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Thread: 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 $\displaystyle x + y \ge 7, 2x + y \le 13, 2x + 3y \le 19$, and give the coordinates of the vertices of S. Prove that, if line $\displaystyle y = kx$ intersects S, then $\displaystyle \dfrac{1}{6} \le k \le \dfrac{5}{2}$.

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



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

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

    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 $\displaystyle x + y \ge 7, 2x + y \le 13, 2x + 3y \le 19$, and give the coordinates of the vertices of S. Prove that, if line $\displaystyle y = kx$ intersects S, then $\displaystyle \dfrac{1}{6} \le k \le \dfrac{5}{2}$.

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



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

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

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

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

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

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

    $\displaystyle y=-2x+13$

    Now the point P lies on the line $\displaystyle y=kx$.

    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, $\displaystyle kx=-2x+13$

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

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

    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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