Question 1073576
Good work on parts 1 and 2.
Your graph should look like
{{{drawing(300,300,-10,40,-10,40,
grid(0),
line(0,20,30,0),line(21,0,0,28)
)}}}
The profit equation asked for in part 3 is the
P(x,y)=10x+15y that you included in the part 2 of your solution. You also shows showed the calculations for the pairs (x,y) for the corners of the,feasible region,
as asked in part 3. I just do not know how they expect the table to be organized. Maybe 3 columns with X, Y, and P(X,Y) as headings.
You found the profit to be maximum at points (0,20) and (12,12).
That maximum profit was $300.
Because the maximum happens at two corners of the feasible region,
It happens all along the edge of the feasible region connecting those w points.
That is because all the equations and inequalities are linear,
and two lines either intersect, or are parallel, or are the same line.
The feasible region boundary equation,
8x + 12y = 240 ,
and the profit equation for any possible profit P,
10x + 15y = P, 
are parallel lines.
For P=0, the profit line passes through the origin.
Each increase in P moves the line, parallel to its previous graph,
and parallel to 8x + 12y = 240.
Eventually, the line would leave the feasible region.
When that happens, you have the maximal feasible profit.
In general, at that point, either the profit line is intersecting a boundary line at a corner,
or it is coinciding with a boundary line all along an edge of the feasible region.
In this case for any conceivable profit,
the profit line had to be parallel to 
8x +12y = 240 , or the same line.
To get that maximum profit, you could make
no small bracelets and 20 large ones, or
3 small bracelets and 18 large ones, or
6 small and 16 large bracelets, or
9 small and 14 large bracelets, or
12 small and 12 large bracelets.
All those (x,y) combinations are points on the line connecting (0,20) and (12,12).