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anonymous

  • one year ago

PLEASE HELP!!!! using the following system of inequalities find the maximum value of f(x,y)=3x+8y x≥0 y≥0 3x+2y≤18 6x+7y≤42

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  1. Nnesha
    • one year ago
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    you need to solve last two equations for y or just use desmos calculator shade the region

  2. anonymous
    • one year ago
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    y≤-3/2x+9 y≤-6/7x+6

  3. anonymous
    • one year ago
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    $$x\le 0\\y\le 0\\y\le -\frac32 x+9\\y\le -\frac67 x+6$$ |dw:1438790372303:dw|

  4. anonymous
    • one year ago
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    the intersection is at $$-\frac32 x+9=-\frac67 x+6\\-21x+14\cdot9=-12x+14\cdot 6\\14\cdot 3=9x\\\frac{14}3=x$$

  5. anonymous
    • one year ago
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    so now we have the following points to test: $$x=0,\qquad y=0\\x=0,\qquad y=6\\x=0,\qquad y=9\\x=14/3,\ \ y=2\\x=6,\qquad y=0\\x=7,\qquad y=0$$

  6. anonymous
    • one year ago
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    oops, ot the \(x=0,y=9\) and \(x=7,y=0\), only the four points that bound the quadrilateral

  7. anonymous
    • one year ago
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    plug these into your objective function and test them: $$f(0,0)=0\\f(0,6)=48\\f\left(\frac{14}3,2\right)=30\\f(6,0)=18$$

  8. anonymous
    • one year ago
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    |dw:1438790959035:dw|

  9. anonymous
    • one year ago
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    so the biggest value happened at \((14/3,2)\) with value \(30\), so \(30\) is our maximum value

  10. anonymous
    • one year ago
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    thank you so much !!!!

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