Multivariable-Calculus · Unit 9 · Video 3 · Interactive Practice

Global Min and Max: x + y + 1/(xy) — Critical Points, Boundary and Infinity

IKey Formulas

Formula	Name	What it tells you
$f_x = 1 - \dfrac{1}{x^2 y}$ , $\ f_y = 1 - \dfrac{1}{x y^2}$	First partials	$x^2 y = 1$ and $x y^2 = 1$ force $x = y = 1$ : the only critical point is $(1, 1)$
$f_{xx} = \dfrac{2}{x^3 y}$ , $\ f_{xy} = \dfrac{1}{x^2 y^2}$ , $\ f_{yy} = \dfrac{2}{x y^3}$	Second partials	At $(1, 1)$ : $A = 2$ , $B = 1$ , $C = 2$
$AC - B^2 = 2 \cdot 2 - 1^2 = 3 > 0$ , $\ A > 0$	Second derivative test	$(1, 1)$ is a local minimum with $f(1, 1) = 3$
$f > x$ , $\ f > y$ , $\ f > \dfrac{1}{xy}$	Boundary and infinity	$f \to \infty$ as $x$ or $y \to 0$ and as $x$ or $y \to \infty$ : no maximum, and the global minimum is $3$

Key Insight: A global minimum or maximum can sit at a critical point, on the boundary, or out at infinity; the critical points alone cannot decide which.

IIVisualization 1 — Where Both Partials Vanish

The curve $f_x = 0$ and the curve $f_y = 0$ cross at exactly one point of the quadrant.

x 2 y 0.5

IIIVisualization 2 — Toward the Axes and Out to Infinity

Along every path toward an axis or outward, one of the three positive terms of $f$ grows without bound.

Path x 1

IVVisualization 3 — The Patch Where $f \le c$

For $c \ge 3$ , the set $f \le c$ is a closed, bounded patch around $(1, 1)$ ; below $3$ it is empty.

Level c 4

💡 A continuous function on a closed, bounded patch reaches a smallest value; here that point lies inside the quadrant, so $f_x = f_y = 0$ there, and it must be $(1, 1)$ .

VQuiz Questions

Problem 1 · Find the Critical Point

Given: $g(x, y) = x + y + \dfrac{8}{xy}$ for $x > 0$ , $y > 0$ — find its only critical point and the value of $g$ there.

Where is the critical point?

What is $g$ at the critical point?

✅ Correct!

x = y

and

y^3 = 8

give

(2, 2)

, where each of the three terms equals

2

❌ Not quite.

(1, 1)

is the critical point for

\frac{1}{xy}

; with

\frac{8}{xy}

the equations become

x^2 y = 8

and

x y^2 = 8

❌ Close, but… after

x = y

the equation is

x^2 \cdot x = x^3 = 8

, not

x^2 = 8

❌ Not quite. Set

g_x = 1 - \frac{8}{x^2 y} = 0

and

g_y = 1 - \frac{8}{x y^2} = 0

, then divide the two equations.

❌ Close, but…

x + y = 4

is only two of the three terms;

\frac{8}{xy} = \frac{8}{4} = 2

must be added.

❌ Not quite. Substitute

(2, 2)

into all three terms:

2 + 2 + \frac{8}{2 \cdot 2}

Show solution

The partial derivatives are

g_x = 1 - \frac{8}{x^2 y}, \qquad g_y = 1 - \frac{8}{x y^2}.

Setting both to zero gives $x^2 y = 8$ and $x y^2 = 8$ . Dividing the first by the second:

\frac{x^2 y}{x y^2} = \frac{x}{y} = 1 \quad\Rightarrow\quad x = y.

Then $x^3 = 8$ , so $x = y = 2$ . The only critical point is $(2, 2)$ , and

g(2, 2) = 2 + 2 + \frac{8}{4} = 6.

Problem 2 · Near the Axis, but $1/(xy)$ Stays Small

Given: $f(x, y) = x + y + \dfrac{1}{xy}$ , followed along the curve $xy = 1$ (so $y = \frac{1}{x}$ ) as $x \to 0$ — find $f(0.1, 10)$ and the behaviour of $f$ along this curve.

What is $f(0.1, 10)$ ?

As $x \to 0$ along $xy = 1$ , what does $f$ do?

✅ Correct! Along

xy = 1

the last term is stuck at

1

, but the

y

term carries

f

to infinity.

❌ Not quite. At

(0.1, 10)

the product is

xy = 1

, so

\frac{1}{xy} = 1

, not

10

❌ Not quite. Add all three terms:

x = 0.1

y = 10

and

\frac{1}{xy} = 1

❌ That is the trap. The last term stays at

1

, but

f

is larger than

y

, and

y = \frac{1}{x}

grows without bound.

❌ Not quite. Write

f

along the curve as a function of

x

alone:

f = x + \frac{1}{x} + 1

Show solution

At $(0.1, 10)$ the product is $xy = 1$ , so

f(0.1, 10) = 0.1 + 10 + 1 = 11.1.

