## 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

1. Nnesha

you need to solve last two equations for y or just use desmos calculator shade the region

2. anonymous

y≤-3/2x+9 y≤-6/7x+6

3. anonymous

$$x\le 0\\y\le 0\\y\le -\frac32 x+9\\y\le -\frac67 x+6$$ |dw:1438790372303:dw|

4. anonymous

the intersection is at $$-\frac32 x+9=-\frac67 x+6\\-21x+14\cdot9=-12x+14\cdot 6\\14\cdot 3=9x\\\frac{14}3=x$$

5. anonymous

so now we have the following points to test: $$x=0,\qquad y=0\\x=0,\qquad y=6\\x=0,\qquad y=9\\x=14/3,\ \ y=2\\x=6,\qquad y=0\\x=7,\qquad y=0$$

6. anonymous

oops, ot the $$x=0,y=9$$ and $$x=7,y=0$$, only the four points that bound the quadrilateral

7. anonymous

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

8. anonymous

|dw:1438790959035:dw|

9. anonymous

so the biggest value happened at $$(14/3,2)$$ with value $$30$$, so $$30$$ is our maximum value

10. anonymous

thank you so much !!!!