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Pre Calc

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1. Describe the linear programming situation for this system of inequalities.

x<(or equal too)1 y>(or equal too)0
3x + y<(or equal too)5

2. Describe the linear programming situation for this system of inequalities where you are asked to find the maximum value of f(x, y) = x + y.

x>(or equal too) 0 y>(or equal too)0
6x + 3y<(or equal too)18 x + 3y<(or equal too)9

  • Pre Calc - ,

    left (or at) of x = 1
    above or on x axis
    below or on y = -3x+5
    which hits x axis at x = 5/3
    and hits the y axis at y = 5
    sketch that
    You see that the vertical line x = 1 hits our sloped line somewhere above the x axis
    find that point
    x = 1
    y = -3(1) + 5 = 2
    so
    we have a corner at (1, 2)
    everything in the upper half plane left of the sloping line and the vertical line x = 1

    It did not say x had to be positive, so it goes left forever. That would not be likely in a real linear programming problem.

  • Pre Calc - ,

    2. Describe the linear programming situation for this system of inequalities where you are asked to find the maximum value of f(x, y) = x + y.

    x>(or equal too) 0 y>(or equal too)0
    6x + 3y<(or equal too)18 x + 3y<(or equal too)9
    =================================
    In first quadrant due to x>/=0 and y>/= 0

    6x + 3y<(or equal too)18
    is
    y </= -2x + 6
    x and y axis intercepts at x = 3 and at y = 6
    below that line

    3 y </= -x + 9
    y </= -(1/3) x + 3
    intercepts at x = 9 and at y = 3
    below that line

    we need the corner where those sloped lines hit
    -(1/3) x + 3 = -2x+6
    -x + 9 = -6 x + 18
    5 x = 9
    x = 9/5
    then y = -(1/3)(9/5) + 3
    = 3 - 3/5 = 12/5

    So we have a corner at (9/5 , 12/5)

    so three corners to test
    (0 , 3) , (9/5 , 12/5) , (3 , 0)
    at (0,3) x+y = 3
    at (9/5,12/5) x+y = 21/5 = 4. something
    at (3,0) x+y = 3 again
    so the max is at the intersection of the sloped lines and is 21/5

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