Difficulty: Intermediate | Prerequisites: Single-variable derivatives, basic integration
This covers the multivariable core of the MAT 211 final: partial derivatives and their economic interpretation (marginal cost), reading level curves, recognising planes in three dimensions, finding critical points for functions of two variables, using the Hessian (second derivative test for multivariable functions), and finding absolute extrema on closed rectangular regions. If you can take ordinary derivatives and understand what a function of one variable looks like, you are ready. These techniques are central to optimisation problems in economics and the sciences.
Partial derivatives hold one variable fixed and differentiate with respect to the other. Critical points of f(x, y) occur where both partial derivatives are zero. The Hessian H = f_xx · f_yy - (f_xy)² classifies them: H > 0 with f_xx > 0 means a local minimum, H > 0 with f_xx < 0 means a local maximum, H < 0 means a saddle point. For absolute extrema on a closed region, check the interior critical points, the boundary critical points, and all corners.
Partial derivative
The derivative of a multivariable function with respect to one variable, treating all other variables as constants. Notation: ∂f/∂x or f_x.
Think of it as: how does the output change when you nudge just one input, holding everything else still?
Marginal cost
The partial derivative of the cost function with respect to a particular product. It represents the approximate cost of producing one additional unit of that product, holding other production levels constant.
In simple terms, this means: if you make one more mountain bike, how much extra does it cost?
Level curve (contour line)
A curve in the xy-plane along which f(x, y) takes a constant value. The set of all points where f(x, y) = c for some constant c.
Think of it as: a topographic map line. Every point on that line sits at the same "height."
Plane (linear function of two variables)
An equation of the form z = ax + by + c is the equation of a plane in three-dimensional space. It is the 3D analogue of a straight line.
In simple terms, this means: a flat surface tilted in space, with no curves.
Critical point (multivariable)
A point (x, y) where both f_x = 0 and f_y = 0. These are candidates for local maxima, local minima, or saddle points.
Think of it as: the tangent plane is perfectly flat (horizontal) at that point.
Hessian (H)
The discriminant used to classify critical points of a function of two variables: H = f_xx · f_yy - (f_xy)².
In simple terms, this means: a single number that tells you whether a critical point is a max, min, or saddle.
Saddle point
A critical point where the function increases in one direction and decreases in another. The Hessian is negative at a saddle point.
Think of it as: a mountain pass or the middle of a Pringle crisp.
Absolute extrema on a closed region
The largest and smallest values of f on a bounded, closed domain. Found by comparing f at interior critical points, boundary critical points, and all corner points.
In simple terms, this means: check everywhere the function could reach a peak or valley, including the edges.
To find ∂f/∂x, treat y as a constant and differentiate with respect to x. To find ∂f/∂y, treat x as a constant and differentiate with respect to y.
Worked example from the exam:
C(x, y) = 50,000 + 500x + 600y - 10xy
Find the marginal cost of producing mountain bikes (y) at x = 20, y = 30.
Step 1: Take the partial derivative with respect to y.
C_y = ∂C/∂y = 600 - 10x
Step 2: Evaluate at x = 20.
C_y(20, 30) = 600 - 10(20) = 600 - 200 = 400 dollars per mountain bike.
Interpretation: at a production level of 20 ten-speed bicycles and 30 mountain bikes, producing one additional mountain bike costs approximately $400.
A level curve is the set of points where f(x, y) = c. On a contour plot, the value printed on each curve is c. To estimate f at a specific point, find which level curve passes through (or near) that point.
Worked example from the exam:
Estimate f(3.5, 0) from the contour plot. At the point (3.5, 0) on the x-axis, the level curves indicate a value between 2 and 3. The point sits roughly on the level curve labelled 2.5.
Answer: B (2.5).
The equation z = 2x + 5y - 7 is linear in x, y, and z with no squared terms, products, or other nonlinearities. Any equation of the form z = ax + by + c defines a plane.
True/False example: z = 2x + 5y - 7 is the equation of a plane.
Answer: True.
For w = f(x, y, z) = 2 + 2x - 4y - 5z, if x, y, and z all increase by 1:
Δw = f(x+1, y+1, z+1) - f(x, y, z)
Since f is linear, Δw = 2(1) - 4(1) - 5(1) = 2 - 4 - 5 = -7.
This works because the constant term (2) cancels out, and each variable's contribution is its coefficient times the change in that variable.
Answer: E (-7).
Set f_x = 0 and f_y = 0 simultaneously and solve.
Worked example from the exam (Q15):
f(x, y) = y³ - 12y - x²
f_x = -2x → set to 0 → x = 0
f_y = 3y² - 12 → set to 0 → 3y² = 12 → y² = 4 → y = ±2
Critical points: (0, 2) and (0, -2).
Once you have a critical point, compute the second partial derivatives and the Hessian:
H = f_xx · f_yy - (f_xy)²
Decision rules:
H > 0 and f_xx > 0 → local minimum
H > 0 and f_xx < 0 → local maximum
H < 0 → saddle point
H = 0 → test is inconclusive
Continuing the worked example:
f_xx = -2, f_yy = 6y, f_xy = 0
At (0, 2): H = (-2)(12) - 0² = -24. Since H < 0, this is a saddle point.
At (0, -2): H = (-2)(-12) - 0² = 24. Since H > 0 and f_xx = -2 < 0, this is a local maximum.
To find the absolute maximum and minimum of f(x, y) on a closed rectangular region:
Find interior critical points by solving f_x = 0 and f_y = 0.
