High School Sequence and Linear Programming Questions for detailed answers

7. The essence is the sum of a series of equal differences, with a total of 50 terms (because the initial value is 1, add 2 each time, and compare with 100, when k is taken to 99, the cycle is ended, it should be noted here that it is added first and then compared).

Then s=(3*1+3*99) 2*50=3*100 2*50=75008, this is a linear programming problem.

The first four inequalities constrain a region, and the next z takes the maximum value in the interval, and here it should be noted that the maximum value must be included at the intersection point (even if there is the same slope, so that the whole change is the maximum value, and the intersection point is also included, it is considered that there is no problem to obtain at the intersection point).

The four intersections here are (0,0)(20,0)(0,25)(10,20), and it is easy to sketch an area.

Then the maximum value of z is obtained at (20,0) and is 100

7, which is a simple proportional sequence s=7500;

8. Since the optimal solution of the convex constraint is always at the vertices, you can bring in 4 vertices to be the answer.

7. k(n)=k(n-1)+2=k(0)+2nk(50)=101>100, the cycle is terminated at this time.

s(n)=s(n-1)+3k(n-1)

s(n-2)+3k(n-2)+3k(n-1)s(0)+3(k(0)+k(1)+…k(n-1))s(0)=0,k(0)=1

s(50)=3(1+3+5+..99)=3*50*50=75008, it is a linear programming problem, and the maximum point appears at (20,0) and z=100

Do you dig your pants and tell Jane to draw a picture first, z=2x+y

If it is seen as a translation of the straight line 2x+y=0, then the drawing can know that if and only when the straight line 2x+y=0 is moved to the passing point (4,8), the value of z is maximum, and the equation for the straight line is 2x+y=16. This point is the focal point of the lines y=x+4 and y=2x. Got it?

Also, it's possible that you're not sure about the area you're planning. I'll show you how. You divide y and x into unequal signs, and y is on the left, and make sure that y's coefficient is positive.

z=x 2+y 2-6x-2y+10=(x-3) 2+(y-1) 2z is the distance from point (x,y) to point (3,1).

By ||x|-|y||1. Available.

When x,y have the same sign: -1 x-y 1

When x,y is different: -1 x+y 1

Draw an image and find the distance from the point (3,1) in the area to get 5 z 17

Gives you a picture at night.

Because , is a range, resulting in a, b is also a range. Finding the maximum and minimum values of (b-3) (a-1) is equivalent to finding the maximum and minimum values of the slopes of points (a, b) and (1, 3). So use linear programming.

The relationship between a and b is Vedic theorem, +a [1,3],a [-3,-1]. α2b. 2b∈[0，2], b∈[0，1]

1.If w=(y-1) (x+1) it should be equal to the slope of the point (-1,1).

2.If it is easier to use as a graph method, select the slope limit value of the xy graph interval to the point (-1,1), and the answer is that w is greater than -1 2 is less than 1.

The real numbers x,y satisfy the group of inequalities. The value range of w=y-1 x+1 is w=y.

1: Can we think of (y-1 x+1) as the slope of the point (to the point (-1,1) or (1,-1)?)

2: Only (x,y) that can be operated in the objective function w=y-1 x+1 is required

z=ax+y, then y=-ax+z, z is the intercept of the line y=-ax+z on the y-axis.

The point (3,1) is the intersection of the straight line x+y=4, x-y=2, and the drawing is known if and only if -a<-1 meets the topic.

Hence a>1

Hope it helps.

Learning progress o ( o thank you.

You draw the target area correctly, treat the objective function as a function of x, y, and only satisfy the slope greater than 1 to get the maximum value at the ) point.

cd is symmetrical with respect to the origin of point d (x,0) then point b (x,1-x 2).

The area of the rectangle is 2x(1-x 2) and the derivative is 2-6x 2 x=1 3 squares.

I have questions about the above answer to the hero, and my answer is as follows:

Suppose that the area enclosed by the parabola and the x-axis is fixed, and the fixed value is p; The coordinates of point c are (x,0);

then there is d(-x,0), a(x,1-x 2).

The area of the rectangular ABCD is: S=2X*(1-X 2)=2X-2X 3;

The value of the whole piece of land y=(p-s)*q+s*3q=(p-2x+2x 3)q+3q(2x-2x 3).

pq+q(4x-4x^3)

For y, the derivative is: dy dx=4-12x 2=0, so x=sqrt(3) 3;

Thus obtaining the coordinates of the cd.

Linear programming problem: Although the equation is not the same as the expression of the drawn line, they are certainly parallel... In the optimization problem, it is only necessary to know the definition domain.

Linear programming questions, why draw graphs? Because in mathematics, many problems are not intuitive, but as soon as the picture is drawn, you can intuitively see the corresponding area, which is convenient to solve the problem and can improve efficiency.

Finally, don't be afraid of trouble when learning math, because math is inherently troublesome!

1. When drawing a straight line, it becomes 2x+y=0, which is to determine the slope of l0, and then it is not to move l up by 50 in the direction of the y axis, which is the line l.

2. Because the graphics are more vivid and intuitive, it is not easy to make mistakes. If you can still do the right thing without the help of graphics, you can also do without drawing.

It's not difficult to understand that if you make a 2x+y+50 line and the shadow part will not intersect, what you ask for is the minimum value of z, then only ensure that 2x+y is the smallest, here is mainly to find the most suitable match, and finding z is the last step.

Drawn below.

The straight line 2x+y=0 is a straight line parallel to the straight line z=2x+y+50. To get the value of z, move the straight line 2x+y=0Finally, the position of Z=2x+Y+50 is determined to calculate the value of X,Y where the minimum value of Z is obtained.

Linear programming requires drawing, and the coordinates of each line with 2 points can be determined. The last value is calculated by determining the slope of the straight line where z is located, and then moving in parallel to determine the intercept of the straight line.

That constant does not affect the slope of the objective function, and the result is the same with or without operations, you try.

50 is the intercept, and the straight line is drawn according to the slope and intercept.

1) Assuming that X trucks of type A and Y trucks of type B are dispatched every day, then 30x+40y 280, because x and y are macro numbers of the whole cover, the desirable values are: x=0, y=7; x=1,y=7;x=2,y=6;x=3,y=5;x=4,y=4;x=5,y=4;x=6,y=3;x=7,y=2;x=8,y=1;x=9,y=1;x=10,y=0.

2) It is known that x,y should satisfy: x 0, y 0,30x+40y 280,z=

3) To minimize the cost, draw the above straight lines on the coordinate axis, you can know that when x=0, y=7, the cost is the smallest, which is 7. Therefore, 0 A-type trucks and 7 B-type trucks should be dispatched per day.

Upstairs didn't notice that "6 Type A trucks and 4 Type B trucks are known".

1) Set high hail type A x vehicles, B type Y vehicles, 0 "wide tremor = x< = 6, 0< = y< = 430x+40y 280 (draw these 3 lines for drawing) x,y rounded.

So the scheme is: (4,4)(5,4)(6,4)(3,6)2)0<=x<=6, 0<=y<=4 30x+40y≥280 ，z=

3) Add this line Z= on the chart, and the point of intersection (4, 4) is the minimum cost point, 4 cars of type A are cautious of failure, and 4 cars of type B.