Multivariable-Calculus · Unit 8 · Video 3 · Interactive Practice

Minimum, Maximum, or Saddle? Writing f as a Sum of Squares Decides

IKey Formulas

Formula	Name	What it tells you
$x^2 + y^2$ , $\ -x^2 - y^2$ , $\ y^2 - x^2$	Model shapes	Local min (bowl), local max (upside-down bowl), saddle at the origin
$(x - y)^2 = x^2 - 2xy + y^2$	Absorbing the cross term	The square that starts with $x^2 - 2xy$
$x^2 - 2xy + 3y^2 + 2x - 2y = (x - y + 1)^2 + 2y^2 - 1$	Completed squares	The video's $f$ written as squares plus a constant
$f(x, y) \ge -1$ , equality only at $(-1, 0)$	Lower bound	The critical point is a minimum with value $-1$

Key Insight: $f_x = f_y = 0$ only says the tangent plane is horizontal; rewriting $f$ as a sum or difference of squares shows which of the three shapes it has.

IIVisualization 1 — Bowl, Cap or Saddle: Slices Through the Origin

A critical point is a saddle when slices through it disagree: some directions rise, others fall.

Surface Direction θ 0°

IIIVisualization 2 — The Squared Form Gives $f \ge -1$

Both forms of $f$ give the same value at every point, and the squared form shows it is never below $-1$ .

x 1 y 1

IVVisualization 3 — The Leftover $y^2$ Coefficient Decides

Replacing $3y^2$ by $cy^2$ keeps the critical point at $(-1, 0)$ ; the sign of $c - 1$ decides its type.

Coefficient c 3

💡 The rewriting is exact only because $f$ is quadratic; for general functions the second-derivative test of the next lecture does this job.

VQuiz Questions

Problem 1 · Read the Squares

Given: $g(x, y) = (x - 2)^2 + 3(y + 1)^2 + 5$ — classify its critical point and give the value of $g$ there.

✅ Correct! Both squares are at least

0

, so

g \ge 5

, with equality only when

x = 2

and

y = -1

❌ Check the signs.

(x - 2)^2 = 0

means

x = 2

, and

(y + 1)^2 = 0

means

y = -1

❌ Not quite. Both squares enter with a plus sign, so

g

is a sum of non-negative terms plus

5

: it can only rise away from the point where both squares vanish.

Show solution

Each square is non-negative: $(x - 2)^2 \ge 0$ and $3(y + 1)^2 \ge 0$ . Therefore

g(x, y) \ge 0 + 0 + 5 = 5 \quad \text{for every } (x, y).

Equality needs both squares to be zero: $x - 2 = 0$ and $y + 1 = 0$ , i.e. $(x, y) = (2, -1)$ , where $g(2, -1) = 5$ .

So $(2, -1)$ is a minimum with value 5 — the bowl shape of $x^2 + y^2$ , shifted and stretched.

Problem 2 · Positive Squares, Hidden Saddle

Given: $f(x, y) = x^2 + 4xy + y^2$ , whose only critical point is $(0, 0)$ — what kind of critical point is it?

✅ Correct!

f = (x + 2y)^2 - 3y^2

is a difference of squares: it rises along

y = 0

and falls along

x = -2y

❌ The cross term matters. Along the line

y = -x

f = x^2 - 4x^2 + x^2 = -2x^2 < 0

, so

f

drops below

f(0, 0) = 0

❌ It can be decided by hand. Complete the square to absorb

4xy

x^2 + 4xy = (x + 2y)^2 - 4y^2

❌ Not quite. Along

y = 0

f = x^2

rises in both directions, so

(0, 0)

cannot be a maximum.

Show solution

The perfect square that starts with $x^2 + 4xy$ is $(x + 2y)^2 = x^2 + 4xy + 4y^2$ . We have only $y^2$ , i.e. $3y^2$ too few:

f(x, y) = (x + 2y)^2 - 3y^2.

Along $y = 0$ : $f = x^2$ , rising in both directions. Along $x = -2y$ : $f = -3y^2$ , falling in both directions.

