Extrema of Functions of Two Variables

Local maximums and minimums of f(x, y) happen only at critical points, where f_x = 0 and f_y = 0. The second derivative test uses D = f_xxf_yy − (f_xy)²: if D > 0, it is a minimum when f_xx > 0 and a maximum when f_xx < 0. If D < 0, it is a saddle point.

Before this lesson: Partial Derivatives, Directional Derivatives and the Gradient, Second Derivative Test and Absolute Extrema

Critical points and saddle points

At the top of a smooth hill, the ground is flat in every direction, so the tangent plane is horizontal and ∇f = ⟨0, 0⟩. The same is true at the bottom of a valley. Points where both partial derivatives are 0 (or one does not exist) are called critical points.

A flat spot is not always a peak or a valley. At a saddle point, the surface curves up in one direction and down in another, like a mountain pass. The second derivative test tells these cases apart.

Second derivative test for two variables

D = f_{xx}(a, b)\,f_{yy}(a, b) - \left[f_{xy}(a, b)\right]^2

• Absolute extrema on a closed, bounded region: a continuous function has an absolute maximum and minimum there. Find them by listing f at every critical point inside, at the best points along each edge, and at the corners. The largest value is the maximum and the smallest is the minimum.
• On each edge, substitute the edge’s equation to get a one-variable function, then use the closed interval method.

Extrema of functions of two variables examples

Example 1: One critical point

Find and classify the critical points of f(x, y) = x² + xy + y² − 3x.

f_x = 2x + y - 3 = 0

f_y = x + 2y = 0

x = -2y

2(-2y) + y - 3 = 0

-3y = 3

y = -1

x = -2(-1) = 2

f_{xx} = 2, \quad f_{yy} = 2, \quad f_{xy} = 1

D = (2)(2) - 1^2 = 3

D \gt 0, \quad f_{xx} \gt 0

f(2, -1) = 4 - 2 + 1 - 6 = -3

Answer: f has a local minimum of −3 at (2, −1).

Example 2: A minimum and a saddle point

Find and classify the critical points of f(x, y) = x³ − 3x + y².

f_x = 3x^2 - 3 = 0

x^2 = 1

x = 1 \text{ or } x = -1

f_y = 2y = 0, \quad y = 0

f_{xx} = 6x, \quad f_{yy} = 2, \quad f_{xy} = 0

D = (6x)(2) - 0^2 = 12x

D(1, 0) = 12 \gt 0, \quad f_{xx}(1, 0) = 6 \gt 0

f(1, 0) = 1 - 3 + 0 = -2

D(-1, 0) = -12 \lt 0

f(-1, 0) = -1 + 3 + 0 = 2

Answer: a local minimum of −2 at (1, 0) and a saddle point at (−1, 0, 2).

Example 3: Two maximums and a saddle

Find and classify the critical points of f(x, y) = 4xy − x⁴ − y⁴.

f_x = 4y - 4x^3 = 0 \Rightarrow y = x^3

f_y = 4x - 4y^3 = 0 \Rightarrow x = y^3

x = (x^3)^3 = x^9

x(x^8 - 1) = 0

x = 0, \ 1, \ -1

(0, 0), \ (1, 1), \ (-1, -1)

f_{xx} = -12x^2, \quad f_{yy} = -12y^2, \quad f_{xy} = 4

D = 144x^2y^2 - 16

D(0, 0) = -16 \lt 0

D(1, 1) = 128 \gt 0, \quad f_{xx}(1, 1) = -12 \lt 0

D(-1, -1) = 128 \gt 0, \quad f_{xx}(-1, -1) = -12 \lt 0

f(1, 1) = 4 - 1 - 1 = 2

Answer: local maximums of 2 at (1, 1) and (−1, −1), and a saddle point at (0, 0).

Example 4: Absolute extrema on a square

Find the absolute maximum and minimum of f(x, y) = x² + y² − 2x − 4y on the square 0 ≤ x ≤ 3, 0 ≤ y ≤ 3.

f_x = 2x - 2 = 0, \quad f_y = 2y - 4 = 0

(1, 2), \quad f(1, 2) = 1 + 4 - 2 - 8 = -5

y = 0: \ g(x) = x^2 - 2x

g'(x) = 2x - 2 = 0 \Rightarrow x = 1, \ g(1) = -1

y = 3: \ g(x) = x^2 - 2x - 3, \ g(1) = -4

x = 0: \ h(y) = y^2 - 4y, \ h(2) = -4

x = 3: \ h(y) = y^2 - 4y + 3, \ h(2) = -1

f(0, 0) = 0, \ f(3, 0) = 3, \ f(0, 3) = -3, \ f(3, 3) = 0

Answer: the absolute minimum is −5 at (1, 2), and the absolute maximum is 3 at the corner (3, 0).

Example 5: Cheapest open box (word problem)

An open-top box must hold 108 cm³. Find the dimensions that use the least cardboard.

V = xyz = 108 \Rightarrow z = \dfrac{108}{xy}

S = xy + 2xz + 2yz

S = xy + \dfrac{216}{y} + \dfrac{216}{x}

S_x = y - \dfrac{216}{x^2} = 0, \quad S_y = x - \dfrac{216}{y^2} = 0

yx^2 = 216, \quad xy^2 = 216

x = y

x^3 = 216 \Rightarrow x = 6, \ y = 6

z = \dfrac{108}{36} = 3

S_{xx} = \dfrac{432}{x^3} = 2, \quad S_{yy} = 2, \quad S_{xy} = 1

D = (2)(2) - 1^2 = 3 \gt 0, \quad S_{xx} \gt 0

S = 36 + 36 + 36 = 108

Answer: a 6 cm by 6 cm base with height 3 cm, using 108 cm² of cardboard.

