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anonymous

  • one year ago

Given the linear programming problem, use the method of corners to determine where the minimum occurs and give the minimum value minimize C=x+3y Subjectto x less than equal to 3 y less than equal to 5 x greater then equal to 0 y greater to equal than 0

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  1. anonymous
    • one year ago
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    Did you draw this?

  2. anonymous
    • one year ago
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    Im not sure how to ... I can read a graph already drawn but creating one myself is a different story.

  3. anonymous
    • one year ago
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    Let's start with this.|dw:1441475021450:dw|

  4. anonymous
    • one year ago
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    \(y \ge 0\) is everything above x-axis|dw:1441475235023:dw|

  5. anonymous
    • one year ago
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    and \(x \ge 0\) is everything right of the y-axis

  6. anonymous
    • one year ago
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    Then for \(y \le 3\) shade everything below the line \(y=3\)|dw:1441475376673:dw|

  7. anonymous
    • one year ago
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    \(x \le 5\) is left of the line x = 5|dw:1441475429193:dw|

  8. anonymous
    • one year ago
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    So your corners are the corners of the black rectangle above, where all 4 inequalities are valid

  9. anonymous
    • one year ago
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    Write the coordinates of those 4 corners then test them all in the constraint \(C=x+3y\) to find the smallest

  10. anonymous
    • one year ago
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    so it would 0,3 and 5,0?

  11. anonymous
    • one year ago
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    yes, those are two of the corners. (0, 0) and (5, 3) are the others

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