Find boundary critical points by restricting f to each edge (substituting one variable's boundary value), then differentiating the resulting single-variable function and setting it to zero.
Evaluate f at all corners.
Compare all values. The largest is the absolute maximum; the smallest is the absolute minimum.
Worked example from the exam (Q13):
f(x, y) = x² - 6x + y² - 4y on 0 ≤ x ≤ 6, 0 ≤ y ≤ 5.
Step 1, interior critical point:
f_x = 2x - 6 = 0 → x = 3
f_y = 2y - 4 = 0 → y = 2
Interior critical point: (3, 2). f(3, 2) = 9 - 18 + 4 - 8 = -13.
Step 2, boundary critical points:
Bottom edge (y = 0): f(x, 0) = x² - 6x. f'= 2x - 6 = 0 → x = 3. Point: (3, 0). f(3, 0) = 9 - 18 = -9.
Top edge (y = 5): f(x, 5) = x² - 6x + 25 - 20 = x² - 6x + 5. f' = 2x - 6 = 0 → x = 3. Point: (3, 5). f(3, 5) = 9 - 18 + 25 - 20 = -4.
Left edge (x = 0): f(0, y) = y² - 4y. f' = 2y - 4 = 0 → y = 2. Point: (0, 2). f(0, 2) = 4 - 8 = -4.
Right edge (x = 6): f(6, y) = 36 - 36 + y² - 4y = y² - 4y. f' = 2y - 4 = 0 → y = 2. Point: (6, 2). f(6, 2) = 4 - 8 = -4.
Step 3, corners:
f(0, 0) = 0, f(0, 5) = 5, f(6, 0) = 0, f(6, 5) = 5.
Step 4, compare all values:
Values: -13, -9, -4, -4, -4, 0, 5, 0, 5.
Absolute minimum of f is -13, occurring at (3, 2).
Absolute maximum of f is 5, occurring at (0, 5) and (6, 5).
Partial derivative of f with respect to x: differentiate normally, treating y as a constant.
Hessian: H = f_xx · f_yy - (f_xy)²
Critical point condition: f_x = 0 and f_y = 0 simultaneously.
For a linear function w = a + bx + cy + dz, the change Δw when all variables increase by 1 is simply b + c + d.
Absolute extrema procedure: interior critical points + boundary critical points + corners → compare f values.
Students often forget that the Hessian requires both H > 0 AND a check on f_xx to distinguish between a local max and a local min. H > 0 alone is not enough.
When finding boundary critical points, students sometimes forget to substitute the boundary constraint before differentiating. You must reduce to a single-variable function first, then find its critical points.
Students confuse the partial derivative ∂f/∂x with the total derivative df/dx. When y depends on x, the total derivative includes a dy/dx term (chain rule).
For Δw problems, students sometimes compute w at the new point instead of the change in w. The question asks for the difference, not the new value.
⚠️ The absolute extrema problem (Q13) is worth 15 points across its sub-parts, making it the highest-value question on the exam. Practise the full procedure.
⚠️ The Hessian test (Q15) is worth 11 points. Memorise the formula and the decision rules cold.
⚠️ Marginal cost (Q12) is straightforward, worth 4 points, and should be quick. Do not waste time here.
⚠️ Level curve reading (Q9) is a visual skill. Practise reading contour plots before exam day.
⚠️ The Δw question (Q10) is a direct computation. For linear functions, the change depends only on the coefficients and the change in each variable, not on the starting point.
True or false: A saddle point has H > 0. (False; saddle points have H < 0)
Fill in the blank: For C(x, y) = 100 + 3x + 7y - 2xy, the marginal cost with respect to x is ______. (3 - 2y)
True or false: z = x² + 3y - 1 is the equation of a plane. (False; x² makes it nonlinear)
Fill in the blank: If f(x, y, z) = 5 + x - 3y + 2z and all variables increase by 1, then Δw = ______. (0, since 1 - 3 + 2 = 0)
True or false: To find absolute extrema on a closed region, you only need to check interior critical points. (False; you also need boundary critical points and corners)
Q: Given f(x, y) = x² + 3xy - y², find f_x and f_y.
A: f_x = 2x + 3y. f_y = 3x - 2y.
Q: Find the critical points of f(x, y) = x² + y² - 4x + 6y.
A: f_x = 2x - 4 = 0 → x = 2. f_y = 2y + 6 = 0 → y = -3. Critical point: (2, -3).
Q: For f(x, y) = x² + y² - 4x + 6y, compute the Hessian at (2, -3) and classify the critical point.
A: f_xx = 2, f_yy = 2, f_xy = 0. H = (2)(2) - 0² = 4. Since H > 0 and f_xx = 2 > 0, the point (2, -3) is a local minimum.
Q: Find the absolute maximum and minimum of f(x, y) = x + y on the region 0 ≤ x ≤ 3, 0 ≤ y ≤ 4.
A: f is linear, so it has no interior critical points (f_x = 1 ≠ 0). Check corners: f(0,0) = 0, f(3,0) = 3, f(0,4) = 4, f(3,4) = 7. Absolute min is 0 at (0,0); absolute max is 7 at (3,4).
The Hessian test is the multivariable analogue of the single-variable second derivative test. Where f''(c) told you the curvature in one dimension, the Hessian captures curvature in two directions at once.
Absolute extrema on closed regions connect to Lagrange multipliers (see Doc 3). When the boundary is a curve rather than the edges of a rectangle, Lagrange multipliers handle the constraint.
Partial derivatives are the building blocks for gradient vectors, directional derivatives, and the multivariable chain rule, all of which appear in more advanced calculus courses.
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