Constraints (1) 6x+3y=24 and (2) 3x+6y=30 intersect, and we find the intersection by solving the system of two equations:
(1) 6x+3y=24
(3)=2×(2)=6x+12y=60
(3)-(1)=9y=36, y=4, so 6x=24-12=12, x=2. Intersection at (2,4)
Both constraints' inequalities tell us that the feasibility region is to the left of these lines, and the requirement is that both x and y are positive. Therefore the axes provide a boundary for the feasibility region. The x and y intercepts are: for constraint (1) x intercept=24/6=4; y intercept=24/3=8; for constraint (2) x intercept=30/3=10; y intercept=30/6=5.
Remember that the feasibility region is to the left of the lines, so we need y intercept 5 rather than 8 because the feasibility region has to be where the constraints overlap. Disregard (0,8). Similarly we disregard x intercept 10 in favour of 4, because, like y intercept 5, 4 is closer to the origin (more left) than 10. The vertices of the feasibility region are (0,5), (2,4), (4,0) and (0,0). We are looking for max p=3x+2y, so we can certainly ignore the origin. Now we plug in the vertices one by one and evaluate p:
(0,5): p=10; (2,4): p=14; (4,0): p=12. Therefore max p is when x=2 and y=4.
A graph helps to visualise this better than text.