How to formula the given linear programming model? Chem Labs uses raw materials I and II to produce two domestic cleaning solutions, A and B.
The daily availabilities of raw materials I and II are 150 and 145 units, respectively. One unit of
solution A consumes 0.5 unit of raw material I and 0.6 unit of raw material II, and one unit of
solution B uses 0.5 unit of raw material I and 0.4 unit of raw material II. The profits per unit of
solutions A and B are 8 and  $10, respectively. The daily demand for solution A lies between
30 and 150 units, and that for solution B between 40 and 200 units. Find the optimal production
amounts of A and B.
My attempt
Let A and B be the no. of units of A and B produced and X and Y be no. of raw materials I and II to be processed respectively. 
The objective function is to maximize the profit, Z.
Z=8A+10B
The objective function is subject to the following constraints
30<=0.5X+0.6Y<=150
40<=0.5X+0.4Y<=200
X<=150
Y<=145
Is this formulation correct? If it is, how can one proceed from this point to find the maximum profit?
 A: The use of raw materials $I$ and $II$ depends on the production of $A$ and $B$, so you don't need the variables $X$ and $Y$. 
You can create a table of given data:
$$\begin{array}{c|c|c|c}
Products&I&II&Profit&Demand\\
\hline
A&0.5&0.6&8&30\le A\le 150\\
B&0.5&0.4&10&40\le B\le 200\\
\hline
Available&\le 150&\le 145&maximize&\end{array}$$
Now we can formulate the LPP: let $A$ and $B$ be the numbers of units of $A$ and $B$, respectively. Then:
$$\pi(A,B)=8A+10B\to \text{max} \ \ \text{subject to}\\
0.5A+0.5B\le 150 \ \ \text{(material I constraint)}\\
0.6A+0.4B\le 145 \ \ \text{(material II constraint)}\\
30\le A\le 150 \ \ \text{(demand for A)}\\
40\le B\le 200 \ \ \text{(demand for B)}\\
$$
You can use graphical or Simplex methods to solve LPP.
Graphical method. 
1) Draw the feasible (green) region from the constraint inequalities:

2) Find the corner points: $A,B,C,D,E,F$.
3) Evaluate the objective (profit) function at the corner points and choose the maximum.
Can you do it?
Answer:

 $\pi(100,200)=2800.$ WolframAlpha answer.

A: We just have to decide how many units of $A$ and $B$ are to be produced. 
You are right that the profit is $8A+10B$ and we want to maximize it.
Now, let's examine the constraint imposed by material I.
$$0.5A+0.5B \le 150$$
Now, let's examine the constraint imposed by material II.
$$0.6A + 0.4B \le 145$$
The demand informations also gives us 
$$30 \le A \le 150$$
and 
$$40 \le B\le 200.$$
Now, we have a $2$-dimensional linear programming problem, you can use a graphical method to solve the problem if you wish.
