URGENT ISSUE SACHOST PROCESS

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The second equation in the inequality group is deformed as x>=y; The third equation is deformed as 2x-2>=y; Subtract the left and right sides of the third equation from the left and right sides of the ground equation to get x-2>=0, which is x>=2;Look at the last subnumerator y-1, because y>=0, so y-1>=-1;The denominator is x+1, because x>=2, so x+1>=3. The result is a>=-1 3

With the inequality group y>=0, x-y>=0, 2x-y-2>=0, linear programming can be carried out and a feasible domain can be drawn. The value range of a=(y-1) (x+1) can be expressed mathematically as the value range of finding the slope w of the straight line where the point (x,y) and the point (-1,1) are located.

So, in fact, the meaning of the problem is to find the range of the slope w of the straight line formed by the point (x,y) and the fixed point (-1,1) in the feasible domain.

Therefore, it is advisable to let the intersection point of the line 2x-y-2=0 and the x-axis be a, that is, a is (1,0) and the fixed point (-1,1) is b. And with the feasible domain, it is not difficult for us to determine:

When the point (x,y) of the feasible domain is at point a, the slope of the line ab is minimal, i.e., the minimum value of w is .

a=(0-1)/(1+1)=-1/2

When the connected straight line is more or less parallel to the straight line y=x, the greater the value of a, but the slope a of the connected straight line can only infinitely approach the slope k=1 of the straight line y=x, and cannot be equal or greater than. So a<1

So to sum up, the answer is that the value range of a is [-1 2,1).

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The range given is represented in coordinates, and a can be seen as the slope.

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