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Math Help - Simultaneous Inequalities

  1. #1
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    Simultaneous Inequalities

    Hi Guys,

    This one is long problem. I figured out most of it, but need a hint with the last proof part.

    Draw a diagram illustrating the region S of the xy-plane which is defined by the simultaneous inequalities x + y \ge 7, 2x + y \le 13, 2x + 3y \le 19, and give the coordinates of the vertices of S. Prove that, if line y = kx intersects S, then \dfrac{1}{6} \le k \le \dfrac{5}{2}.

    The Point P lies on y = kx and is in the region S. Prove that, when \dfrac{1}{6} \le k \le \dfrac{3}{5}, the maximum value for the y-coordinate of P is \dfrac{13k}{2 + k} and find the corresponding expression when \dfrac{3}{5} \le k \le \dfrac{5}{2}



    I completed the initial parts of the problem. Got the 3 vertices of the region S, A = (6,1), B=(5,3), C = (2,5)

    Then since y = kx must pass through A or C, got max and min slopes and hence proved, \dfrac{1}{6} \le k \le \dfrac{5}{2}.

    I am stuck at the next part of the problem. Any ideas? Thanks for your help!
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  2. #2
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    Quote Originally Posted by mathguy80 View Post
    Hi Guys,

    This one is long problem. I figured out most of it, but need a hint with the last proof part.

    Draw a diagram illustrating the region S of the xy-plane which is defined by the simultaneous inequalities x + y \ge 7, 2x + y \le 13, 2x + 3y \le 19, and give the coordinates of the vertices of S. Prove that, if line y = kx intersects S, then \dfrac{1}{6} \le k \le \dfrac{5}{2}.

    The Point P lies on y = kx and is in the region S. Prove that, when \dfrac{1}{6} \le k \le \dfrac{3}{5}, the maximum value for the y-coordinate of P is \dfrac{13k}{2 + k} and find the corresponding expression when \dfrac{3}{5} \le k \le \dfrac{5}{2}



    I completed the initial parts of the problem. Got the 3 vertices of the region S, A = (6,1), B=(5,3), C = (2,5)

    Then since y = kx must pass through A or C, got max and min slopes and hence proved, \dfrac{1}{6} \le k \le \dfrac{5}{2}.

    I am stuck at the next part of the problem. Any ideas? Thanks for your help!
    Dear mathguy80,

    Since, \dfrac{1}{6} \le k \le \dfrac{3}{5}, the line lies in between the points A and B.

    Consider the line which goes through A and B. Its equation is,

    \frac{y-1}{x-6}=\frac{1-3}{6-5}=-2

    y=-2x+13

    Now the point P lies on the line y=kx.

    Since k>0, it is clear that y would be a maximum when the point P lies on AB (Please refer to your diagram).

    Therefore, kx=-2x+13

    x=\frac{13}{k+2}

    Therefore, y=\frac{13k}{k+2}

    Hope this will help you.
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  3. #3
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    Hey @Sudharaka. I was away for a while and forgot about this thread! Your solution checks out. Thanks and sorry for the late reply.
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  4. #4
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    Quote Originally Posted by mathguy80 View Post
    Hey @Sudharaka. I was away for a while and forgot about this thread! Your solution checks out. Thanks and sorry for the late reply.
    You are welcome.
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