Topic 12 / 17advanced

Optimization

Find local maxima, minima, and saddle points of $f(x, y)$ using critical points and the second derivative test. Find absolute extrema on closed bounded regions.

To optimize a function of two variables, find **critical points** where $\nabla f = 0$ , then classify them with the second derivative test. For absolute extrema on a closed, bounded region, also check the boundary.

Critical points

Set

f_{x} = 0

and

f_{y} = 0

simultaneously. Solutions are critical points. Local extrema can only occur at critical points or where partials don't exist.

Second derivative test

At a critical point

(a, b)

, compute the discriminant

D = f_{xx} f_{y y} - (f_{x y})^{2} .

D > 0, f_{xx} > 0

: local minimum. -

D > 0, f_{xx} < 0

: local maximum. -

D < 0

: saddle point. -

D = 0

: test inconclusive.

Key takeaways

First check the sign of $D$ .
If $D > 0$ , then $f_{xx}$ tells max vs. min.
$D < 0$ ⇒ saddle, regardless of $f_{xx}$ .

Absolute extrema on closed bounded regions

Continuous function on closed bounded

\Rightarrow

absolute max and min exist (Extreme Value Theorem). Method: 1. Find interior critical points (set

\nabla f = 0

). 2. Find extrema on each piece of the boundary (parametrize, reduce to 1D problem). 3. Compare all candidate values; pick the largest and smallest.

Strategy tips

Don't forget to check **corners** of the boundary as candidates.
Lagrange multipliers are useful for constrained optimization on smooth boundaries.

Lagrange multipliers (for advanced constraint problems)

To extremize

f (x, y, z)

subject to

g (x, y, z) = c

, set

\nabla f = λ \nabla g

and solve together with the constraint. Each solution is a candidate; compare values.

Worked examples

Example 1

Find and classify the critical points of

f (x, y) = 3 x y - x^{2} y - x y^{2}

1
Partials.
$f_{x} = 3 y - 2 x y - y^{2} = y (3 - 2 x - y), f_{y} = 3 x - x^{2} - 2 x y = x (3 - x - 2 y)$
2
Critical points: solve simultaneously.
3
Case 1: $y = 0, x = 0$ . Point $(0, 0)$ .
4
Case 2: $y = 0$ and $3 - x - 2 y = 0 \Rightarrow x = 3$ . Point $(3, 0)$ .
5
Case 3: $x = 0$ and $3 - 2 x - y = 0 \Rightarrow y = 3$ . Point $(0, 3)$ .
6
Case 4: $3 - 2 x - y = 0$ and $3 - x - 2 y = 0$ . Solve: $x = y = 1$ . Point $(1, 1)$ .
7
Second partials.
$f_{xx} = - 2 y, f_{y y} = - 2 x, f_{x y} = 3 - 2 x - 2 y$
8
Classify $(0, 0)$ : $D = (0) (0) - 9 = - 9 < 0$ → saddle. $(3, 0)$ : $D = (0) (- 6) - (3 - 6 - 0)^{2} = - 9 < 0$ → saddle. $(0, 3)$ : similar saddle. $(1, 1)$ : $f_{xx} = - 2, f_{y y} = - 2, f_{x y} = - 1$ , $D = 4 - 1 = 3 > 0$ and $f_{xx} = - 2 < 0$ → local **max**.

Answer

Saddle points at

(0, 0), (3, 0), (0, 3)

. Local max at

(1, 1)

with

f (1, 1) = 3 - 1 - 1 = 1

Example 2

Find absolute max and min of

f (x, y) = x^{2} + 2 x y + 3 y^{2}

on the disk

x^{2} + y^{2} \leq 1

1
Interior: $f_{x} = 2 x + 2 y = 0, f_{y} = 2 x + 6 y = 0$ . Subtract: $- 4 y = 0 \Rightarrow y = 0$ , then $x = 0$ . Critical point $(0, 0)$ with $f = 0$ .
2
Boundary: parameterize $x = cos t, y = sin t$ .
$f = cos^{2} t + 2 cos t sin t + 3 sin^{2} t = 1 + sin 2 t + 2 sin^{2} t = 2 + sin 2 t - cos 2 t$
3
Express as $A + R sin (2 t - ϕ)$ .
$= 2 + 2 sin (2 t - π /4)$
4
Range: $2 - 2 \leq f \leq 2 + 2$ .
5
Combine with interior value 0. Min = 0, max = $2 + 2$ .

Answer

Absolute min: 0 at

(0, 0)

. Absolute max:

2 + 2

on the boundary.

Example 3

Find the points on the surface

x y^{2} z^{3} = 2

closest to the origin.

1
Minimize $D^{2} = x^{2} + y^{2} + z^{2}$ subject to $g (x, y, z) = x y^{2} z^{3} - 2 = 0$ via Lagrange multipliers.
2
Set $\nabla (D^{2}) = λ \nabla g$ .
$(2 x, 2 y, 2 z) = λ (y^{2} z^{3}, 2 x y z^{3}, 3 x y^{2} z^{2})$
3
Three equations + constraint, four unknowns. Manipulate to find ratios.
4
From the equations: $x : y /2 : z /3 = 1 : 1 : 1$ (after careful manipulation), giving $y^{2} = 2 x^{2}$ , $z^{2} = 3 x^{2}$ .
5
Substitute into constraint.
$x \cdot 2 x^{2} \cdot (3 x^{2})^{3/2} = 2 \Rightarrow 2 x^{3} \cdot 33 x^{3} = 2 \Rightarrow x^{6} = \frac{1}{3 3}$
6
Solve and verify minimum (full algebra omitted; symmetric multiple solutions exist).

Answer

Closest points have

x^{2} = (33)^{- 1/3}

y^{2} = 2 x^{2}

z^{2} = 3 x^{2}

(with appropriate signs). Compute numerically as needed.

Interactive visualizations

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Surface

f (x, y) = 3 x y - x^{2} y - x y^{2}

. Look for the local max near

(1, 1)

and saddles at

(0, 0), (3, 0), (0, 3)

Formulas in this topic

Critical points

f_{x} = 0, f_{y} = 0

Solve simultaneously.

Discriminant

D = f_{xx} f_{y y} - (f_{x y})^{2}

Used in second derivative test.

Lagrange multiplier condition

\nabla f = λ \nabla g

For constraint g(x, y, z) = c.

Critical points

Second derivative test

Absolute extrema on closed bounded regions

Lagrange multipliers (for advanced constraint problems)

Worked examples

Interactive visualizations

Formulas in this topic

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