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.