Two products A and B can be made on either of the two extrusion machines M1 or M2. The time in minutes taken by these products according to the machine used, are as given below. Assume a day has 8 working hours.
for A: M1 takes 8 min and M2 takes 10 min
for B: M1 takes 6 min and M2 takes 6 min
If A and B are produced in the ratio 3:1, then what is the maximum total of products A and B that can be produced in a day?
a. 120. b. 160. c. 116 d. none of these
If A is produced only on M1 and B only on M2, then what is the unutilised time on M2 in a day, if A and B are produced in 1:1 ratio?
a. 20 min. b. 60 min. c. 150 min d. none of these
I tried solving the first one and somehow got 120 by using two variables x and y and eliminating options by some assumptions. I get the second answer to be 120 min but the answer says it's 150 min. It would be helpful if somebody will elaborate on how these problems are to be solved. Are there any specified methods or steps. Also, while searching for some help on the internet, i found things
related to linear programming. Is that related to this problem?
Best Answer
You can set this up as a Linear Program (although I'm not sure if it is the only way to solve it).
Let $a1, a2$ be the number of A produced on $M1$ and $M2$ respectively.
Let $b1, b2$ be the number of B produced on $M1$ and $M2$ respectively.
The setup is:
maximize:
$\quad\quad Z = a1 + a2 + b1 + b2$
subject to:
$\quad\quad a1 + a2 - 3b1 - 3b2 = 0$
$\quad\quad 8a1 + 0a2 + 6b1 + 0b2 <= 480$
$\quad\quad 0a1 + 10a2 + 0b1 + 6b2 <= 480$
You can use the online calculator at http://comnuan.com/cmnn03/cmnn03004/ with the following setup:
max
$1 \quad 1 \quad 1 \quad 1$
$1 \;\;\; 1 \;\; -3 \; -3 \quad \;\; 0 \quad 0$
$8 \;\;\; 0 \quad \;\; 6 \quad \;\; 0 \;\;-1 \quad 480$
$0 \;\;10 \quad \; 0 \quad \;\; 6 \;\; -1 \quad 480$
Hope this helps.
Further Clarification:
You have 2 machines $M1$ and $M2$. Each machine is available for 480 minutes each.
Let $a1$ and $a2$ be the number of units of A produced on $M1$ and $M2$ respectively.
Let $b1$ and $b2$ be the number of units of B produced on $M1$ and $M2$ respectively.
1) The time used to produce $a1$ units on $M1$ is 8*a1
2) The time used to produce $b1$ units on $M1$ is 6*b1
3) The time used to produce $a2$ units on $M2$ is 10*a2
4) The time used to produce $b2$ units on $M2$ is 6*b2
Since $M1$ can only be run for 480 minutes, the time used to produce a1 units of A and b1 units of B cannot exceed 480 minutes. That is, any values of a1 and b1 that need more than 480 minutes to produce is an "infeasible" solution (i.e. invalid).
That's the reason for the equation: $\quad\quad 8a1 + 0a2 + 6b1 + 0b2 <= 480$
This ensures that only "valid" solutions are considered.