Minimize cost based on constraints given There are three different types of printers A, B, and C that are used for printing documents. A shop hires the mentioned type of printers to print 1500 documents in a day. One printer each of types A, B, and C can print 40, 60, and 75 documents in a day, respectively. The fixed cost of hiring one printer each of types A, B, and C is Rs. 100, Rs. 150, and Rs. 180, respectively. In addition, for every document printed by the printer of types A, B, and C an extra cost of Rs. 4, Rs. 5, and Rs. 3 is levied, respectively.
If the shop hired 10 printers of type B, then how many printers of type C must the shop hire in order to minimize the cost?
I am not able to understand how to frame an equation of such a big question. It's quite a tricky task for me. That's why I am finding this question really tough.
 A: Well, since you don't know how many printers of each type are hired, it only makes sense to give them names, i.e. use variables for these unknown quantities.
For example, let's say $x$ printers of type A have been hired. How many documents will they print in one day? Since each printer of type A prints $40$ documents in a day, $x$ printers of type A will print $40x$ documents in a day. And what is the associated cost? For one printer of type A we have the fixed costs of Rs. $100$ plus $40$ pages at Rs. $4$ each for a total of Rs. $100+40\cdot4=260$; therefore $x$ of them will cost us Rs. $260x$ per day.
Do the same for printers of types B and C. In fact, for type B you don't even need a variable, since you're given that there are $10$ of them. Then set up two expressions: one representing the total cost and one representing the total number of documents. Now your goal is to minimize the total cost function under the constraint that the total number of documents is $1500$.
A: Following Vikram's comment:
$$z=\frac{900-40x}{75}=\frac{180-8x}{15}=12-\frac{8x}{15}$$
Then
$$Cost =260x+405\frac{180-8x}{15}+4500=44x+9360$$
Cost is minimum for $x=0$
We need to check that the corresponding $z$ is a positive integer:
$$x=0 \Rightarrow z=12$$
When $y=10 $ Cost is minimum for $x=0$ and $z=12$ and Cost $ =9360$
