Question

Find the absolute extrema of the function over the region R. (In this case, R contains the boundaries.) Use a computer algebra system to confirm your results. (There are multiple correct answers.) f(x, y) = (2x − y)2 R: The triangular region in the xy-plane with vertices (2, 0), (0, 1), and (1, 2)

Answer 1

Answer:

Absolute minima  

all the points along the line y = 2x for 2/5 ≤ x ≤ 4/5

minimum value f=0

Absolute maximum (2,0)

maximum value f=16

Step-by-step explanation:

Let's first find the extrema inside the triangle

Since  

$\bf \frac{\partial f}{\partial x}=4(2x-y)\\\frac{\partial f}{\partial y}=-2(2x-y)$

we see that the partial derivatives vanishes when y=2x.

When y=2x, f(x,y)=0.

Hence all the points along the segment of the line y=2x inside the triangle are minimums (see picture)

Minima (value=0)

all the points along y=2x with 2/5 ≤ x ≤ 4/5

There are no more extrema points in the interior so the rest must lie in the border.

The segment that joins (2,0) with (0,1) is part of the line

y = -x/2+1 with 0≤ x≤ 2

at all that points

$\bf f(x,y)=f(x,-x/2+1)=(2x-(-x/2+1))^2=(5x/2-1)^2$ (0≤ x≤ 2)

this is a parabola which attains its maximum at x=2 and we have a local maximum at (2,0)

The segment that joins (0,1) with (1,2) is part of the line

y = x+1 with 0≤ x≤1

at all that points

$\bf f(x,y)=f(x,x+1)=(2x-(x+1))^2=(x-1)^2$ (0≤ x≤ 1)

this is a parabola which attains its maximum at x=0 and we have a local maximum at (0,1)

The segment that joins (1,2) with (2,0) is part of the line

y = -2x+4 with 1≤ x≤2

at all that points

$\bf f(x,y)=f(x,-2x+4)=(2x-(-2x+4))^2=(4x-4)^2$ (1≤ x≤ 2)

this is a parabola which attains its maximum at x=2 and we have a local maximum at (2,0)

Since

f(0,1) = 1 and f(2,0) = 16

the absolute maximum is reached at (2,0)

We have then as a summary

Absolute minima  

all the points along the line y = 2x for 2/5 ≤ x ≤ 4/5

minimum value f=0

Absolute maximum (2,0)

maximum value f=16