Maxima, Minima and Saddle Points

Q: How do you find critical points of f(x, y)?

Solve f_x = 0 and f_y = 0 together. Points where a partial derivative does not exist are also critical points.

Q: What is the second derivative test for two variables?

At a critical point, compute D = f_xx f_yy − (f_xy)². If D > 0 and f_xx > 0 it is a local minimum; if D > 0 and f_xx < 0, a local maximum; if D < 0, a saddle point; if D = 0 the test says nothing.

Q: What is a saddle point?

A critical point where the function increases in some directions and decreases in others, like the middle of a horse's saddle or a mountain pass.

Q: How do you find absolute extremes on a closed region?

Evaluate f at every critical point inside the region and find the extremes of f along the boundary. The largest and smallest of all those values are the absolute maximum and minimum.

Q: What does D = 0 mean?

The test is inconclusive. x⁴ + y⁴ has D = 0 at the origin and a minimum there, while x³ + y³ has D = 0 and neither.

Quick answer

At a local maximum or minimum of a smooth function of two variables both partial derivatives are zero, so extremes lie among the critical points, where ∇f = 0. At a saddle point, though, the surface rises one way and falls another. The second derivative test sorts them with D = f_xx f_yy − f_xy²: D > 0 gives a minimum or maximum depending on the sign of f_xx, and D < 0 gives a saddle. On a closed, bounded region the absolute extremes also have to be sought on the boundary.

What you'll learn

Find the critical points of a function of two variables
Classify critical points with the second derivative test
Recognize saddle points
Find absolute extremes on a closed, bounded region

Where the surface levels out

At the top of a smooth hill, every slice through the summit also has a top there, so its slope is zero. In particular the slices in the $x$ and $y$ directions level out, and both partial derivatives vanish. Points where

\nabla f = \mathbf{0}, \quad\text{that is,}\quad f_x = 0 \text{ and } f_y = 0

or where a partial derivative fails to exist, are the critical points. Every local maximum or minimum of a smooth function is one.

The converse fails. For $f(x, y) = x^2 - y^2$ the origin is a critical point, yet the slice along the $x$ -axis curves up and the slice along the $y$ -axis curves down. The origin is a saddle point: a low point in one direction and a high point in another.

slice y = 0: z = x²
slice x = 0: z = −y²

Two slices of z = x² − y² through the origin

The second derivative test

At a critical point $(a, b)$ , compute

D = f_{xx}(a, b)\,f_{yy}(a, b) - \bigl(f_{xy}(a, b)\bigr)^2

$D > 0$ and $f_{xx} > 0$ : local minimum.
$D > 0$ and $f_{xx} < 0$ : local maximum.
$D < 0$ : saddle point.
$D = 0$ : the test gives no answer.

Why D decides

Near a critical point the gradient term vanishes, so the second-order behavior takes over. Writing $A = f_{xx}$ , $B = f_{xy}$ and $C = f_{yy}$ at the point, a small step $(h, k)$ changes $f$ by about $\tfrac{1}{2}\left(Ah^2 + 2Bhk + Ck^2\right)$ . Completing the square, as for a quadratic,

Ah^2 + 2Bhk + Ck^2 = A\left(h + \frac{B}{A}k\right)^2 + \frac{AC - B^2}{A}\,k^2

When $D = AC - B^2 > 0$ , both coefficients share the sign of $A$ , so every step changes $f$ the same way. If $A > 0$ every step goes up, a minimum; if $A < 0$ every step goes down, a maximum. When $D < 0$ the two coefficients have opposite signs, and some steps raise $f$ while others lower it. The sign of $D$ says whether the surface curves the same way in every direction; the sign of $f_{xx}$ then says which way.

Worked examples

Example 2 — One minimum, one saddle

Classify the critical points of $f(x, y) = x^3 - 3x + y^2$ .

$f_x = 3x^2 - 3 = 0$ gives $x = \pm 1$ , and $f_y = 2y = 0$ gives $y = 0$ . With $f_{xx} = 6x$ , $f_{yy} = 2$ and $f_{xy} = 0$ :

Point	$D = 12x$	$f_{xx}$	Type	Value
$(1, 0)$	$12$	$6$	local minimum	$-2$
$(-1, 0)$	$-12$	$-6$	saddle point	$2$

The saddle has the larger value, and the function has no maximum at all: it grows without limit as $x$ grows.

Example 3 — The cheapest box

An open box, with no lid, must hold $32$ cubic feet. Find the dimensions that use the least material.

With base $x$ by $y$ and height $z = \tfrac{32}{xy}$ , the surface area is

S = xy + 2xz + 2yz = xy + \frac{64}{y} + \frac{64}{x}

$S_x = y - \tfrac{64}{x^2} = 0$ and $S_y = x - \tfrac{64}{y^2} = 0$ . Subtracting shows $x = y$ , and then $x^3 = 64$ , so $x = y = 4$ and $z = 2$ . The area is $16 + 16 + 16 = 48$ square feet. The test confirms a minimum: at $(4, 4)$ , $S_{xx} = \tfrac{128}{x^3} = 2$ , $S_{yy} = 2$ and $S_{xy} = 1$ , so $D = 2 \cdot 2 - 1 = 3 > 0$ with $S_{xx} > 0$ .

