Chapter 31 — Quiz

10 questions covering critical points, the Hessian second-derivative test, saddle points, absolute extrema on closed regions, and Lagrange multipliers. Work each by hand, then reveal the answer.

1. Find all critical points of $f(x,y) = x^2 + y^2 - 6x + 2y + 4$.

Answer $f_x = 2x - 6 = 0 \Rightarrow x = 3$; $f_y = 2y + 2 = 0 \Rightarrow y = -1$. The lone critical point is $(3, -1)$. (Completing the square, $f = (x-3)^2 + (y+1)^2 - 6$, an upward bowl, so this is a minimum.) See Section 31.2.

2. State the discriminant $D$ used in the second-derivative test, and the rule that distinguishes a saddle point from a local extremum.

Answer $D = f_{xx}f_{yy} - f_{xy}^2 = \det(H_f)$. If $D < 0$ the critical point is a **saddle point**; if $D > 0$ it is a local extremum (a minimum when $f_{xx} > 0$, a maximum when $f_{xx} < 0$); if $D = 0$ the test is inconclusive. See Sections 31.3–31.4.

3. Classify the critical point of $f(x,y) = x^2 + 4xy + y^2$ at the origin.

Answer $f_x = 2x + 4y$, $f_y = 4x + 2y$; both vanish only at $(0,0)$. Then $f_{xx} = 2$, $f_{yy} = 2$, $f_{xy} = 4$, so $D = (2)(2) - 4^2 = 4 - 16 = -12 < 0$. The origin is a **saddle point**. See Section 31.4.

4. The function $f(x,y) = x^3 + y^3 - 3xy$ has critical points $(0,0)$ and $(1,1)$. Classify each.

Answer $f_{xx} = 6x$, $f_{yy} = 6y$, $f_{xy} = -3$. At $(0,0)$: $D = (0)(0) - 9 = -9 < 0$ → **saddle**. At $(1,1)$: $D = (6)(6) - 9 = 27 > 0$ with $f_{xx} = 6 > 0$ → **local minimum**, value $f(1,1) = 1 + 1 - 3 = -1$. This is Example 2 of Section 31.4.

5. Why does a saddle point have no analog in single-variable calculus?

Answer A curve has only one direction to move along, so a flat spot is either a peak or a valley. A surface has infinitely many directions at once: at a saddle, $f$ increases along some directions through the point and decreases along others (e.g., $f = x^2 - y^2$ curves up along the $x$-axis and down along the $y$-axis). That mixed behavior requires more than one independent direction. See Section 31.2.

6. List the three steps for finding absolute extrema of a continuous $f$ on a closed, bounded region $R$.

Answer (1) Find all critical points in the **interior** and list $f$ there. (2) Find the extreme values of $f$ on the **boundary** (parametrize and use single-variable calculus, or Lagrange multipliers). (3) **Compare** all collected values: the largest is the absolute maximum, the smallest the absolute minimum. The Extreme Value Theorem guarantees both exist. See Section 31.5.

7. Find the absolute maximum and minimum of $f(x,y) = xy$ on the closed disk $x^2 + y^2 \le 1$.

Answer Interior: $\nabla f = \langle y, x\rangle = \mathbf 0$ only at $(0,0)$, where $f = 0$. Boundary: with $x = \cos\theta$, $y = \sin\theta$, $f = \tfrac12\sin 2\theta$, ranging over $[-\tfrac12, \tfrac12]$. Comparing $\{0, \tfrac12, -\tfrac12\}$: **maximum $\tfrac12$**, **minimum $-\tfrac12$**. The interior point is a saddle decoy. See Section 31.5.

8. Write the Lagrange multiplier condition for optimizing $f(x,y)$ subject to $g(x,y) = 0$, and explain its geometric meaning.

Answer $\nabla f = \lambda\nabla g$, together with the constraint $g = 0$. Geometrically, at a constrained extremum the level curve of $f$ is **tangent** to the constraint curve, so their perpendiculars — the gradients — are parallel. The scalar $\lambda$ is the proportionality factor (the Lagrange multiplier). See Section 31.8.

9. Use Lagrange multipliers to maximize $f(x,y) = xy$ subject to $x + y = 10$ (with $x, y > 0$).

Answer $\nabla f = \langle y, x\rangle$, $\nabla g = \langle 1, 1\rangle$, so $y = \lambda$ and $x = \lambda$, giving $x = y$. The constraint $x + y = 10$ then gives $x = y = 5$, and $f = 25$. Checking another feasible point, $f(1,9) = 9 < 25$, confirms a **maximum** of $25$ at $(5,5)$. See Section 31.8.

10. In the consumer's utility-maximization problem, what does the Lagrange multiplier $\lambda$ represent, and what is the "equal marginal utility per dollar" condition?

Answer Dividing the Lagrange equations $U_x = \lambda p_x$, $U_y = \lambda p_y$ gives $\dfrac{U_x}{p_x} = \dfrac{U_y}{p_y} = \lambda$: at the optimum the **last dollar spent on each good yields the same additional utility**. The multiplier $\lambda$ is the common value — the **marginal utility of income** (the shadow price of the budget), the extra utility one more dollar of budget would buy. See Section 31.10.

Scoring Guide

Score Interpretation
9–10 Excellent. You can find, classify, and constrain-optimize with confidence. Move on to multiple integrals (Chapter 32).
7–8 Solid. Revisit any missed items — most likely the $D=0$ edge cases or the boundary step of closed-region problems.
5–6 Partial. Re-read Sections 31.4 (the test) and 31.8 (Lagrange), then redo Parts B and D of the exercises.
0–4 Review the full chapter, focusing on the worked examples in Sections 31.4 and 31.8 before reattempting.

Self-check. Questions 1–5 target the unconstrained test (Sections 31.2–31.4); 6–7 the closed-region method (Section 31.5); 8–10 Lagrange multipliers and their economic meaning (Sections 31.8, 31.10). A lopsided score tells you exactly which third to review.