Answer:
Part 5) The maximum value of z is 110.7
Part 6) The maximum value of z is 6
Part 7) The minimum value of z is 120
Part 8 The minimum value of z is 3
Step-by-step explanation:
Part 5) Maximizes function z subject to constraints
[tex]z=5x+7y[/tex]
[tex]2x+3y\leq 45[/tex] ----> constraints A
[tex]x-3y\leq 2[/tex] ---> constraints B
[tex]x\geq 0[/tex] ---> constraints C
[tex]y\geq 0[/tex] ---> constraints D
using a graphing tool
The solution is the shaded area
The vertices of the shaded area are
(0,0),(2,0),(15.7,4.6) and (0,15)
see the attached figure N 1
Substitute the value of x and the value of y of each vertex in the objective function z and then compare the results
For (0,0) ----> [tex]z=5(0)+7(0)=0[/tex]
For (2,0) ----> [tex]z=5(2)+7(0)=10[/tex]
For (15.7,4.6) ----> [tex]z=5(15.7)+7(4.6)=110.7[/tex]
For (0,17) ----> [tex]z=5(0)+7(15)=105[/tex]
therefore
The maximum value of z is 110.7
Part 6) Maximizes function z subject to constraints
[tex]z=2x+y[/tex]
[tex]y\leq x+2[/tex] ----> constraints A
[tex]y\leq -x+3[/tex] ---> constraints B
[tex]x\geq 0[/tex] ---> constraints C
[tex]y\geq 0[/tex] ---> constraints D
using a graphing tool
The solution is the shaded area
The vertices of the shaded area are
(0,0),(3,0),(0.5,2.5) and (0,2)
see the attached figure N 2
Substitute the value of x and the value of y of each vertex in the objective function z and then compare the results
For (0,0) ---->[tex]z=2(0)+(0)=0[/tex]
For (3,0) ----> [tex]z=2(3)+(0)=6[/tex]
For (0.5,2.5) ----> [tex]z=2(0.5)+(2.5)=3.5[/tex]
For (0,2) ----> [tex]z=2(0)+(2)=2[/tex]
therefore
The maximum value of z is 6
Part 7) Minimize function z subject to constraints
[tex]z=2x+2y[/tex]
[tex]x+2y\geq 80[/tex] ----> constraints A
[tex]3x+2y\geq 160[/tex] ---> constraints B
[tex]5x+2y\geq 200[/tex] --> constraints C
[tex]x\geq 0[/tex] ---> constraints D
[tex]y\geq 0[/tex] ---> constraints E
using a graphing tool
The solution is the shaded area
The vertices of the shaded area are
(0,100),(20,50),(40,20) and (80,0)
see the attached figure N 3
Substitute the value of x and the value of y of each vertex in the objective function z and then compare the results
For (0,100) ---->[tex]z=2(0)+2(100)=200[/tex]
For (20,50) ----> [tex]z=2(20)+2(50)=140[/tex]
For (40,20) ----> [tex]z=2(40)+2(20)=120[/tex]
For (80,0) ----> [tex]z=2(80)+2(0)=160[/tex]
therefore
The minimum value of z is 120
Part 8) Minimize function z subject to constraints
[tex]z=7x+3y[/tex]
[tex]3x-y\geq -2[/tex] ----> constraints A
[tex]x+y\leq 9[/tex] ---> constraints B
[tex]x-y=-1[/tex] --> constraints C
[tex]x\geq 0[/tex] ---> constraints D
[tex]y\geq 0[/tex] ---> constraints E
using a graphing tool
The solution is the line segment (0,1) to (4,5)
see the attached figure N 4
Substitute the value of x and the value of y of each endpoint in the objective function z and then compare the results
For (0,1) ---->[tex]z=7(0)+3(1)=3[/tex]
For (4,5) ---->[tex]z=7(4)+3(5)=43[/tex]
therefore
The minimum value of z is 3
