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Back mathematics
My personal suggestion is to find the next example to explain it in detail or the textbook example has a format. The following is for reference only ~

Solution: First, the equations are "expanded" (for example, 0≤x+y≤20 is divided into two, which is convenient for drawing later).

Square y≥x- 10

Cheng y≥-x

Group y ≤-x+20

Enclose y≥0

No y≤ 15

Then let y=x- 10, y=-x, y=-x+20, y=0, y= 15 and draw as follows.

Note: the inequality in this step will be plotted after the equation, and at the same time, it is necessary to substitute special points according to the original inequality to find out which direction the value range is! It is also an error-prone point, such as y ≥-x. If you make an image of y=-x and substitute it into a special point (not unique, such as point 1, 0), you will get 0≥- 1 in the case of inequality, so the value is taken in the upward part of the function image (individuals mark the function relationship with directional arrows on the sketch, indicating that

After drawing and judging the range of values, a closed graphic area will be formed, as shown in the following figure (the picture comes from the geometric sketchpad ~)

Then set the objective function z=2x+3y, and then convert it into y=-2/3x+ 1/3z.

Then make a function image with y=-2/3x. In the process of solving this problem, the maximum and minimum values are actually obtained by moving the image of this objective function in this closed area. This is generally to take the points (a, b, c, d, o) that make up the polygon angle.

In fact, we can know the range of values according to z=2x+3y. In this way, it is easy to know that the maximum value of point A is 55 and the minimum value of point O is 0.

It is important to note that some topics may be mixed ≥, ≤, >,