Anonymous
Anonymous asked in Science & MathematicsMathematics · 6 months ago

Math help?

Maximize the function

f(x, y) = 6x + 7y

in the region determined by the following constraints.

3x+ 2y ≤ 18

3x + 4y≥ 12

x≥0

y≥0

3 Answers

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  • 6 months ago
    Favorite Answer

    Maximize f=6x+7y

    constraints:

    3x+2y<=18

    3x+4y>=12

    x, y>=0

    (a) Graphical solutions:

    (i) draw lines

    3x+2y=18----(1)

    3x+4y=12----(2)

    7y+6x=0=>y=-6x/7----(3)

    on a piece of the graph paper.

    (ii) Find out the feasible region R

    which is bounded by (1), (2) &

    x=0, y=0.

    (iii) Push the line (3) parallely to the

    farthest corner of R which should be

    (0,9) with the help of the set squares.

    (iv) The max.f=6(0)+7(9)=63.

    (b) The numerical method:

    (i) Using the M-technique with M=1000.

    Rewrite the constraints as

    3x+2y+S1=18

    3x+4y-S2+R=12

    x, y>=0

    Maximize f-3006x-4007y+1000S2=-12000.

    After 2 iterations, get

    x=0

    y=9

    S1=0

    S2=24

    R=0

    Max. f=63

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  • 6 months ago

    Please check answers already provided here: https://au.answers.yahoo.com/question/index?qid=20...

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  • 6 months ago

    It's a linear function, so it will be maximized on the boundary; specifically at one of the vertices when the bounary is a polygon. Write the first two inequalities as equations in intercept form:

    x/6 + y/9 = 1 . . . . divide by 18 to get x-intercept = 6 and y-intercept = 9

    x/4 + y/3 = 1 . . . . divide by 12 to get x-intercept = 4, y intercept = 3

    The >= direction of the 2nd inequality, and the fact both intercepts of the second boundary line are inside the region defined by the 1st inequality, make it easy. All corners of the boundary are on the axes. Evaluate the function at (6, 0), (0, 9) , (4, 0) and (0, 4) to find the maximum location and value. I get f(0, 9) = 63.

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