On Extremum Problems

In one sentence

To find the extreme values of $z = f (x, y)$ on a region $D$ , think of $f (x, y) = c$ as a family of curves (level curves). The value $c$ can be taken by $z$ only if its level curve has a point in common with $D$ . So:

z_{m a x} = the last c before the level curve leaves D .

For a linear expression $z = a x + b y$ the level curves are a family of parallel lines; when $D$ is a convex polygon, the last thing they touch is always a vertex.

Example: the extreme values of z = 2x + y on triangle ABC

Step 1 / 5

A level line
$A (- 2, 1)$ , $B (3, 4)$ , $C (4, - 2)$ . At $c = 0$ the line $2 x + y = 0$ crosses the triangle, which means $z = 0$ is attained.

$2 x + y = 0$
Push it up and to the right
The last point
Gone
The other direction

Predict first

On the same triangle $A (- 2, 1)$ , $B (3, 4)$ , $C (4, - 2)$ , what is the maximum of $z = x - y$ ?

Choose an answer first; then you can see the result.

Stop 1 · A linear problem in three variables

$x + y + z = 2$ , $0 \leq x \leq 1$ , $0 \leq y \leq 1$ , $z \geq 0$ . Find the extreme values of $F = x + 2 y + 3 z$ .

First use the equation to eliminate one variable: $z = 2 - x - y$ and $F = 6 - (2 x + y)$ , and the conditions become $0 \leq x \leq 1$ , $0 \leq y \leq 1$ ( $z = 2 - x - y \geq 0$ then holds automatically). The problem is now: the extreme values of $2 x + y$ on a square.

2 x + y

has minimum

0

(0, 0)

and maximum

3

(1, 1)

. So

F

has maximum

6

(

x = y = 0, z = 2

) and minimum

3

(

x = y = 1, z = 0

Stop 2 · A linear expression on a conic: the level line is tangent

Now let the region be the ellipse $\frac{x ^{2}}{16} + \frac{y ^{2}}{9} \leq 1$ , with the objective still $z = x + y$ . At the moment the level line leaves the ellipse, it is tangent to it.

Push

c

5

: the line and the ellipse have only one point left in common.

One problem, several solutions

The point $(x, y)$ lies on the ellipse $\frac{x ^{2}}{16} + \frac{y ^{2}}{9} = 1$ . Find the maximum of $x + y$ .

Tangent level line: discriminant 0

Substitute $y = c - x$ : $9 x^{2} + 16 (c - x)^{2} = 144$ , that is $25 x^{2} - 32 c x + 16 c^{2} - 144 = 0$ . Tangency means $Δ = 0$ : $(32 c)^{2} = 100 (16 c^{2} - 144)$ , $1024 c^{2} = 1600 c^{2} - 14400$ , $c^{2} = 25$ , so the maximum is $c = 5$ .

CostSubstitute, tidy up, compute one discriminant.

The eccentric angle as parameter

$x + y = 4 cos φ + 3 sin φ = 5 sin (φ + φ_{0}) \leq 5$ , where $tan φ_{0} = \frac{4}{3}$ . Equality holds when $cos φ = \frac{4}{5}$ and $sin φ = \frac{3}{5}$ , at the point of tangency $(\frac{16}{5}, \frac{9}{5})$ .

CostOne line, and you get the point of tangency for free.

Stop 3 · When the level curves are not lines

The level curves of $z = x^{2} + y^{2}$ are circles centered at the origin, and $z$ is the squared distance to the origin; the level curves of $z = \frac{y - 1}{x + 2}$ are lines through $(- 2, 1)$ , and $z$ is the slope. First recognize what the level curves are, then see where they last touch the region. That may be a vertex, the middle of an edge (see the Mistake Museum exhibits below), or a point of tangency.

Practice

Exercise 1Basic

$x \geq 0$ , $y \geq 0$ , $3 x + 4 y \leq 12$ . Find the maximum of $z = 3 x + 2 y$ .

Exercise 2Basic

On the same region, find the minimum of $z = x - 2 y$ .

Exercise 3Advanced

The point $(x, y)$ lies on the ellipse $\frac{x ^{2}}{4} + y^{2} = 1$ . Let $M$ be the maximum of $x + 2 y$ . Find $M^{2}$ .

Exercise 4Challenge

$x + y \geq 2$ , $x \leq 3$ , $y \leq 3$ . Find the minimum of $x^{2} + y^{2}$ .

In one sentence

Example: the extreme values of z = 2x + y on triangle ABC

Stop 1 · A linear problem in three variables

Stop 2 · A linear expression on a conic: the level line is tangent

Tangent level line: discriminant 0

The eccentric angle as parameter

Stop 3 · When the level curves are not lines

Practice

Mistake Museum · Exhibits from this stop

Comparing only the vertices when the objective is not linear

“The bigger the intercept, the bigger z”