A math problem of linear programming, solved?

If the production of A x kg and the production of B y kg are made, the optimization model is:

max = 600x + 400y

2y <=100,2x + 3y <= 120,x>=0;

y>=0;

LINGO solves**:

max = 600*x + 400*y;

4*x + 2*y <= 100;

2*x + 3*y <= 120;

Solution: Global Optimal Solution Found

objective value:

infeasibilities:

total solver iterations: 2variable value reduced costx

yrow slack or surplus dual price is x=,y=35, the target is the largest, 18500

Set up the production of XKG A and YKG B.

4x+2y≤100

2x+3y≤120

z=600x+400y

Draw. 4x+2y=100

2x+3y=120

x=0, y=0 four straight lines Find the linear programming interval.

z=600x+400y is 600x+400y=0 translation.

1) According to the constraints, draw the feasible solution domain:

2) Draw an objective function diagram

Let the value of the objective function be zero, and the slope can be obtained, and a straight line through the origin is made according to the slope, 2x+y=0.

If the feasible solution domain is in the first quadrant and the slope of the objective function contour is negative) if the problem is given to find the maximum, move the objective function contour parallel to the point where it intersects with the last feasible solution domain, which is the optimal solution of the problem; If the problem is given to find the minimum value, the objective function contour is moved parallel to the point where it intersects first with the feasible solution domain, which is the optimal solution to the problem.

So, when panning to point a, a maximum value of 9

In addition, when panning to point b, a minimum value of 3

The freight rate of each air conditioner is 20 yuan for a type A car, and the freight rate for each type B car is 30 yuan, so a type A car can use 4 cars, a total of 80 air conditioners, and another 20 units are completed by 2 type B cars, and the total freight is 4 * 400 + 2 * 300 = 2200 yuan minimum freight.

As shown in the figure, the red line is the objective function, and the shaded part is the solution range of the objective function. From the diagram, we can see that the objective function reaches a minimum value of -1 at the point (m-1,1), i.e., (m-1)-1=-1 gives m=1

If you need 10 grams of raw material X for type A and 10 grams for raw material Y, the total cost of Z yuan is: Z=3X+2Y

Protein: 5x+7y>=35

Iron: 10x+4y>=40

x=y=3 is the most cost-effective.

z = 28 grams of A and 30 grams of B are the most cost-effective.

Solution: Set up a raw material x grams, B raw material y grams, then.

5x/10+7y/10≥35（1）

10x/10+4y/10≥40（2）

x≥0y≥0

The total cost is z=3x 10+2y 10

The minimum value is obtained when z crosses the intersection of two straight lines (1) and (2).

Since the picture can't be uploaded, it won't come with it.

Suppose the number of bags in the two packages is x, and y is 35x+24y>=106 (constraints), and the number of costs is: z=140x+120y, and the minimum requirements (objective function) are all possible as follows:

x = 0 1 2 3 4

y = 5 3 2 1 0

z = 600 500 520 540 560, it will cost at least 500 yuan (buy 1 bag for a large package, buy 3 bags for a small package).

I often have this kind of question in high school, and it's nothing more than this kind of thinking.

To give you a** this detail.

Or the math problem of two rivers to lay sewers is the most cost-effective.

1) Solution: You can draw a picture first, I don't have word here, I can't draw.

After drawing, it is obvious that the intersection of 2x-y>=0 and y>=x is in the first quadrant.

y>=-x+ b can be seen as a horizontal line of y>=-x, which is gradually pushed upwards to cover the upper right.

As can be seen from the figure, it is obvious that the intersection of y>=-x+ b and 2x-y>=0 is the point taken, and the minimum value is taken.

The simultaneous equations obtain the intersection point (b 3, 2b 3), z=2x+y=4b3=3, and the solution is: b=9 4.

2) Solution: f(x)=sinx(sinx-cosx)=(sinx) 2-sinxcosx=(1-cos2x) 2-(sin2x) 2=1 2-(cos2x+sin2x) 2, put forward the root number 2 in parentheses, and judge its monotonically decreasing interval in the form of sin(2x+4).

The answer is: -3 8+k x 8+k

1. Make a feasible domain. There are two cases: b>0 and <0.

In the second one, the brackets are first dismantled, and the double angle formula is used respectively. and then the differential product. Mobile phone inconvenience.

1) According to the slope of the image and the equation, it can be concluded that the point through which z=2x+y obtains the minimum value is the intersection of the equation y=x and y=-x+b, and the intersection point is (b 2, b 2), then 3b 2 = 2, b = 2

2)f(x)=(sinx) 2-sinxcosx=-1 2[2(cosx) 2-2]-1 2sin2x=-1 2cos2x-1 2sinx+1 2=- 2 2sin(2x+ 4)+1 2 then the monotonically decreasing interval of f(x) is [-3 8+k, 3 8+k](k z).

Hope it helps!

1. First draw 2x-y=0, y=x, y=-x+ b with three straight lines, find the general feasible domain, and then find the intersection coordinates of these three straight lines, and then draw the straight line 2x+y=0 (of the dashed line), translate the straight line 2x+y=0, from which you can get b<0, you can find the translated straight line through the intersection of y=-x+ b and 2x-y=0, and obtain the minimum value.

2、f(x)=sinx(sinx-cosx)=（sinx）^2-sinxcosx=(1-cos2x)/

Root number two) sin (2x+4).

The solution is then obtained according to the monotonic interval of sin(2x+4).