anonymous
  • anonymous
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
Mathematics
katieb
  • katieb
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Nnesha
  • Nnesha
you need to solve last two equations for y or just use desmos calculator shade the region
anonymous
  • anonymous
y≤-3/2x+9 y≤-6/7x+6
anonymous
  • anonymous
$$x\le 0\\y\le 0\\y\le -\frac32 x+9\\y\le -\frac67 x+6$$ |dw:1438790372303:dw|

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anonymous
  • anonymous
the intersection is at $$-\frac32 x+9=-\frac67 x+6\\-21x+14\cdot9=-12x+14\cdot 6\\14\cdot 3=9x\\\frac{14}3=x$$
anonymous
  • anonymous
so now we have the following points to test: $$x=0,\qquad y=0\\x=0,\qquad y=6\\x=0,\qquad y=9\\x=14/3,\ \ y=2\\x=6,\qquad y=0\\x=7,\qquad y=0$$
anonymous
  • anonymous
oops, ot the \(x=0,y=9\) and \(x=7,y=0\), only the four points that bound the quadrilateral
anonymous
  • anonymous
plug these into your objective function and test them: $$f(0,0)=0\\f(0,6)=48\\f\left(\frac{14}3,2\right)=30\\f(6,0)=18$$
anonymous
  • anonymous
|dw:1438790959035:dw|
anonymous
  • anonymous
so the biggest value happened at \((14/3,2)\) with value \(30\), so \(30\) is our maximum value
anonymous
  • anonymous
thank you so much !!!!

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