A company makes three products X, Y and Z. Each product requires processing by three machines A, B and C. The time required to produce one unit of each product is shown below.
Product X:
Machine A: 1
Machine B: 2
Machine C: 2
Product Y:
Machine A:2
Machine B: 2
Machine C: 2
Product Z:
Machine A:2
Machine B: 1
Machine C:4
The machines are available for 200, 525 and 350 hours each month. How many units of each product can be manufactured per month if all three machines are utilized to their full capacity?

I saw this same question yesterday , tried it and got nowhere.
The way I see it :
for machine A : x + 2y + 2z = 200
for machine B : 2x + 2y + z = 525
for machine C : 2x + 2y + 4z = 350
solving these I got
x = 266 1/3
y = 25
z = 58 1/3
these answers satisfy the above equations, but of course make no sense.... (can't produce negative number of products)
Perhaps somebody else could look at this and see an error in my thinking.
Or ...., the problem could be flawed.

The problem makes no sense. With constraints (machine time), and costs (time on each machine for each product), this would make sense if there was a profit function, which there is not.

This is an incomplete linear programming problem I suspect.


A company has three machines A, B and C which all produce the same two parts, X and
Y. of all the parts produced, machine A produces 60%, machine B produces 30%, and
machine C produces the rest. 40% of the parts made by machine A are part X, 50% of the
parts made by machine B are part X, and 70% of the parts made by machine C are part X.
A part produced by this company is randomly sampled and is determined to be an X part.
With the knowledge that it is an X part, find the probabilities that the part came from
machine A, B or C.

@bobpursley why are you so mean