First check for the critical points of f by checking where the first-order derivatives vanish.
[tex]\dfrac{\partial f}{\partial x}=4x-4=0\implies x=1[/tex]
[tex]\dfrac{\partial f}{\partial y}=2y-4=0\implies y=2[/tex]
Notice how the point (1, 2) lies on the line y = 2x ; at this point, we get a value of f(1, 2) = -5 (MIN).
Next, check the points where the boundary lines intersect, which occurs at the points (0, 0), (0, 2), and (1, 2). We already checked the last one. We find f(0, 0) = 1 (MAX) and f(0, 2) = -3.
Now check on the boundary lines themselves. If x = 0, then
[tex]f(0,y)=y^2-4y+1=(y-2)^2-3[/tex]
which has a maximum value of -3 when y = 2 (so we get the same critical point as before).
If y = 2, then
[tex]f(x, 2)=2x^2-4x-3=2(x-1)^2-5[/tex]
with a maximum of -5 when x = 1.
If y = 2x, then
[tex]f(x,2x)=6x^2-12x+1=6(x-1)^2-5[/tex]
with the same maximum of -5 when x = 1.
This question is based on the absolute maximum and absolute minimum.
We get this by differentiating the terms.
Given:
f(x,y) = [tex]2x^{2} - 4x + y^2 - 4y +1[/tex], bounded by the lines x=0,y=2,y=2x in the first quadrant,bounded by the lines x=0,y=2,y=2x in the first quadrant.
We need to determined the absolute maximum and absolute minimum of the function.
Now, partial differentiating wrt x and y.
[tex]\dfrac{\partial f}{ \partial x} = 4x -4 = 0 \Rightarrow x= 1 \\\dfrac{\partial f}{ \partial y} = 2y - 4 = 0 \Rightarrow y = 2[/tex]
Now, point (1, 2) lies on the line y = 2x ; at this point, we get a value of
f(1, 2) = -5 (MIN).
Next, check the points where the boundary lines intersect, which occurs at the points (0, 0), (0, 2), and (1, 2).
Now, find f(0, 0) = 1 (MAX) and f(0, 2) = -3.
Now check on the boundary lines themselves.
If x = 0, then we get,
[tex]f(0,y) = y^2 - 4y +1 = ( y-2)^2 -3\\[/tex]
which has a maximum value of -3 when y = 2 (so we get the same critical point as before).
If y = 2, then we get,
f(x,2) = [tex]2x^2-4x -3 = 2(x-1)^2 -5[/tex] with a maximum of -5 when x = 1.
If y = 2x, then we get,
f(x,2x) = [tex]6x^2 -12x +1 = 6(x-1)^2 -5[/tex] with the same maximum of -5 when x = 1.
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