Absolute Extrema of two variables?

[tak13]

Homework Statement

Given function f(x,y) = 3x² - 2xy + 2y
a) find the critical point(s) of f,
b) sketch the region R: the triangular region in the xy-plane with vertices (0,0) (0,6) and (3,6)
c) find the absolute maximum and absolute minimum of f over R

Homework Equations

The Attempt at a Solution

I did all of them, just want you guys to check if I made any mistake:

a)fx= 6x - 2y
fy= -2x + 2

6x - 2y = 0 ; 6 = 2y ; y = 3
-2x + 2 = 0 ; x = 1

critical points (1,3)

b)
[PLAIN]http://img844.imageshack.us/img844/4607/68009750.jpg c) f(1,3) = 3*1 - 2*1*3 + 2*3
= 3 - 6 + 6
= 3

x= 0 0=<y=<6
y= 6 0=<x=<3
y=2x 0=<x=<3x= 0 0=<y=<6
f(0,y) = 2y
f(0,0) = 0
f(0,6) = 12y= 6 0=<x=<3
f(x,6) = 3x² -12x + 12
f(0,6) = 0 - 0 + 12 = 12
f(3,6) = 9 - 36 +12 = -15y=2x 0=<x=<3
f(x,y)=g(x,y)= 3x² -2x(2x) + 2(2x)
= 3x² -4x² + 4x
f(0) = 0
f(3) = 27 - 36 + 12 = 3

Absolute max is 12 at f(0,6) and absolute min is -15 at f(3,6)

Thanks in advance!

 

[Mark44]

Homework Statement

Given function f(x,y) = 3x² - 2xy + 2y
a) find the critical point(s) of f,
b) sketch the region R: the triangular region in the xy-plane with vertices (0,0) (0,6) and (3,6)
c) find the absolute maximum and absolute minimum of f over R

Homework Equations

The Attempt at a Solution

I did all of them, just want you guys to check if I made any mistake:

a)fx= 6x - 2y
fy= -2x + 2

6x - 2y = 0 ; 6 = 2y ; y = 3
-2x + 2 = 0 ; x = 1

critical points (1,3)

b)
[PLAIN]http://img844.imageshack.us/img844/4607/68009750.jpg


c) f(1,3) = 3*1 - 2*1*3 + 2*3
= 3 - 6 + 6
= 3

x= 0 0=<y=<6
y= 6 0=<x=<3
y=2x 0=<x=<3


x= 0 0=<y=<6
f(0,y) = 2y
f(0,0) = 0
f(0,6) = 12


y= 6 0=<x=<3
f(x,6) = 3x² -12x + 12
f(0,6) = 0 - 0 + 12 = 12
f(3,6) = 9 - 36 +12 = -15

Mistake above. f(3, 6) = 3.

y=2x 0=<x=<3
f(x,y)=g(x,y)= 3x² -2x(2x) + 2(2x)
= 3x² -4x² + 4x

Along the line y = 2x, f(x, y) = f(x, 2x) = g(x)

f(0) = 0
f(3) = 27 - 36 + 12 = 3

Absolute max is 12 at f(0,6) and absolute min is -15 at f(3,6)

Thanks in advance!

Check along the boundary lines again for the minimum.

 

[tak13]

Along the line y = 2x, f(x, y) = f(x, 2x) = g(x)

So for this one, it is just about the way I write it right? not something wrong the the results?

Check along the boundary lines again for the minimum.

So I checked again, due the the error I made above, the absolute minimum now is 0 at f(0,0) and f(2,6). is that correct?

Thanks for reply!

 

[Mark44]

Yeah, the first was just correcting how you wrote it. As for the second, the smallest value I got was 0, too.

I didn't say it before -- except for the small error, your work was pretty good!