question_answer

A librarian has to accommodate two different types of books on a shelf. The books are 6 cm and 4 cm thick and weight 1 kg and \[1\frac{1}{2}\] kg each, respectively. The shelf is 96 cm long and at most can support a weight of 21 kg. How should the shelf be filled with the books of two types in order to include the greatest number of books? Make it as an LPP and solve it graphically.

Answer:

Let the number of two types of books be x and y, respectively. The given data can be written in tabular from, which is as follows

Types of books	Thickness (in cm)	Weight (in kg)
x	6	1
y	4	\[1\frac{1}{2}=\frac{3}{2}\]
	at most 96	at most 21

            The required LPP is maximize \[Z=x+y\]             Subject to the constraints                         \[6x+4y\le 96\]                         \[3x+2y\le 48\] [dividing both sides by 2] ... (i)             and       \[x+\frac{3}{2}y\le 21\] \[\Rightarrow \] \[2x+3y\le 42\]     ? (ii)             [multiplying both sides by 2]             and       \[x,\,\,y\ge 0\] On considering the inequalities as equations, we get             \[3x+2y=48\]                            ? (iii) and       \[2x+3y=42\]                            ? (iv) Table for line \[3x+2y=48\] is

x	0	16
y	24	0

\[\therefore \] It passes through the points (0, 24) and (16, 0). On putting (0, 0) in \[3x+2y\le 48,\] we get \[0\le 48\] So. the half plane is towards the origin.              and table for line \[2x+3y=42\] is

x	0	21
y	14	0

\[\therefore \] It passes through the points (0, 14) and (21, 0). On putting (0, 0) in \[2x+3y\le 42,\] we get                         \[0\le 42\]                                   [true] So, the half plane is towards the origin. Now, on multiplying Eq. (iii) by 2 and Eq. (iv) by 3 and then subtracting Eq. (iv) from Eq. (iii), we get \[\begin{matrix}    6x+4y=96  \\    6x+9y=126  \\    ---  \\    -\,5y=-\,30  \\ \end{matrix}\] \[\Rightarrow \]   y = 6 On putting y = 6 Eq. (iii), we get \[3x+2(6)=48\] \[\Rightarrow \] \[3x=48-12=36\]\[\Rightarrow \]\[x=12\] So, the point of intersection is (12, 6).           Now, the graph of above LPP is as follows From the graph, OABC is the feasible region. The corner points of feasible region are O (0, 0), A (0, 14), B (12, 6) and C (16, 0), respectively Now, the values of Z at corner points are shown in the following table:

Corner points	\[\mathbf{Z =x+ y}\].
O (0, 0)	\[Z=0+0=0\]
A (0, 14)	\[Z=0+14=14\]
B (12, 6)	\[Z=12+6=18\][maximum]
C (16, 0)	\[Z=16+0=16\]

From the table, maximum value of Z is 18. Hence, the maximum number of books is 18 and number of books of I type is 12 and books of II type is 6.