this changes your inequalities and equations as follows:
maximize x + 2y, such that:
x <= 5
y <= 5
2x + 2y = 12
x,y >= 0
you would graph the equations.
those would be:
x = 5
y = 5
2x + 2y = 12
x = 0
y = 0
you would then find the region of feasibility by looking at the areas on the graph that satisfy the inequalities and equations.
those would be:
x >= 5
y >= 5
2x + 2y = 12 (this is an equation and not an inqquality)
x >= 0
y >= 0
your graph will look like this:
you can see that your feasible region is the area on the graph that is:
on or below the line y = 5
on or to the left of the line x = 5
on or above the line y = 0
on or to the right of the line x = 0
on the line 2x + 2y = 12
what this says is that your solution has to:
be on the line of 2x + 2y = 12
have an x-coordinate that is greater than or equal to 0 and less than or equal to 5.
have a y-coordinate that is greater than or equal to 0 and less than or equal to 5.
your max/min balue should be at the corner points of the feasible region.
those corner points would be at (x,y) = (1,5) and (x,y) = (5,1)
i plotted some other points on the line just to show you that the max/min value is indeed on the corner points of the feasible region.
the values i got for x + 2y at the following coordinates are:
(1,5) = 11
(2,4) = 10
(3,3) = 9
(4,2) = 8
(5.1) = 7
your maximum solution is at the coordinate of (1,5).
all of your constraints have to be satisfied.
at the point (1,5):
x <= 5 is satisfied because x = 1
y <= 5 is satisfied because y = 5
x >= 0 is satisfied because x = 1
y >= 0 is satisfied because y = 5
2x + 2y = 12 is satisfied because 2*1 + 2*5 = 2 + 10 = 12
your solution is that the maximum value is when x = 1 and y = 5.
that is one of the corner points of the feasible region.