微積分絕對極值問題非常急(20點)跪求神人

(x,y)=x^2-2xy+2y,find the absolute maximum points and the absolute minimum points on D={(x,y):0<=x<=2,0<=y<=3x}
第一階偏導數找出 stationary points: (1,1).
f(1,1) = 1第2階導數測驗可得知此連相對極值都不是, 而是一個鞍點.



絕對極值可能在 stationary point 或邊界.三個邊界:
(1) y=0, x in [0,2], f(x,0) = x^2, min.=0=f(0,0), max.=4=f(2,0)(2) x=2, y in [0,6], f(2,y) = 4-4y+2y = 4-2y, min. = -8=f(2,6), max.=4=f(2,0)(3) y=3x, x in [0,2], f(x,3x) = x^2-2x(3x)+2(3x) = -5x^2+6x = h(x) (say)h'(x) = 6-10x, critical point: x = 3/5
h(0) = 0, h(2) = -20+12 = -8
(這兩個值已在 (1), (2) 中出現, h(0)=f(0,0), h(2)=f(2,6))
h(3/5) = f(3/5,9/5) = -5(3/5)^2+6(3/5) = 9/5比較以上各結果,
abs.max. = 4 = f(2,0)
abs.min. = -8 = f(0,6)

2014-06-16 17:40:38 補充：
修正: 最後結論 abs.min. = -8 = f(2,6)


又: f(3/5,9/5) = 9/5 是在邊界 y=3x 上的一個相對及絕對極大,
卻不是 f(x,y) 在指定區域的絕對極大.

到下面的網址看看吧

▶▶http://candy5660601.pixnet.net/blog

到下面的網址看看吧

▶▶http://candy5660601.pixnet.net/blog

到下面的網址看看吧

▶▶http://candy5660601.pixnet.net/blog

到下面的網址看看吧

▶▶http://candy5660601.pixnet.net/blog

老怪物的解法才是對的

麻辣是錯的

D={(x,y):0<=x<=2,0<=y<=3x} => y=3x, (0,0)<=(x,y)<=(2,6)F(x,y)=x^2-2xy+2y=x^2-2x*3x+2*3x=x^2-6x^2+6x=-5x^2+6x=-5[x^2-6x/5+(6/10)^2]+36*5/100=-5(x-3/5)^2+9/5max=F(3/5,9/5)=9/5min=F(2)=-5*(2-3/5)^2+9/5=9/5-5*(7/5)^2=9/5-49/5=-8

2014-06-16 07:23:04 補充：
max=F(3/5,9/5)=9/5

改為max=F(3/5)=9/5