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I'm supposed to find the absolute extreme values taken on by f on the set of D.
f (x, y) = 2x^2+y^2−4x−2y+2, D = {(x, y) : 0 ≤ x ≤ 2,0 ≤ y ≤ 2x}.
Could someone please teach me how to do this. The book doesn't really do it for me. I need to figure out how to find the extreme values.
f (x, y) = 2x^2 + y^2 − 4x − 2y + 2
df / dx = 4x -4
df / dy = 2y -2
A) set df /dx = 0
4x -4 = 0
4x = 4
x = 1
B) set df / dy = 0
2y -2 = 0
2y = 2
y = 1
You also need to 'walk' yourself around the boundary, which reduces this to a calculus one problem.
BUT you should always reduce these...
$\displaystyle 2(x^2-2x)+y^2-2y+2 = 2(x^2-2x+1)-2+(y^2-2y+1)-1+2 = 2(x-1)^2+(y-1)^2-1$.
That clearly shows that the GLOBAL min is -1 when x=y=1.
And your max will occur somewhere on the boundary.
