這里用普通的做法,可以適用于普遍情形,但可能顯得復(fù)雜.
(x-2)²+y²=1是C(2,0)為圓心,半徑為1的圓,1 ≤ x ≤ 3
(1) z = y/x,y = xz
(x-2)²+x²z² =1
z² = f(x) = (-x² + 4x - 3)/x²
f'(x) = (6 - 4x)/x³ = 0
x = 3/2
f(3/2) = 1/3
此時(shí)z = y/x最大,為√(1/3) = √3/3 (不考慮z < 0)
(2)
z =x²+y²
y² = z - x²
(x-2)²+ z - x² =1
z = 4x - 3
x = 1,z = 1
x = 3,z = 9
z的范圍:[0,9]
(3) z = y - x
y = x + z
(x-2)²+ (x + z)² =1
z = -x - √(-x² + 4x - 3) (最小值,不考慮+√(-x² + 4x - 3) )
z' = -1 + (x - 2)/√(-x² + 4x - 3) = 0
(x - 2)/√(-x² + 4x - 3) = 1,須x > 2
解得x = 2 + √2/2 (舍去x = 2 - √2/2 < 2)
此時(shí)z = y - x最小,為-2 - √2