Can't find absolute minimums and maximums!
Find the absolute maximum and minimum values of $f(x,y) = y^{2}+x^{2}-4x+9$ on the set D where D is the closed triangular region with vertices $(8,0), (0,6)$, and $(0,-6)$.
To find critical points:
$F_x = 0$
$F_y = 0$
$F_x$ (first order derivative with respect to x) $ = 2x - 4$
$F_y = 2y$
Setting them equal to $0$, to find critical points:
$2x - 4 = 0$
$2y = 0$
critical point: $(2,0)$
...
Now, the boundary of the triangle can be expressed in 3 lines:
Left side of triangle, $x = 0$:
Function can be expressed by the one variable function
$f(0, y) = y^2 + 9$
Absolute maximums:
You evaluate those at the given points, and I think also supposed to evaluate at the critical point? (derivative equal to $0$ etc)
$(0,6), (0,-6)$
Absolute minimums:
??? (8,0) <- why not at these points? If y = 0, then that's when it has the absolute minimum, no?
...
Upper right side of the triangle:
between points: $(0,6)$ and $(8,0)$
$y= \frac{-3}{4} x + 6 $
Function can be expressed by the one variable function
$f(x, \frac{-3}{4} x + 6 ) = \ (\frac{-3}{4} x + 6)^2 + x^2 - 4x + 9$
Absolute maximums:
$(0, 6)$
Absolute minimums:
(8,0) INCORRECT, WHY???
...
Lower right side of the triangle:
between points: $(0,-6)$ and $(8,0)$
$y = \frac{3}{4} x - 6$
Function can be expressed by the one variable function
(3/4 x - 6)^2 + x^2 - 4x + 9
Absolute maximums:
????
Absolute minimums:
????
p.s. here's the screenshot