Answer
Therefore, the absolute maximum is 1 at (0, 0)
and the absolute minimum is 5 at (1, 2).
Work Step by Step
(i) On OA, 2 f x y f y y y ( , ) (0, ) 4 1 on
0 2; (0, ) 2 4 0 2; y f y y y
f (0, 0) 1 and (0, 2) 3 f
(ii) On AB, 2 f x y f x x x ( , ) ( , 2) 2 4 3 on
0 1; x ( , 2) 4 4 0 1; f x x x
f (0, 2) 3 and (1, 2) 5 f
(iii) On OB, 2 f x y f x x x x ( , ) ( , 2 ) 6 12 1 on 0 1; x endpoint values have been found above;
f x x x x ( , 2 ) 12 12 0 1 and y 2, but (1, 2) is not an interior point of OB
(iv) For interior points of the triangular region, ( , ) 4 4 0 xf x y x and ( , ) 2 4 0 1 yf x y y x and
y 2, but (1, 2) is not an interior point of the region. Therefore, the absolute maximum is 1 at (0, 0)
and the absolute minimum is 5 at (1, 2).
