Source: Stewart's Multivariable Calculus, §14.7 | Homework Set 5
Difficulty: Intermediate-Advanced | Prerequisites: Partial derivatives (§14.3), gradient vector (§14.6), single-variable max/min from Calc I
Section 14.7 is the multivariable version of finding maxima and minima from Calc I, but with a richer landscape: functions of two variables can have saddle points (neither max nor min), which have no single-variable analogue. The section covers two main tasks: classifying critical points using the second derivative test, and finding absolute extrema on closed bounded regions by checking both the interior and the boundary. If you can find partial derivatives and solve systems of equations, you have the prerequisites. This is one of the most exam-heavy sections in the course.
Critical points are where both partial derivatives are zero (or undefined). The second derivative test uses the discriminant D = fxx fyy - (fxy)² to classify them as local max, local min, or saddle point. For absolute extrema on a closed region, evaluate f at every interior critical point and at every critical point along the boundary, then pick the largest and smallest values.
Critical point
A point (a, b) where fx(a, b) = 0 and fy(a, b) = 0, or where one of these partial derivatives does not exist.
Think of it as: a flat spot on the surface, where the tangent plane is horizontal. Could be a hilltop, a valley bottom, or a saddle.
Second derivative test (discriminant D)
D = fxx(a, b) fyy(a, b) - [fxy(a, b)]². If D > 0 and fxx > 0, local minimum. If D > 0 and fxx < 0, local maximum. If D < 0, saddle point. If D = 0, the test is inconclusive.
In simple terms, this means: D tells you whether the surface curves the same way in all directions (a bowl or a dome) or curves differently in different directions (a saddle).
Saddle point
A critical point where f is a local max in one direction and a local min in another. The surface near the point looks like a horse saddle or a mountain pass.
Think of it as: you are at a point that is the lowest on a ridge but the highest in the valley. D < 0 confirms it.
Absolute (global) maximum / minimum
The largest or smallest value of f on the entire domain (or on a specified closed bounded region). These can occur at interior critical points or on the boundary.
In simple terms, this means: the overall winner, not just a local champion. On a closed region, you must check everywhere.
Closed bounded region
A region that includes its boundary and fits inside some large circle. The Extreme Value Theorem guarantees that a continuous function on such a region has an absolute max and an absolute min.
Think of it as: a finite region with no gaps or missing edges, like a triangle or a disc.
Compute fx and fy.
Set fx = 0 and fy = 0 simultaneously.
Solve the resulting system. Each solution (a, b) is a critical point.
At each critical point (a, b), compute:
fxx(a, b), fyy(a, b), fxy(a, b)
D = fxx fyy - (fxy)²
Classification:
D > 0 and fxx > 0: local minimum
D > 0 and fxx < 0: local maximum
D < 0: saddle point
D = 0: inconclusive (need other methods)
f(s, t) = 8s⁴ + t² + st - 2s² - s³
Step 1: partial derivatives.
fs = 32s³ + t - 4s - 3s² = 0
ft = 2t + s = 0, so t = -s/2
Step 2: substitute t = -s/2 into the first equation.
32s³ - s/2 - 4s - 3s² = 0
32s³ - 3s² - 9s/2 = 0
s(32s² - 3s - 9/2) = 0
s(64s² - 6s - 9) = 0 (multiplying inside by 2)
So s = 0, or 64s² - 6s - 9 = 0.
Using the quadratic formula on 64s² - 6s - 9 = 0:
s = (6 ± √(36 + 2304)) / 128 = (6 ± √2340) / 128 = (6 ± 6√65) / 128 = (3 ± 3√65) / 64
This gives three critical points. For each, find t = -s/2.
Step 3: classify each. Compute the second-order partials:
fss = 96s² - 4 - 6s, ftt = 2, fst = 1
D = fss · ftt - (fst)² = 2(96s² - 4 - 6s) - 1
Evaluate D and fss at each critical point to classify as local max, local min, or saddle.
Find all critical points of f in the interior of the region.
Find the extreme values of f on each piece of the boundary (parametrise each edge, reduce to single-variable optimisation).
Compare all values from steps 1 and 2. The largest is the absolute maximum; the smallest is the absolute minimum.
This works because of the Extreme Value Theorem: a continuous function on a closed bounded set always attains its max and min.
f(x, y) = x³ - 12y - y³ + 3x on the triangle with vertices (2, 3), (2, -2), (-2, 2).
Step 1: interior critical points.
fx = 3x² + 3 = 0 gives x² = -1, which has no real solutions. So there are no critical points in the interior.
Step 2: boundary. The triangle has three edges.
Edge 1: x = 2, -2 ≤ y ≤ 3.
f(2, y) = 8 - 12y - y³ + 6 = 14 - 12y - y³
f'(y) = -12 - 3y² = 0 gives y² = -4, no real solutions. So check endpoints only: f(2, -2) = 14 + 24 + 8 = 46, f(2, 3) = 14 - 36 - 27 = -49.
Edge 2: from (2, -2) to (-2, 2). Parametrise: x = 2 - 4t, y = -2 + 4t for 0 ≤ t ≤ 1. Substitute into f, differentiate with respect to t, set equal to zero, and evaluate at critical t values and endpoints.
