Raymond B. answered 07/25/19
Math, microeconomics or criminal justice
Let A=number of Model A manufactured daily. Let B=number of Model B manufactured daily.
Profit = 50A+55B
A+B=50
A=50-B Profit = 50(50-B)+55B = 2500-50B+50B = 2,500+5B is at a maximum by maximizing B at 30.
B=50-A Profit = 50A +55(50-A) = 50A+2750-55A = 2,750-5A is at a maximum by minimizing A at 20
you do as many Model B as possible, which is 30. That means A=20 is the maximum possible for Model A.
Maximum profit is 50(20)+55(30)= 1,000+1,650=$2,650
You want maximum number of B possible because it is more profitable.