Firstly, draw three constrained linear images in the same coordinate system (that is, L 1:x-y=- 1 and other three linear images).

Then understand the region represented by the constraint.

x-y & gt； =- 1 is a column, indicating the straight line and the right side of the straight line.

what do you think? You can select any point on the straight line L 1

At the same horizontal position, the ordinate is the same,

The right abscissa of this point is larger than the original abscissa, so the right point should be >; =- 1.

Similarly, the other two straight lines can also be viewed left and right.

It is concluded that the intersection area of the three constraints is a closed triangle.

Then make an image of L2 line: 4x+y = 0,

The image of the objective function Z=4x+y is a straight line parallel to L2,

Where z is the intercept of this straight line on the y axis.

(it can be changed to y=-4x+Z, and if the y coefficient is-1, then z is the reciprocal of the intercept).

Translate L2, pass through every point in the target area, and get the point with the largest intercept.

Drawing design

X-y+ 1=0 or less

X+y- 1=0, 3x-y-3=0 or more.

This is a triangle.

y=-4x+z

When a straight line and a triangle with a slope of -4 have a common point, their maximum intercept.

Obviously, there is a maximum when x-y+ 1 = 0 and 3x-y-3 = 0 intersect (2,3).

Substituting that point is what you want.

So z=4x+y max = 1 1.

Method two,

Combine these three conditions into three equations and solve them.

The three points to be solved are respectively substituted into z = 4xy to find the maximum value.

This method reduces the steps of drawing, saves time and has a good effect.