GraphsNrelations trouble

[abzzzz]

I can never seem to find the feasible region! I don't know how to find it and it will never stick in my head

Can someone give me an exam, like inequalities/inequations   and show me what and where to shade? My teacher mentioned something about plugging (0,0) into the equation, and if it's true you shade a particular region?

[sarah_h]

http://www.vcaa.vic.edu.au/vcaa/vce/studies/mathematics/further/pastexams/2009/2009furmath2-w.pdf
question 4
you have to find the feasible region to answer it
basically to find the feasible region you can choose any number on the graph either above or below the line and see if the numbers fit the equation, if the selected point fits the equation then it considered to be withing the feasible region. If the point does not fit the equation then it is not a part of the feasible region. the easiest point to test is (0,0)

so for the exam above:
1. for the first equation  x + y ≤ 10, you could use the point (0,0)      let x=0 and y=0
                                       0+0 ≤  10     
                                            0≤  10
this is true therefore you want the area that (0,0) is within, which is below the line

2. for the equation  3x + 5y ≤ 41, you could use the point (4,2)          let x=4 and  y=2
                                3x4+5x2≤41
                                   12+10≤41
                                         22≤41
this is true therefore the point (4,2) is within the feasible region which is below the line

[abzzzz]

So if what you have plugged into the equation is true, the region to be shaded is always below that line? And if it were false, above the line?

And if x ≤ 10, if you were to use 0 as x, you would need the region from the left of 10?

[abeybaby]

say I told you that x-2y≤3. rearranging: y≥(3-x)/2=1.5-0.5x
so if i graph the line y=1.5-0.5x, then for every point on that line, the y value EQUALS 1.5-0.5x. but I dont want it to equal 1.5-0.5x, i want the y value to be BIGGER than (or equal to) that. so I shade everything above that line.

Say I told you 2x+2y<1, then y<0.5-x. so for every point on the line y=0.5-x, the y value IS 0.5-x. I want it to be LESS than (but not equal to) 0.5-x, so i'll shade everything under it, but I wont include the actual line.

then find the intersection.

[sarah_h]

if you always use the point (0,0) then that's true
but what i was trying to say was if the point chosen fits the equation then it is the feasible region which is the are you shade.
if the point chosen doesnt fit the equation then it isn't the feasible region and you would shade the other side of the line.

so for the question from before: http://www.vcaa.vic.edu.au/vcaa/vce/studies/mathematics/further/pastexams/2009/2009furmath2-w.pdf

for the equation  x + y ≤ 10, you could use a point above the line such as (9,10)
                            9+10≤ 10
                                 19≤10
this is not true hence it the point (9,10) is not a part of the feasible region. the feasible region would be below the line, this is the area you would shade
for the equation  3x + 5y ≤ 41, you could use the point (9,10) again
                     3x10+5x10 ≤ 41
                              30+50≤41
                                    80≤41
this is not true hence the point (9,10) is not a part of the feasible region. you would shade the feasible region which is below the line.

And if x ≤ 10, if you were to use 0 as x, you would need the region from the left of 10?

and yeah that's true