For the following question that has appeared in one of my homework questions, I have:
A fisheries laboratory needs to prepare $300$ litres of salt solution to replenish their tropical tank with salt solution of concentration $10$ gm/litre.
They have a supply of fresh water, a supply of local sea water of $6$ gm/litre and a supply of concentrated salt solution of $36$ gm/litre.
Let $x, y, z$ be the quantity in litres of the fresh water, sea water and concentrated solution used to make the mixture.
a) Set up two linear equations for finding the values of $x, y$ and $z$.
b) Noticing that the values of the variables must be greater than or equal to $0$, describe the range of feasible values for $x, y$ and $z$.
c) How much of each type of water should be used to make the mixture if they want to use as much of the local sea water as possible?
For part a, I think the two linear equations are as follows :
$x+y+z = 300$ (since the final solution needs to be $300$ L)
$6y+36z = 3000$ (as the final solution needs $3000$gm of salt)
For part b, I have stated that $0 \le z \le 300, y = 500-5z, x = 4z-200$.
How would I approach part c as from part a it is clear that the system has infinitely many solutions.
The general solution of the linear system (restrictions) can be expressed in the form $$ \left(\frac{650}{3}-\frac{5y}{6}, y, \frac{250}{3}-\frac y6\right), \quad y \in \mathbb{R} $$
Since the admissible solutions must have positive components, we must have $$ \frac{650}{3}-\frac{5y}{6}\ge 0, \quad y \ge 0, \quad \frac{250}{3}-\frac y6 \ge 0, $$ which leads to the conclusion that $y \in [0, 260]$. Setting $y$ to its maximum value (260), the other variables are fixed at $x = 0$ and $z=40$.