The figure shows the graphical model of a linear program The

The figure shows the graphical model of a linear program. The large numbers on the right (1, 2, and 3) indicate the constraints. The feasible region is shown in white, and the infeasible region is shaded. The variables are restricted to nonnegative values. The small numbers (0, 1, 2, and 3) indicate four feasible corner points: 0, 1, 2, 3. Three objective functions are under consideration, as indicated by the three lines labeled A, B, and C. The arrows represent the directions of increasing objective function. Objective B is parallel to constraint 3. In each case, specify the location of the optimal solution. If there is more than one optimal solution, characterize all of them. Maximize A Maximize B Maximize C Minimize A Minimize B Minimize C Drop x_1 lesstanorequalto 0 and minimize C

Solution

Concepts that we will use :

Step 1: Since the objective function is Z = ax + by, draw a dotted line for the equation ax + by = k, where k is any constant. Sometimes it is convenient to take k as the LCM of a and b.

Step 2: To maximise Z draw a line parallel to ax + by = k and farthest from the origin. This line should contain at least one point of the feasible region. Find the coordinates of this point by solving the equations of the lines on which it lies.

a) Max A : Move objective function A farther from origin (0,0) parallel we will get corner points 1 or 2 or 3 as soultion as we will meet them.

b)Max B :Move objective function B farther from origin (0,0) parallel we will get corner points 2 or 3 as soultion as we will meet them and since this function is parallel to constraint 3 we may have infinitely many solution..

c)Max C Move objective function C farther from origin (0,0) parallel we will get corner points 3 as soultion as we will meet them.

d)Min A Move objective function B towards origin (0,0) parallel we are getting 0 as corner point nearest to 0 hence 0 is solution.

e) )Min B Move objective function B towards origin (0,0) parallel we are getting 0 and 1 as corner point nearest to 0 hence 0 or 1 is solution.

f) Min C Move objective function C towards origin (0,0) parallel we are getting 0 and 1 and 2 as corner point nearest to 0 hence 0 or 1 or 2 is solution.

g)Min C Move objective function C towards origin (0,0) parallel we are getting 0 and 1 and 2 as corner point nearest to 0 hence 0 or 1 or 2 is solution.(No affect of x₁ > 0 removal)

Step 1: Since the objective function is Z = ax + by, draw a dotted line for the equation ax + by = k, where k is any constant. Sometimes it is convenient to take k as the LCM of a and b.
Step 2: To maximise Z draw a line parallel to ax + by = k and farthest from the origin. This line should contain at least one point of the feasible region. Find the coordinates of this point by solving the equations of the lines on which it lies.
To minimise Z draw a line parallel to ax + by = k and nearest to the origin. This line should contain at least one point of the feasible region. Find the co-ordinates of this point by solving the equation of the line on which it lies.
a) Max A : Move objective function A farther from origin (0,0) parallel we will get corner points 1 or 2 or 3 as soultion as we will meet them. b)Max B :Move objective function B farther from origin (0,0) parallel we will get corner points 2 or 3 as soultion as we will meet them and since this function is parallel to constraint 3 we may have infinitely many solution.. c)Max C Move objective function C farther from origin (0,0) parallel we will get corner points 3 as soultion as we will meet them.
d)Min A Move objective function B towards origin (0,0) parallel we are getting 0 as corner point nearest to 0 hence 0 is solution. e) )Min B Move objective function B towards origin (0,0) parallel we are getting 0 and 1 as corner point nearest to 0 hence 0 or 1 is solution. f) Min C Move objective function C towards origin (0,0) parallel we are getting 0 and 1 and 2 as corner point nearest to 0 hence 0 or 1 or 2 is solution. g)Min C Move objective function C towards origin (0,0) parallel we are getting 0 and 1 and 2 as corner point nearest to 0 hence 0 or 1 or 2 is solution.(No affect of x1 > 0 removal)