Question 173706
Solve by graphing.
3x + 4y ≤ 12
x + 3y ≤ 6
x ≥ 0
y ≥ 0

:
Plot the 1st two equations, put the equations in the slope/intercept form
3x + 4y =< 12
4y = 12 - 3x
y = {{{12/4}}} - {{{3/4}}}x
y = {{{-3/4}}}x + 3
:
Calculate two points (substitute the values for x and find y):
 x | y
-------
 0 | 4
 4 | 1
Graph should look like this:
{{{ graph( 300, 200, -4, 5, -4, 6, -.75x+4) }}}
:
Do the same with the 2nd equation
x + 3y => 6
3y = 6 - x
y = {{{6/3}}} - {{{x/3}}}
y = -{{{1/3}}}x + 2
Calculate two points (substitute the values for x and find y):
 x | y
-------
 0 | 2
 3 | 1
Graph should look like this:
{{{ graph( 300, 200, -4, 5, -4, 6, -.33x+2) }}}
:
The last two equations tell you that only positive values for x and y are considered
:
Put both graphs on the same coordinate system:
{{{ graph( 300, 200, -4, 5, -4, 6, -.75x+4, -.33x+2) }}}
Look at this graph
The 1st equation:3x + 4y =< ; 12, the purple line
Area of feasibility is at or below this line
:
The 2nd equation: x + 3y => 6, green line
Area of feasibility is at or above this line
:
x => 0
y => 0
As I said before, this means only positive values for x & y, inside the area of feasibility are considered.
:
The solution is the triangular area between the lines and to the right of the y axis.