Edge 3: from (-2, 2) to (2, 3). Parametrise similarly and repeat.
Step 3: Compare all values found. The largest is the absolute maximum, the smallest is the absolute minimum.
The key takeaway: even when there are no interior critical points, you must check every boundary edge carefully. The extrema live on the boundary.
Look for closed contour loops where the values increase inward. The innermost loop surrounds the peak. On the homework contour map (Problem 1), maxima appear where contour values climb to their highest enclosed value.
Closed loops where the values decrease inward. The innermost contour surrounds the lowest point.
Saddle points sit where contour lines from two different "peaks" or two different "valleys" nearly meet, creating an hourglass or figure-eight pattern. The value increases in one direction and decreases in another. On the contour map, look for points between two maxima or between two minima where the contours pinch together.
To hike from (-0.7, 0.3) to (0.7, 0.2) with the least total climbing, trace a path that follows contour lines as closely as possible, avoiding unnecessary crossings. Every time you cross from one contour to a higher one, that elevation difference adds to your cumulative gain. Descents add zero to cumulative gain but going back up costs you. The optimal path threads through low passes (saddle points) rather than going over peaks.
D = f_{xx}(a,b)\,f_{yy}(a,b) - \bigl[f_{xy}(a,b)\bigr]^2Classification at a critical point (a, b) where ∇f = 0:
D > 0, fxx > 0: local minimum
D > 0, fxx < 0: local maximum
D < 0: saddle point
D = 0: test inconclusive
Optimising a function on a constrained domain models countless engineering and economics problems: minimising cost subject to material constraints, maximising profit within a feasible region, or finding the best location for a facility given geographic boundaries. Contour-map reading is essentially what hikers, pilots, and meteorologists do professionally when interpreting topographic or weather charts.
Students often forget to check the boundary when finding absolute extrema on a closed region. The boundary almost always contributes the answer.
A critical point where D = 0 is not automatically a saddle point. The test simply gives no information there, and you need other methods (graphing, higher-order tests).
Students sometimes apply the second derivative test at points where fx ≠ 0 or fy ≠ 0. The test only applies at critical points.
On contour maps, students sometimes identify a saddle point as a minimum because the value there is lower than the surrounding peaks. A saddle is lower than some neighbours and higher than others.
⚠️ The second derivative test is tested on almost every exam. Memorise the discriminant D = fxx fyy - (fxy)² and the classification rules.
⚠️ Absolute extrema on a closed region is a multi-step problem that often appears as a long-answer question. Budget time for the boundary work.
⚠️ You will very likely be asked to identify maxima, minima, and saddle points from a contour map. Practise reading contour maps visually.
⚠️ When the boundary is a triangle, you get three edges, each parametrised differently. This is where most students lose marks by forgetting an edge or making algebra errors.
True or False: If D > 0 at a critical point, it must be either a local max or local min. True.
Fill in the blank: D = fxx fyy - ____ . Answer: (fxy)²
True or False: A saddle point is a point where D = 0. False - a saddle has D < 0. D = 0 is inconclusive.
True or False: When finding absolute extrema on a closed region, you only need to check the interior critical points. False - you must also check the boundary.
Fill in the blank: The Extreme Value Theorem requires the region to be ____ and ____ . Answer: closed and bounded
Q: Find and classify the critical points of f(x, y) = x² + y² - 4x + 6y + 13.
A: fx = 2x - 4 = 0 gives x = 2. fy = 2y + 6 = 0 gives y = -3. Critical point: (2, -3). fxx = 2, fyy = 2, fxy = 0. D = (2)(2) - 0 = 4 > 0, and fxx = 2 > 0, so (2, -3) is a local minimum. f(2, -3) = 4 + 9 - 8 - 18 + 13 = 0.
Q: Find the absolute max and min of f(x, y) = x² + y² on the square [0, 1] × [0, 1].
A: Interior: fx = 2x = 0, fy = 2y = 0 gives (0, 0), which is a corner. Boundary: on each edge, f reduces to a single-variable function. Checking all corners: f(0,0) = 0, f(1,0) = 1, f(0,1) = 1, f(1,1) = 2. Absolute min = 0 at (0,0), absolute max = 2 at (1,1).
Q: On a contour map, you see two closed loops of increasing value centred at two different points, with contour lines pinching together between them. What is at the pinch point?
A: A saddle point. The value is a local max along the line connecting the two peaks but a local min in the perpendicular direction.
Critical points are where ∇f = 0, which connects directly to §14.6 (gradient). The next section (§14.8, Lagrange multipliers) handles optimisation with constraints, building on everything here. In linear algebra, the second derivative test is related to the eigenvalues of the Hessian matrix: D > 0 with fxx > 0 means both eigenvalues are positive (positive definite), which is a concept that carries through to advanced optimisation and machine learning.
Critical points multivariable, second derivative test two variables, discriminant D, Hessian determinant, saddle point, local maximum, local minimum, absolute extrema closed region, Extreme Value Theorem multivariable, contour map maxima minima saddle, boundary optimisation, parametrise boundary, MATH 215, Stewart §14.7, Calculus III