Example 4 — Absolute extremes on a rectangle

Find the absolute maximum and minimum of $f(x, y) = x^2 - 2xy + 2y$ on the rectangle $0 \le x \le 3$ , $0 \le y \le 2$ .

A continuous function on a closed, bounded region reaches both extremes, at critical points or on the boundary. The only critical point is $(1, 1)$ , where $f = 1$ . On the four edges:

Edge	$f$ becomes	Least	Greatest
$y = 0$	$x^2$	$0$ at $(0, 0)$	$9$ at $(3, 0)$
$x = 3$	$9 - 4y$	$1$ at $(3, 2)$	$9$ at $(3, 0)$
$y = 2$	$(x - 2)^2$	$0$ at $(2, 2)$	$4$ at $(0, 2)$
$x = 0$	$2y$	$0$ at $(0, 0)$	$4$ at $(0, 2)$

The absolute maximum is $9$ at $(3, 0)$ and the absolute minimum is $0$ , reached at both $(0, 0)$ and $(2, 2)$ .

Common mistakes

Practice problems

Find and classify the critical point of $f(x, y) = x^2 + y^2 + 4x - 2y$ .

Answer

A local minimum at $(-2, 1)$ , with value $-5$

Full solution

$f_x = 2x + 4$ and $f_y = 2y - 2$ vanish at $(-2, 1)$ . $D = 4 > 0$ and $f_{xx} = 2 > 0$ .
Classify the critical point of $f(x, y) = xy$ .

Answer

A saddle at the origin

Full solution

$f_x = y$ and $f_y = x$ vanish only at $(0, 0)$ . $D = 0 \cdot 0 - 1^2 = -1 < 0$ . Along $y = x$ the function is $x^2$ , and along $y = -x$ it is $-x^2$ .
Find and classify the critical point of $f(x, y) = 4 - x^2 - y^2 + 2x$ .

Answer

A local maximum at $(1, 0)$ , with value $5$

Full solution

$f_x = -2x + 2$ and $f_y = -2y$ vanish at $(1, 0)$ . $D = 4 > 0$ and $f_{xx} = -2 < 0$ .
Find and classify the critical points of $f(x, y) = x^3 - 12x + y^2 - 4y$ .

Answer

A local minimum at $(2, 2)$ with value $-20$ , and a saddle at $(-2, 2)$

Full solution

$f_x = 3x^2 - 12 = 0$ gives $x = \pm 2$ , and $f_y = 2y - 4 = 0$ gives $y = 2$ . $D = 6x \cdot 2$ , which is $24$ at $x = 2$ and $-24$ at $x = -2$ .
Compute $D$ at the origin for $f(x, y) = x^4 + y^4$ . What kind of point is it?

Answer

$D = 0$ ; it is a minimum.

Full solution

All second partial derivatives vanish at the origin, so the test says nothing. But $x^4 + y^4 > 0$ everywhere else, so the origin is the lowest point.
Show that $f(x, y) = x^2 + xy + y^2$ has a minimum at the origin.

Answer

$D = 2 \cdot 2 - 1 = 3 > 0$ and $f_{xx} = 2 > 0$ .

Full solution

$f_x = 2x + y$ and $f_y = x + 2y$ both vanish only at $(0, 0)$ , and the test gives a local minimum.
At a critical point, $f_{xx} = -3$ , $f_{yy} = -2$ and $f_{xy} = 3$ . Classify it.

Answer

A saddle point

Full solution

$D = 6 - 9 = -3 < 0$ . Both $f_{xx}$ and $f_{yy}$ are negative, yet the mixed term is large enough to make some directions rise.
A box with a lid has volume $8$ . Show that the cube, $2$ by $2$ by $2$ , has the least surface area.

Answer

$S = 2xy + \tfrac{16}{x} + \tfrac{16}{y}$ has its critical point at $x = y = 2$ .

Full solution

With $z = \tfrac{8}{xy}$ , $S = 2xy + 2xz + 2yz = 2xy + \tfrac{16}{y} + \tfrac{16}{x}$ . $S_x = 2y - \tfrac{16}{x^2} = 0$ and $S_y = 2x - \tfrac{16}{y^2} = 0$ give $x = y$ and $x^3 = 8$ . The area is $24$ .
Why must a continuous function on the square $0 \le x \le 1$ , $0 \le y \le 1$ have an absolute maximum?

Answer

The square is closed and bounded.

Full solution

The extreme value theorem guarantees both an absolute maximum and an absolute minimum for a continuous function on a closed, bounded region.
A student finds the critical points of $x^2 - 2xy + 2y$ on the rectangle of Example 4, gets only $(1, 1)$ , and reports $1$ as both the maximum and the minimum. What went wrong?

Hint

Evaluate $f$ at the corner $(3, 0)$ .

Answer

The boundary was skipped. The maximum is $9$ at $(3, 0)$ and the minimum is $0$ .

Full solution

On a closed region, extremes often occur on the edge, where the gradient is not zero. Checking all four edges, as in Example 4, finds the true extremes.

Frequently asked questions

How do you find critical points of f(x, y)?