Attempt:
First, I call $x$ be the number of shafts produced per year and $y$ the number of frames produced per year. We have that each machine works at most $4500$ hours. we can place all of our data in the following table:
Thus, we want to maximize $f(x,y) = x+y$ subject to
$$ 0.6 x + 0.8 y \leq 90 000 $$
$$ 0.3 x + 0 y \leq 22 500 $$
$$ 0.4 x + 0.6 y \leq 45 000 $$
$$ 0x + 0.2 y \leq 135 00 $$
$$ 0x + 0.3 y \leq 27 000 $$
and obviously $x \geq 0$ and $y \geq 0 $
Is this the correct formulation?
Best Answer
I suppose your model is not correct. Solving the model with an LP-solver yields the solution $x=75000$ and $y=25000$. I assume, however, that for every frame you also need a shaft. Hence, you need to add the following constraint:
$$x=y$$
Solving the problem with the additional constraint yields the solution $x=y=45000$. Please note that luckily the number of shafts/frames in the solution is an integer. This need not happen always. Thus, you should also add the constraints $x\in \mathbb{Z}$ and $y \in \mathbb{Z}$ in general, even though these were not needed here.