Some directions rise and others fall, so $(0, 0)$ is a saddle point — like $y^2 - x^2$ , not like $x^2 + y^2$ .

Problem 3 · Complete the Squares, Then Read Off the Minimum

Given: $f(x, y) = x^2 + 2xy + 2y^2 - 4y$ — rewrite $f$ as squares plus a constant and find its minimum.

Which rewriting is correct?

Where is the minimum, and what is its value?

✅ Correct!

f = (x + y)^2 + (y - 2)^2 - 4 \ge -4

, with equality only at

(-2, 2)

❌ Check the constant.

(y - 2)^2 = y^2 - 4y + 4

adds an extra

4

that

f

does not have, so it must be subtracted.

❌ Check the expansion.

(x + 2y)^2 = x^2 + 4xy + 4y^2

has cross term

4xy

, not

2xy

; the square starting with

x^2 + 2xy

(x + y)^2

❌ Check the signs.

y - 2 = 0

gives

y = 2

, and then

x + y = 0

gives

x = -2

❌ Both squares must vanish. At

(0, 2)

(x + y)^2 = 4

, so

f(0, 2) = 0

, not

-4

❌ Check the value. The squares are at least

0

, so the smallest value is the constant term,

-4

Show solution

Step 1: absorb the cross term. $(x + y)^2 = x^2 + 2xy + y^2$ , so

f = (x + y)^2 + y^2 - 4y.

Step 2: complete the square in $y$ . $(y - 2)^2 = y^2 - 4y + 4$ has one extra $4$ , so subtract it:

f = (x + y)^2 + (y - 2)^2 - 4.

Step 3: read off the bound. Both squares are at least $0$ , so $f \ge -4$ . Equality needs $y - 2 = 0$ and $x + y = 0$ , i.e. $(x, y) = (-2, 2)$ .

Check with the original formula:

f(-2, 2) = 4 - 8 + 8 - 8 = -4.

So the minimum is at $(-2, 2)$ with value $-4$ .

Problem 4 · When Does It Become a Saddle?

Given: $h(x, y) = x^2 + 6xy + c\,y^2$ , with a critical point at $(0, 0)$ — for which values of $c$ is $(0, 0)$ a saddle point?

✅ Correct!

h = (x + 3y)^2 + (c - 9)y^2

is a difference of squares exactly when

c - 9 < 0

❌ A positive $c$ can still give a saddle. For

c = 3

, along

x = -3y

h = 9y^2 - 18y^2 + 3y^2 = -6y^2 < 0

, while along

y = 0

h = x^2 > 0

❌ Square the half-coefficient. The square that absorbs

6xy

(x + 3y)^2

, which contains

9y^2

, not

3y^2

❌ That is the minimum case. For

c > 9

h

is a sum of squares, so

h \ge 0

with equality only at the origin.

Show solution

Absorb the cross term: $(x + 3y)^2 = x^2 + 6xy + 9y^2$ , so

h(x, y) = (x + 3y)^2 + (c - 9)\,y^2.

Along $y = 0$ , $h = x^2$ rises. Along the line $x = -3y$ , the first square is zero and $h = (c - 9)y^2$ .

$c > 9$ : sum of squares, so $(0, 0)$ is a minimum.
$c < 9$ : $h$ falls along $x = -3y$ but rises along $y = 0$ , so $(0, 0)$ is a saddle point.
$c = 9$ : $h = (x + 3y)^2$ , which is $0$ along the whole line $x = -3y$ .

So the origin is a saddle exactly when $c < 9$ .

Solved: 0 / 4

Minimum, Maximum, or Saddle? Writing f as a Sum of Squares Decides

IKey Formulas

IIVisualization 1 — Bowl, Cap or Saddle: Slices Through the Origin

IIIVisualization 2 — The Squared Form Gives f≥−1f \ge -1f≥−1

IVVisualization 3 — The Leftover y2y^2y2 Coefficient Decides

VQuiz Questions

IIIVisualization 2 — The Squared Form Gives $f \ge -1$

IVVisualization 3 — The Leftover $y^2$ Coefficient Decides