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( 1) 16/3
The straight line y=x-z 1+ 1 has a slope of 1 and an intercept of 1-z 1. Maximize z 1 and minimize the intercept.
It can be seen that the intercept at (8/3, -5/3) is the smallest, which makes z 1 the largest.
Bring in the solution: z 1= 16/3.
(2)(﹣∞,- 1 1/8]∪[ 1/2,∞)
Z2=(y-2)/(x-0), which represents the slopes of points m and (0,2).
Wherein the minimum value of the positive part of the slope is at the slope of point m and (0,2) on the straight line of x-2y+4=0 that satisfies the condition,
The slope is (2-0)/[0-(-4)]= 1/2,
Where the maximum value of the negative part of the slope is the slope of the straight line formed by (8/3, -5/3) and (0,2),
The slope is (-5/3-2)/(8/3-0) =-11/8,
Therefore, the range of the slope z2 is: (-∞,-11/8] ∩ [1/2, ∞).
(3)37
z3=(x-2)？ +(y- 1)？ The meaning of the expression is the square of the distance from point m to (2 1) that satisfies the condition.
When the maximum value of z3 is taken, the distance from the point M that meets the conditions to (2 1) is the farthest.
It is known that when m is (-4,0), z3 is the largest.
z3=(2+4)？ +( 1-0)? =37
I hope my answer is helpful to you. . I wish you a happy study. . .