Answer:
Max = (6,0); min = (-2, 4)
Step-by-step explanation:
1. Summarize the constraints
[tex]\text{Constraints} = \begin{cases}(a)\qquad 2x - y & \leq 12\\(b)\qquad 4x+ 2y & \geq 0\\(c) \qquad x + 2y & \leq 6\\ \end{cases}[/tex]
2. Optimization equation
z = 5x + 2y
3. Graph the constraints to identify the feasible region
See the figure below.
The "TRUE" regions for each graph are the shaded areas to the side of the line indicated by the arrows.
The "feasibility region" is the dark green area where all three areas overlap and all three conditions are satisfied.
5. Determine the points of intersection among the constraints
(i) Constraints (a) and (b)
[tex]\begin{array}{rcr}2x - y & = & 12\\4x + 2y & = & 0\\4x - 2y & = & 24\\8x&=&24\\x & = & \mathbf{3}\\6 - y & = & 12\\-y & = &6\\y & = & \mathbf{-6}\\\end{array}\\[/tex]
The lines intersect at (3,-6).
(ii) Constraints (a) and (c)
[tex]\begin{array}{rcr}2x - y & = & 12\\x + 2y & = & 6\\4x - 2y & = &24\\5x & = & 30\\x & = & \mathbf{6}\\6 + 2y & = & 6\\2y & = &0\\y & = & \mathbf{0}\\\end{array}[/tex]
The lines intersect at (6,0).
(iii) Constraints (b) and (c)
[tex]\begin{array}{rcr}4x+ 2y &= & 0\\x + 2y &=& 6\\3x & = & -6\\x & = & \mathbf{-2}\\-2 +2y & = & 6\\2y & = &8\\y & = & \mathbf{4}\\\end{array}[/tex]
The lines intersect at (-2,4).
6. Determine the x- and y-intercepts of the feasible region
The five black dots at (3,-6), (6,0), and (-2,4) are the vertices of the polygon that represents the feasible region.
Each vertex is a possible maximum or minimum of z.
7. Calculate the maxima and minima
Calculate z at each of the vertices.
(i) At (-2,4)
z = 5x + 2y = 5(-2) + 2(4) = -10 + 8 = 2
(ii) At (3,-6)
z = 5(3) + 2(-6) = 15 - 12 = 3
(iii) At (6,0)
z = 5(6)+ 2(0) = 30 + 0 = 30
The maximum of z occurs at (6,0).
The minimum of z occurs at (-2, 4).
