Step 1:

Let the no of items manufactured by the firm of type A be x

Let the no of items manufactured by the firm of type B be y

It is said that the firm can manufacture items of both type is at most 24

$\Rightarrow x+y\leq 24$

The maximum time available for both A and B is 16.

Where A takes 1 hour and B takes $\large\frac{1}{2}$ an hour.

$\therefore x+\large\frac{1}{2}$$y\leq 16$

The profit of one unit of item A is Rs.300 and that of B is Rs160

Maximum profit to be attained is given by $Z=300x+160y$

Clearly $x,y\geq 0$

Step 2:

Now let us solve the above problem graphically :

Let us draw the lines $x+y=24$ and $x+\large\frac{1}{2}$$y=16$ on the graph.

Clearly the shaded portion is the feasible region.

The corner points are OACD where coordinates are $O(0,0),A(16,0),C(8,16),D(0,24)$

Step 3:

Let us obtain the value of objective function as follows :

At the points $(x,y)$ the value of the objective function subjected to $Z=300x+160y$

At $O(0,0)$ the value of the objective function $Z=0$

At $A(16,0)$ the value of the objective function $Z=300\times 8+160\times 0=4800$

At $C(8,16)$ the value of the objective function $Z=300\times 8+160\times 16=4960$

At $D(0,24)$ the value of the objective function $Z=300\times 0+160\times 24=3840$

Step 4:

Clearly the maximum value is $Z=4960$ at $C(8,16)$

Hence 8items of type A and 16 items of type B are required to acquire a maximum profit.