left (or at) of x = 1
above or on x axis
below or on y = -3x+5
which hits x axis at x = 5/3
and hits the y axis at y = 5
You see that the vertical line x = 1 hits our sloped line somewhere above the x axis
find that point
x = 1
y = -3(1) + 5 = 2
we have a corner at (1, 2)
everything in the upper half plane left of the sloping line and the vertical line x = 1
It did not say x had to be positive, so it goes left forever. That would not be likely in a real linear programming problem.
2. Describe the linear programming situation for this system of inequalities where you are asked to find the maximum value of f(x, y) = x + y.
x>(or equal too) 0 y>(or equal too)0
6x + 3y<(or equal too)18 x + 3y<(or equal too)9
In first quadrant due to x>/=0 and y>/= 0
6x + 3y<(or equal too)18
y </= -2x + 6
x and y axis intercepts at x = 3 and at y = 6
below that line
3 y </= -x + 9
y </= -(1/3) x + 3
intercepts at x = 9 and at y = 3
below that line
we need the corner where those sloped lines hit
-(1/3) x + 3 = -2x+6
-x + 9 = -6 x + 18
5 x = 9
x = 9/5
then y = -(1/3)(9/5) + 3
= 3 - 3/5 = 12/5
So we have a corner at (9/5 , 12/5)
so three corners to test
(0 , 3) , (9/5 , 12/5) , (3 , 0)
at (0,3) x+y = 3
at (9/5,12/5) x+y = 21/5 = 4. something
at (3,0) x+y = 3 again
so the max is at the intersection of the sloped lines and is 21/5
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