Ask all kinds of students and teachers, how to do this question? Thank you very much, I must be grat

Updated on Financial 2024-08-15
6 answers
  1. Anonymous users2024-02-16

    Solution: Set up to rent X trains of type A and Y trains of type B, according to the relationship in the question, there are:

    x+y=16

    18x+16y≥266

    10x+11y≥169

    Draw the above three equations to establish the coordinate axis, and you will get a region.

    Because the fuel cost = 1500x+1200y, so let 1500x+1200y be the smallest, that is, on the line of x+y=16, and the integer solution closest to the origin in the region is the solution, it can be found that when taking 18x+16y=269 and 10x+11y=169 near the intersection point, it should be the closest to the origin in the three regions, and the nearest integer interprets x=5, y=11, so the solution is to rent 5 cars of the first type, 11 cars of type B, the minimum is 5 1500 + 11 1200 = 20700 yuan.

  2. Anonymous users2024-02-15

    Solution: (1) Set up to rent X trucks of type A, and rent (16-x) trucks of type B, according to the title, 18x+16(16x) 266

    10x+11(16−x)≥169②

    By , x 5, by , x 7, so, 5 x 7, x is a positive integer, x = 5 or 6 or 7, therefore, there are 3 car rental schemes: scheme 1: group A type of trucks 5, type B 11 trucks.

    Plan 2: Group 6 type A trucks and 10 type B trucks. Option 3:

    There are 7 trucks of type A and 9 trucks of type B. (2) Method 1: From (1), rent X trucks of type A and rent trucks of type B for (16-x), and the total fuel cost of the two trucks is Y yuan.

    According to the title, y=1500x+1200(16-x), =300x+19200, 300 0, when x=5, y has a minimum value, y min=300 5+19200=20700 yuan;

  3. Anonymous users2024-02-14

    <> rotate it yourself and watch it. Thank you.

  4. Anonymous users2024-02-13

    After conversion you get (a+b) 2 (a-b) 2=2, then (a+b) (a-b)=+sqrt(2).

  5. Anonymous users2024-02-12

    Proof: (1).

    Prove the monotonicity of a function by definition].

    Take x1, x2 r, and x1 x2

    then f(x1) f(x2) a [2 (2 x1 1)] a [2 (2 x2 1)].

    2(2 x1 2 x2)] [(2 x1 1)(2 x2 1)] y 2 x increments on ( and x1 x2 2 x1 2 x2

    2^x1)-(2^x2)<0

    (2 x1 1) (2 x2 1) 0

    f(x1)-f(x2)<0

    i.e. f(x1) f(x2).

    f(x) is an increment function on (.

    2) f(x) is an odd function, then f(0) a [2 (2 0 1)] a 1 0

    A 1 is tested and f(x) is an odd function when a 1.

    3) f(t 2+2) + f (t 2-tk) > 0 is constant and immediately has f(t 2+2) >-f(t 2-tk) = f(tk-t 2) and t 2 is obtained by the increase function t 2+2>tk-t 2

    2t^2-tk+2>0

    Then there is discriminant = k 2-16<0

    i.e. get -4

  6. Anonymous users2024-02-11

    Big brother, I can't see clearly, but I can't see it clearly.

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