compare all candidates. Why distractors fail: Option A ignores the boundary, where extrema frequently occur. Option B ignores interior critical points. Option D misapplies the Second Derivative Test, which classifies local extrema at critical points but cannot be applied at every point.

Answer: A point where $f_x = 0$ and $f_y = 0$ simultaneously

Definition of a critical point: A critical point of $f(x, y)$ is a point $(a, b)$ in the domain where both first partial derivatives vanish: $f_x(a, b) = 0$ and $f_y(a, b) = 0$. Why the correct answer works: Option B directly states the definition: both $f_x$ and $f_y$ equal zero simultaneously. Why distractors fail: Option A confuses the function value being zero with the derivatives being zero. Option C describes part of the Second Derivative Test, not the definition of a critical point. Option D describes the Lagrange multiplier condition for constrained optimization.

Answer: Two multipliers: $\nabla f = \lambda_1 \nabla g_1 + \lambda_2 \nabla g_2$ with $g_1 = 0$ and $g_2 = 0$

Multiple constraints require multiple multipliers: With $k$ constraints, the Lagrange condition is $\nabla f = \lambda_1 \nabla g_1 + \lambda_2 \nabla g_2 + \cdots + \lambda_k \nabla g_k$, along with all constraint equations. Why the correct answer works: With two constraints, we need two multipliers and the system $\nabla f = \lambda_1 \nabla g_1 + \lambda_2 \nabla g_2$, plus $g_1 = 0$ and $g_2 = 0$. Why distractors fail: Option A uses only one multiplier, which is insufficient. Option C incorrectly sets up two separate equations rather than a linear combination. Option D suggests direct substitution, which works in some cases but does not answer the question about the Lagrange multiplier method.

Answer: $24$

Set up Lagrange conditions: $\nabla C = \lambda \nabla g$ with $g(x,y) = x + y$: $4x = \lambda$ and $2y = \lambda$. Solve the system: From $4x = 2y$, we get $y = 2x$. Substituting into the constraint: $x + 2x = 6$, so $x = 2$ and $y = 4$. Evaluate: $C(2, 4) = 2(4) + 16 = 8 + 16 = 24$. Why distractors fail: Option A ($18$) could result from using $x = y = 3$, ignoring the different coefficients. Option C ($36$) is $C(0, 6)$, a boundary evaluation. Option D ($12$) results from an arithmetic error.

Answer: The critical point is a saddle point

Interpreting the discriminant: When $D < 0$, the Hessian matrix is indefinite, meaning $f$ increases in some directions and decreases in others near the critical point. Why the correct answer works: A negative discriminant means the critical point is a saddle point — it is neither a local max nor a local min. Why distractors fail: Options A and B require $D > 0$ (with additional sign conditions on $f_{xx}$). Option D applies when $D = 0$, not when $D < 0$.

Answer: Saddle point at $(-2, -1)$

Find the critical point: $f_x = 2x + 4 = 0 \Rightarrow x = -2$; $f_y = -2y - 2 = 0 \Rightarrow y = -1$. The critical point is $(-2, -1)$. Compute second derivatives and discriminant: $f_{xx} = 2$, $f_{yy} = -2$, $f_{xy} = 0$. So $D = (2)(-2) - 0^2 = -4$. Classify using D: Since $D = -4 < 0$, the critical point is a saddle point. Why distractors fail: Options A and B require $D > 0$, which is not the case. Option D applies when $D = 0$.

Answer: Local minimum

Compute second partial derivatives: $f_{xx} = 6x$, $f_{yy} = 6y$, $f_{xy} = -3$. Evaluate at (1, 1): At $(1, 1)$: $f_{xx} = 6$, $f_{yy} = 6$, $f_{xy} = -3$. So $D = (6)(6) - (-3)^2 = 36 - 9 = 27 > 0$. Classify: Since $D > 0$ and $f_{xx} = 6 > 0$, the point $(1, 1)$ is a local minimum. Why distractors fail: Option A would require $f_{xx} < 0$. Option C would require $D < 0$. Option D would require $D = 0$.

Answer: Absolute max is $4$ at $(2, 0)$ and $(0, 2)$; absolute min is $0$ at $(0, 0)$ and $(2, 2)$

Find interior critical points: $f_x = 2x - 2y = 0$ and $f_y = -2x + 2 = 0$ give $x = 1$, $y = 1$. $f(1,1) = 1 - 2 + 2 = 1$. This is in the interior. Evaluate at corners: $f(0,0) = 0$, $f(2,0) = 4$, $f(0,2) = 0 - 0 + 4 = 4$, $f(2,2) = 4 - 8 + 4 = 0$. Check boundary edges: On each edge, reduce to a one-variable function and find any additional critical points. After analysis, no boundary critical point yields a value outside the range $[0, 4]$. Conclusion: The absolute maximum is $4$, attained at both $(2, 0)$ and $(0, 2)$. The absolute minimum is $0$, attained at both $(0, 0)$ and $(2, 2)$. The interior critical point $(1,1)$ gives $f = 1$, which is neither the max nor the min. Why distractors fail: Option A claims the min is $-1$, which is incorrect ($f(1,1) = 1$). Option B correctly identifies both max locations but incorrectly claims the min is $1$ at $(1,1)$, missing the lower value of $0$. Option D gives an incorrect minimum value of $-2$.

Answer: Incorrect, the point could be a saddle point or a degenerate case; further testing is needed

Critical points are candidates, not conclusions: A critical point is a necessary condition for a local extremum (for differentiable functions), but it is not sufficient. The point could be a saddle point or a degenerate case. Why the correct answer works: Option B correctly states that additional analysis (such as the Second Derivative Test) is required to classify the critical point. Why distractors fail: Option A is false — saddle points are critical points that are not extrema. Option C adds an irrelevant condition; continuity alone doesn't guarantee an extremum. Option D is false; there is no requirement that $f_{xy} = 0$.