Common mistakes

Related methods

One-variable second derivative test

With one variable, f″ alone decides, as in the second derivative test from Calculus 1. With two variables, D also checks that f bends the same way in every direction.

Constraints and Lagrange multipliers

In Example 5, the constraint xyz = 108 could be solved for z. When a constraint is hard to solve, or a boundary is a curve, the next lesson’s method, Lagrange multipliers, handles it directly.

Practice problems

1. Find and classify the critical points of f(x, y) = x² + y² − 6x + 2y.
2. Find and classify the critical points of f(x, y) = xy − x² − y² + 3y.
3. Find and classify the critical points of f(x, y) = x² − y² + 4x + 2y.
4. Find and classify the critical points of f(x, y) = x³ + y³ − 3xy.
5. Find and classify the critical points of f(x, y) = y³ − 12y + x².
6. Show that the second derivative test fails for f(x, y) = x⁴ + y⁴ at (0, 0), and decide what kind of point it is.
7. Find and classify the critical point of f(x, y) = e^(−(x² + y²)).
8. Find the absolute maximum and minimum of f(x, y) = 2x + y − x² on the rectangle 0 ≤ x ≤ 2, 0 ≤ y ≤ 1.
9. Find the point on the plane x + y + z = 6 closest to the origin.
10. A company makes x hundred phone cases of one kind and y hundred of another each day, with daily profit P(x, y) = 36x + 42y − x² − y² − xy, in hundreds of dollars. How many of each should it make to earn the most?

f_x = 2x - 6 = 0, \quad x = 3

f_y = 2y + 2 = 0, \quad y = -1

D = (2)(2) - 0 = 4 \gt 0, \quad f_{xx} = 2 \gt 0

f(3, -1) = 9 + 1 - 18 - 2 = -10

f_x = y - 2x = 0, \quad f_y = x - 2y + 3 = 0

x - 2(2x) + 3 = 0 \Rightarrow x = 1, \ y = 2

D = (-2)(-2) - 1^2 = 3 \gt 0, \quad f_{xx} = -2 \lt 0

f(1, 2) = 2 - 1 - 4 + 6 = 3

f_x = 2x + 4 = 0, \quad f_y = -2y + 2 = 0

D = (2)(-2) - 0 = -4 \lt 0

f_x = 3x^2 - 3y = 0, \quad f_y = 3y^2 - 3x = 0

x = x^4 \Rightarrow x(x^3 - 1) = 0

f_{xx} = 6x, \quad f_{yy} = 6y, \quad f_{xy} = -3

D(0, 0) = 0 - 9 = -9 \lt 0

D(1, 1) = 36 - 9 = 27 \gt 0, \quad f_{xx} = 6 \gt 0

f_x = 2x = 0, \quad f_y = 3y^2 - 12 = 0

D = (2)(6y) - 0 = 12y

D(0, 2) = 24 \gt 0, \quad f_{xx} = 2 \gt 0

D(0, -2) = -24 \lt 0

f_{xx} = 12x^2, \quad f_{yy} = 12y^2, \quad f_{xy} = 0

D(0, 0) = 0

f(x, y) \ge 0 = f(0, 0)

f_x = -2xe^{-(x^2 + y^2)}, \quad f_y = -2ye^{-(x^2 + y^2)}

f_{xx}(0, 0) = -2, \quad f_{yy}(0, 0) = -2, \quad f_{xy}(0, 0) = 0

D = 4 \gt 0, \quad f_{xx} \lt 0

f_y = 1 \ne 0

y = 0: \ 2x - x^2; \quad y = 1: \ 2x + 1 - x^2

x = 0: \ f = y; \quad x = 2: \ f = y

f(0, 0) = 0, \ f(2, 0) = 0, \ f(0, 1) = 1, \ f(2, 1) = 1

d^2 = x^2 + y^2 + (6 - x - y)^2

2x - 2(6 - x - y) = 0, \quad 2y - 2(6 - x - y) = 0

x = y = 2, \quad z = 2

D = (4)(4) - 2^2 = 12 \gt 0, \quad f_{xx} = 4 \gt 0

P_x = 36 - 2x - y = 0, \quad P_y = 42 - x - 2y = 0

42 - x - 2(36 - 2x) = 0 \Rightarrow 3x = 30

x = 10, \quad y = 16

D = (-2)(-2) - (-1)^2 = 3 \gt 0, \quad P_{xx} \lt 0

P(10, 16) = 360 + 672 - 100 - 256 - 160 = 516

Extrema of two-variable functions FAQ

How do you find critical points of a function of two variables?

Set both first partial derivatives equal to 0 and solve the system. Also include any points where a partial derivative does not exist.

What is a saddle point?

It is a critical point that is neither a maximum nor a minimum: the surface goes up in some directions and down in others. The second derivative test shows it with D < 0.

What does D = 0 mean in the second derivative test?

The test is inconclusive. The point could be a maximum, a minimum or a saddle, so you have to look at the function itself.

How do you find absolute extrema on a closed region?

Evaluate f at interior critical points, at the best points along each edge and at the corners. The largest and smallest of these values are the absolute maximum and minimum.

Summary

• Critical points: solve f_x = 0 and f_y = 0 together.
• D = f_xxf_yy − (f_xy)² at each critical point.
• D > 0 and f_xx > 0: minimum. D > 0 and f_xx < 0: maximum. D < 0: saddle. D = 0: no conclusion.
• On a closed, bounded region, compare interior critical points, edges and corners.
• In word problems, use the constraint to remove a variable first.

Previous lesson: Directional Derivatives and the Gradient

Next lesson: Lagrange Multipliers

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