Absolute Extreme Values

• Feb 27th 2008, 01:14 PM
Absolute Extreme Values

Find the absolute extreme values taken on by f(x,y) = 2x^2 -4x + y^2 -4y +3 on the triangular region bounded by the lines x = 0, y = 2, y=2x

Steps I have done:

gradient of f = (4x -4 , 2y -4) = 0 so then x = 1, y =2 to give the point (1,2)

Now the three lines make up a triangle which would then have three boundaries

B1 = { (x,y): 0 <= x <= 1, y =2 }
parameterize to: (t,2) t is in [0,2]
plugging back into f(t) gives me an equation, which I derive and solve for t to get the extreme value. Gives me (1,2)

B2 = {(x,y) : 0 <= x <=1, y = 2x}
parameterize to: (t,2t) t is in [0,2]
plugging back into f(t) gives me an equation, which I derive and solve for t to get the extreme value. Gives me (1,2)

How do I parameterize the last boundary which is the upward line on the y axis? I know the boundary = {(x,y): x = 0, 0 <= y <= 2 }
• Feb 28th 2008, 03:27 AM
mr fantastic
Quote:

Originally Posted by powerader

Find the absolute extreme values taken on by f(x,y) = 2x^2 -4x + y^2 -4y +3 on the triangular region bounded by the lines x = 0, y = 2, y=2x

Steps I have done:

gradient of f = (4x -4 , 2y -4) = 0 so then x = 1, y =2 to give the point (1,2)

Now the three lines make up a triangle which would then have three boundaries

B1 = { (x,y): 0 <= x <= 1, y =2 }
parameterize to: (t,2) t is in [0,2]
plugging back into f(t) gives me an equation, which I derive and solve for t to get the extreme value. Gives me (1,2)

B2 = {(x,y) : 0 <= x <=1, y = 2x}
parameterize to: (t,2t) t is in [0,2]
plugging back into f(t) gives me an equation, which I derive and solve for t to get the extreme value. Gives me (1,2)

How do I parameterize the last boundary which is the upward line on the y axis? I know the boundary = {(x,y): x = 0, 0 <= y <= 2 }

$\displaystyle x = 0$
$\displaystyle y = t, \, 0 \leq t \leq 2$