Question

2. Consider the following linear model where C1 has not yet been defined. Max s.t. z = C1x1 + x2 X1 + x2 = 6 X1 + 2.5x2 < 10 X1 > 0, x2 > 0 Use the graphical approach that we covered to find the optimal solution, x*=(x1, xỉ) for all values of -00 < ci so. Hint: First draw the feasible region and notice that there are only a few corner points that can be the optimal solution. Also remember that if the objective function line reaches a maximum and it includes two adjacent corner points, both of these points and all of the points that create the line that connects them are optimal. There is literally an infinite number of optimal solutions but only one optimal objective function value because if you plug in the values of xi and X2 you get the same (max) value. Now, ci controls that slope of the objective function line and this is the critical concept. Go to any of the corner points and some values of cı that make that point the optimal solution. Then ask yourself how much can the slope increase (or decrease) that will move the optimal solution to an adjacent corner point. That range is the answer to this question.

Answer 1

Solution:

First we have to find the roots of each equation by applying x₁=0 & x₂=0.

Consider x₁ + x₂ =6 as equation1 and x₁ + 2.5x₂ =10 as equation 2.

In the below attachement, we can find the roots of the equation.

Max z = C, x, + H2 sit x, & X2 s6 X, +2.522 £ 10 First, we write these constrains as equation 2, + x2 = 6 x, & 2.5x = 10 substitute x = in egn. . ota2 = 6 = X = 6. (2,30,X2=6] A(0, 6) a sub. da so in egn. - 2, to = 6 = a = 6. [x, = 6,22 = 0 B (6,0) sub. I, 50 in egn & of 2.522 22 [2, = 0, X2 = 4) 10 2.5 c(0,4) sub. Dg=0 in egn (2,= 10; 42=0 x,to =10 x = 0 D (10,0)
  
Now, draw the feasible feasible region using these point 9 8 ។ LA(0,6) 6 c(0,4) a, fX₂ = 6 3 &G (3,3 Hit 2.582=10 2 Feasible -Region BC6,-) D(10,0) > ) (0,0) I 2 3 4 5 6 7 8 9 lo VIII egn. O region. - egn. @ region. Here, less than (4) mean for each constraints, towards the origin. The common region for both equation (marked as is called feasible Region [OCGB] the region is

table for finding maximum value. Here apply the feasible region [OC GB] points to max z, we get points value then a maximum value. Max 2=ca,+ 22 0(0,0) C.lolto 4 C,co) + 4 c(0,4) G (3,3) C, (3) +3 3+34, IB (6,0) 6, (6) to bC, A then 1 Here, If = 1; 3+3C, ebe, is same. otherwise for all e, >1 ; 6c, is maximum. i Max 2 = 60,to = 60 (x1=6, K₂ = 0)

Therefore the optimum solution is max z = 6c_1; for all the values of -∞ ≤c₁≤∞ Since x₁ = 6, x₂ = 0