Find the absolute extrema of f(x,y)=2x^2-4x+y^2-4y+1 on the region of the first quadrant bounded by the lines

Find the absolute extrema of f(x,y)=2x^2-4x+y^2-4y+1 on the region of the first quadrant bounded by the lines x=0, y=2x and y=2. YOur proceddure is (1) find all critical points in the interior of the region (2) find all critical points on the boundary lines; for this, you will substiture for one of the variables, thus reducing f toa function of one variable, and then find its critical point(s) but discard if they're outside the region, and (3) list eact corner point. Present your list of all candidate points, giving (x, y, z) and suitable descriptions of each and then pick the winners

1) Critical points insider the region:

Set the first partial derivatives of f equal to 0:
f_x = 4x - 4, f_y = 2y - 4

Setting these to 0 yields (x, y) = (1, 2).
Note that this is on the boundary line y = 2; so we'll ignore it (see below).
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2) Critical points on boundary lines.
i) x = 0 for y in [0, 2]:
==> g(y) = f(0, y) = y^2 - 4y + 1
g'(y) = 2y - 4 = 0 ==> y = 2

So, the critical point is (x, y) = (0, 2).
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ii) y = 2 for x in [0, 1]:
==> g(x) = f(x, 2) = 2x^2 - 4x - 3
g'(x) = 4x - 4 = 0 ==> x = 1

So, the critical point is (x, y) = (1, 2).
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iii) y = 2x for x in [0, 1]:
==> g(x) = f(x, 2x) = 6x^2 - 12x + 1
g'(x) = 12x - 12 ==> x = 1

So, the critical point is (x, y) = (1, 2).
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3) The corner points of the region are (x, y) = (0, 0), (1, 2), (0, 2).
(Note that the critical points from 2 are also corner points!)
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4) Testing all of the points for extrema:
f(0, 0) = 1 <---Maximum
f(1, 2) = -5 <---Minimum
f(0, 2) = -3

I hope this helps!