Along $y = \frac{1}{x}$ the function is $f = x + \frac{1}{x} + 1$ . As $x \to 0$ the term $\frac{1}{x}$ (which is $y$ ) goes to infinity, so $f \to \infty$ . Approaching the $y$ -axis does not require the last term to blow up: the bound $f > y$ does the work instead, exactly as at the video's point $(0.2, 5)$ , where $f = 6.2$ .

Problem 3 · Classify with the Second Derivative Test

Given: $h(x, y) = 2x + y + \dfrac{4}{xy}$ for $x > 0$ , $y > 0$ — find its critical point and classify it.

Where is the critical point?

What does the test give there?

✅ Correct!

A = 4

B = 1

C = 1

give

AC - B^2 = 3 > 0

with

A > 0

: a local minimum,

h(1, 2) = 6

❌ Close, but… check

h_x

(2, 1)

2 - \frac{4}{4 \cdot 1} = 1 \ne 0

. The coefficient

2

sits on

x

, which makes

x

the smaller coordinate.

❌ Not quite. Solve

x^2 y = 2

and

x y^2 = 4

; dividing gives

\frac{x}{y} = \frac{1}{2}

❌ Close, but… with

AC - B^2 > 0

the sign of

A

decides, and

A = 4 > 0

means a minimum.

❌ Close, but… the test subtracts the square of the mixed partial:

4 \cdot 1 - 1^2 = 3

, not

4 + 1

❌ Not quite. Evaluate

h_{xx} = \frac{8}{x^3 y}

h_{xy} = \frac{4}{x^2 y^2}

h_{yy} = \frac{8}{x y^3}

(1, 2)

Show solution

Critical point.

h_x = 2 - \frac{4}{x^2 y} = 0 \;\Rightarrow\; x^2 y = 2, \qquad h_y = 1 - \frac{4}{x y^2} = 0 \;\Rightarrow\; x y^2 = 4.

Dividing, $\frac{x}{y} = \frac{1}{2}$ , so $y = 2x$ . Then $x^2 (2x) = 2$ gives $x^3 = 1$ , so $x = 1$ and $y = 2$ .

Second derivative test.

A = h_{xx} = \frac{8}{x^3 y} = \frac{8}{2} = 4, \qquad B = h_{xy} = \frac{4}{x^2 y^2} = \frac{4}{4} = 1, \qquad C = h_{yy} = \frac{8}{x y^3} = \frac{8}{8} = 1.

AC - B^2 = 4 \cdot 1 - 1^2 = 3 > 0, \qquad A = 4 > 0,

so $(1, 2)$ is a local minimum, with $h(1, 2) = 2 + 2 + 2 = 6$ . Since $h > 2x$ , $h > y$ and $h > \frac{4}{xy}$ , the same boundary-and-infinity argument as in the video makes it the global minimum.

Problem 4 · A Closed Square Changes the Answer

Given: $f(x, y) = x + y + \dfrac{1}{xy}$ , now only on the closed square $\tfrac{1}{2} \le x \le 2$ , $\tfrac{1}{2} \le y \le 2$ (edges included) — find the maximum of $f$ on the square.

✅ Correct! The square is closed and bounded, so a maximum exists; it is on the boundary, at the corner nearest the origin, where

\frac{1}{xy} = 4

❌ Close, but… at

(2, 2)

the sum

x + y

is largest, but

\frac{1}{xy}

is only

\frac{1}{4}

; near the origin the reciprocal term wins.

❌ Not quite.

(1, 1)

is still the critical point, but it is the minimum; the maximum must be on the boundary of the square.

❌ Not quite. On the open quadrant

f

is unbounded, but this square is closed and bounded and stays away from the axes, so

f

reaches a largest value.

Show solution

The only critical point, $(1, 1)$ , is the minimum $f = 3$ . The maximum is on the boundary. Check the edges.

Edge $x = \tfrac{1}{2}$ : $f = \tfrac{1}{2} + y + \tfrac{2}{y}$ , which decreases to $y = \sqrt{2}$ and then increases, so its largest values are at the ends: $f(\tfrac{1}{2}, \tfrac{1}{2}) = 5$ and $f(\tfrac{1}{2}, 2) = \tfrac{7}{2}$ .

Edge $x = 2$ : $f = 2 + y + \tfrac{1}{2y}$ , with ends $f(2, \tfrac{1}{2}) = \tfrac{7}{2}$ and $f(2, 2) = \tfrac{17}{4}$ .

The edges $y = \tfrac{1}{2}$ and $y = 2$ give the same values by the symmetry $f(x, y) = f(y, x)$ . Comparing the candidates:

\max\left(5,\ \tfrac{7}{2},\ \tfrac{17}{4}\right) = 5 = f\left(\tfrac{1}{2}, \tfrac{1}{2}\right).

Including the boundary is what makes the maximum exist; on the open quadrant the same function has none.

Solved: 0 / 4

Global Min and Max: x + y + 1/(xy) — Critical Points, Boundary and Infinity

IKey Formulas

IIVisualization 1 — Where Both Partials Vanish

IIIVisualization 2 — Toward the Axes and Out to Infinity

IVVisualization 3 — The Patch Where f≤cf \le cf≤c

VQuiz Questions

IVVisualization 3 — The Patch Where $f \le c$