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Math

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Submitted By finaldawn
Words 793
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Question 1
Z=Sales-Cost
Z=820-360
Z=460
Y=Sales-Cost
Y=580-210
Y=370

Let Z= Number of units of product Z produced and sold.
Let Y= Number of units of product Y produced and Sold.
Maximize
460Z+370Y 1) 12Z+8Y<=600 hours of assembly time 2) 10Z+5Y<=450 hours of quality control testing time 3) Y>=10 units 4) Z>=0.25(Z+Y)
Z>=0.25Z+0.25Y
Z-0.25Z-0.25Y>=0
0.75Z-0.25Y>=0
Question 2
To graph the constraints we find the X and Y axis intercepts by replacing the opposite by 0

1) 12Z+8(0)=600
12Z=600
Z=600/12
Z=50
| 12(0)+8Y=600
8Y=600
Y=600/8
Y=75 | 2) 10Z+5(0)=450
10Z=450
Z=450/10
Z=45
| 10(0)+5Y=450
5Y=450
Y=450/5
Y=90 | 3) Y>=10 | | 4) 0.75Z+0.25(0)=0
0=0If replace Z by 0 we will also get 0 therefore we use (0,0) as one point | We use an arbitrary number, such as 20 for Z.0.75(20)-0.25Y=015-0.25y=0
Y=-15/-0.25Y=60
Check figure:
0.75(20)-0.25(60)>=0
15-15>=0
0>=0
Yes!! This works! |

To find optimal solution we find the corners found within the feasible region and establish which point optimizes profits. =460(0)+370(10)
=$3,700
| =460(16)+370(50)
=7,360+18,500
=$25,860This point is the optimal solution. | =460(40)+370(10)
=18,400+3,700
=$22,100 | |
460Z+370Y

Therefore, to maximize profits we must produce 16 units of Z and 50 units of Y for a total contribution margin of $25,860.
To graph the optimal solution:
460Z+370(0)=25,860
Z=25,860/460
Z=56.22
460(0)+370Y=25,860
Y=25,860/370
Y=69.89
Question 3
Z=Sales-Cost
Z=730-360
Z=370
Y=Sales-Cost
Y=670-210
Y=460

Let Z= Number of units of product Z produced and sold.
Let Y= Number of units of product Y produced and Sold.
Maximize
370Z+460Y 1) 12Z+8Y<=600 hours of assembly time 2) 10Z+5Y<=450 hours of quality control testing time 3) Z>=10 units 4)

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