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
[tex] \bf \frac{\partial f}{\partial x}=4(2x-y)\\\frac{\partial f}{\partial y}=-2(2x-y)[/tex]
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
[tex] \bf f(x,y)=f(x,-x/2+1)=(2x-(-x/2+1))^2=(5x/2-1)^2[/tex] (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
[tex] \bf f(x,y)=f(x,x+1)=(2x-(x+1))^2=(x-1)^2[/tex] (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
[tex] \bf f(x,y)=f(x,-2x+4)=(2x-(-2x+4))^2=(4x-4)^2[/tex] (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
