Graph the area bounded by $\displaystyle y<\frac{1}{2}x+6$, $\displaystyle x+3y\geq12$, $\displaystyle x\geq0$, $\displaystyle x\leq12$.

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- Oct 19th 2009, 12:43 PMElectroNerdShading for multiple functions on a graph
Graph the area bounded by $\displaystyle y<\frac{1}{2}x+6$, $\displaystyle x+3y\geq12$, $\displaystyle x\geq0$, $\displaystyle x\leq12$.

- Oct 19th 2009, 08:42 PMmathaddict
HI

Firstly , graph this 4 first lines first .

$\displaystyle y=\frac{1}{2}x+6$ take the area below this line , <

$\displaystyle y=-\frac{1}{3}x+4 $ take the area above this line , >=

x=0 take area to the right of the line , >=

x=12 take are to the left of the line , <=

Then , the 'quadrilateral' as a result of the intersection of these lines is most likely to be the region bounded . I am not saying it's but normally it is . So now you will have to check .