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A problem in the 2008 math college entrance examination (you can just say ideas)
When x+y is less than or equal to 1, that is, the areas of x and y are right triangles.

At this time, ax+by is less than or equal to 1, and the deformation is a triangle of x/(1/a)+y/(1/b) < =1,and the intercepts are1/b.

When x is greater than or equal to 0, y is greater than or equal to 0, and x+y is less than or equal to 1, there is always ax+ less than or equal to 1, that is, x+y.

Therefore,1/a >; = 1, 1/b >= 1, that is, a.

The area is 1