i could not understand the solution can u explain please (part b and c)

2. Consider the following model: maxz=15x1+12x2s.t.2x1+x2100x1+x280x140x1,x20 (a) Draw the feasible region and find an optimal solution using the graphical method. The optimal solution is at the corner point (20,60) and its objective function value is equal to 1020 . (b) What is the smallest possible value for the objective function coefficient of x1 for which the solution found in part (a) remains optimal? The smallest possible value for the objective function coefficient of x1 for which the point (20,60) remains optimal is 12 (which is the case when the slope of the iso-profit lines is equal to -1 , and therefore they are parallel to the line x1+x2=80 ). (c) What is the largest possible value for the objective function coefficient of x2 for which the solution found in part (a) remains optimal? The largest possible value for the objective function coefficient of x2 for which the point (20,60) remains optimal is 15 (which is again the case when the slope of the iso-profit lines is equal to -1 , and therefore they are parallel to the line x1+x2=80 )