I am trying to learn Linear Programming. However, I don’t know how to solve the following problem. Maybe you can help, because I am curious to the right approach and solution for this problem. It involves Newtonian mechanics.
A company uses a crane for lifting (see picture). The crane has $2$ hoists. They can work at the same time. Both ends can take $10000 \,\rm N$. If the two hoist both pick 1 artefact at the same time, what is the maximum force that the first hoist can take?
The force on hoist $1$ is represented by $x$.
The force on hoist $2$ is $9000 \,\rm N$.
The position of hoist $1$ is fixed.
The position of hoist $2$ is not fixed, as long as it is to the right of the first hoist.
The distance between both hoist is represented by $y$.
A force $z$ down at position $a$ from the left will exert a force of $\frac{40-a}{40}z$ on the left support and $\frac{a}{40}z$ on the right.
The force distribution of hoist 2 does not change that the force acts on the crane effectively at the same center position. Thus you get inequalities $$ 10\,000 \ge \frac{35}{40}⋅x+\frac{32-y}{40}⋅9000\\ 10\,000 \ge \frac{5}{40}⋅x+\frac{8+y}{40}⋅9000 $$ you get the geometry constraint $$ 0\le y\le 29 $$ and you want of course maximize $x$.
Graphically the maximum is at $y=29$ with $x=\frac{40⋅10\,000-3⋅9000}{35}=10657.142857$