Simplification of two formulas in conditions

y≥0.25x+0.75

y≦-0.6x+5

draft

y=0.25x+0.75

y=-0.6x+5

There is also a condition given by the title, x≥ 1, so draw x= 1 (a straight line perpendicular to the horizontal axis).

These three formulas

Draw it in two-dimensional coordinates

The area between the three lines is the scope of the subject condition.

The figure is a triangle, and the rightmost corner is the intersection of two straight lines.

Z=2x+y can be changed to y=-2x+z, where z is unknown.

Analysis y =-2x+z.

It is found that it is a straight line with a slope of -2, but the intercept is unknown.

Then we can imagine what y=-2x+z looks like on the coordinate map. It is a straight line with the same inclination angle to the lower right, and will move up and down in parallel when Z takes different values.

At this time, the range we drew in front is useful. The straight line y=-2x+z must have a short segment, at least one point of which is within the triangle we draw.

That is, y=-2x+z moves up and down in parallel on the diagram, to a position, and then goes out of the triangle a little bit upward, and the maximum value is obtained at this point Z. Write down x and y at this time and substitute z=2x+y to get the maximum value of z.

In the same way, y=-2x+z moves to a position, then moves down a little, and a triangle is formed, so it can't move down any more. At this point, z gets the minimum value. Write down x and y at this time and substitute z=2x+y to get the minimum value of z.

In this problem, the maximum value of the rightmost angle of the triangle is the minimum value of the lower left corner.

The rightmost angle is the intersection of two inequalities in the condition, x=5, y=2, so the maximum value is 5*2+2= 12.

The lower left corner is the intersection of x= 1 and y=0.25x+0.75, x= 1, y= 1, and the minimum value is 1*2+ 1=3.