[tex]f(x,y)=2x^2-8x+y^2-8y+2[/tex]
[tex]f_x(x,y)=4x-8=0\implies x=2[/tex]
[tex]f_y(x,y)=2y-8=0\implies y=4[/tex]
There is one critical point at (2, 4), but this point happens to fall on one of the boundaries of the region. We'll get to that point in a moment.
Along the boundary [tex]x=0[/tex], we have
[tex]f(0,y)=y^2-8y+2=(y-4)^2-14[/tex]
which attains a maximum value of
[tex]f(0,4)=-14[/tex]
Along [tex]y=44[/tex], we have
[tex]f(x,44)=2x^2-8x+1586=2(x-2)^2+1578[/tex]
which attains a maximum of
[tex]f(2,44)=1578[/tex]
Along [tex]y=2x[/tex], we have
[tex]f(x,2x)=6x^2-24x+2=6(x-2)^2-22[/tex]
which attains a maximum of
[tex]f(2,4)=-22[/tex]
So over the given region, the absolute maximum of [tex]f(x,y)[/tex] is 1578 at (2, 44).
The absolute maximum of f(x, y) at (2, 44) is 1578.
From the information given, it should be noted that there is only one critical point which is at (2,4).
Along the boundary x = 0, we'll have:
f(0, y) = y² - 8y + 2 = (y - 4)² - 14. This will give a maximum value of -14.
Also, along y = 44, we'll have f(x, 44) = 2x² - 8x + 1586 = 2(x - 2)² + 1578. This will attain a maximum of f(2, 44) = 1578.
Therefore, absolute maximum of f(x, y) at (2, 44) is 1578.
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