Question

Find the absolute maximum and minimum values of the following function in the closed region bounded by the triangle with vertices​ (0,0), ​(0,44​), and ​(22​,44​) in the first quadrant. f left parenthesis x comma y right parenthesis equals 2 x squared minus 8 x plus y squared minus 8 y plus 2f(x,y)=2x2−8x+y2−8y+2 Determine the absolute maximum value of f. Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice. A. On the given​ region, the​ function's absolute maximum is nothing. B. On the given​ region, the function has no absolute maximum value. The function assumes this value at nothing. ​(Type an ordered pair. Use a comma to separate answers as​ needed.) Determine the absolute maximum value of f. Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice. A. On the given​ region, the​ function's absolute minimum is nothing. B. On the given​ region, the function has no absolute minimum value. The function assumes this value at nothing.

Answer 1

$f(x,y)=2x^2-8x+y^2-8y+2$

$f_x(x,y)=4x-8=0\implies x=2$

$f_y(x,y)=2y-8=0\implies y=4$

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 $x=0$, we have

$f(0,y)=y^2-8y+2=(y-4)^2-14$

which attains a maximum value of

$f(0,4)=-14$

Along $y=44$, we have

$f(x,44)=2x^2-8x+1586=2(x-2)^2+1578$

which attains a maximum of

$f(2,44)=1578$

Along $y=2x$, we have

$f(x,2x)=6x^2-24x+2=6(x-2)^2-22$

which attains a maximum of

$f(2,4)=-22$

So over the given region, the absolute maximum of $f(x,y)$ is 1578 at (2, 44).