Linear Programming — Class 12 Maths Solution

Let the person takes $x$ units of tablet X and Y units of tablet Y.

So, from the given information, We have the following conditions as per the question,

$6x + 2y \ge 18 \Rightarrow 3x + y \ge 9$ …….(i)

$3x + 3y \ge 21 \Rightarrow x + y \ge 7$ ……(ii)

and $2x + 4y \ge 16 \Rightarrow x + 2y \ge 8$ …….(iii)

Also, we know that here, $x \ge 0,y \ge 0$

The price of each tablet of X and Y is Rs.2 and Rs.1, respectively.

So, the corresponding LPP is minimise $Z = 2x + y$,

subject to $3x + y \ge 9$,$x + y \ge 7$, $x + 2y \ge 8$, $x \ge 0$, $y \ge 0$

From the shaded graph,

we see that for the shown unbounded region,

We have the following conditions as per the question, coordinates of corner points A, B, C and D as (8,0),(6,1),(1,6), and (0,9), respectively.

[on solving $x + 2y = 8$ and $x + y = 7$,

we get $x = 6,$ $y = 1$ and on solving $3x + y = 9$ and $x + y = 7$,

we get $x = 1,y = 6$]

Thus, we see that 8 is the minimum value of Z at the corner point (1,6) .

Here, we see that the feasible region is unbounded.

Therefore, 8 may or may not be the minimum value of Z.

To decide this issue, we graph the inequality
$2x + y < 8$ ……(v)

and check whether the resulting open half has points in common with feasible region or not.

If it has common point, then 8 will not be the minimum value of Z, otherwise 8 will be the minimum value of Z.

Thus, from the graph it is clear that, it has no common point.

Therefore, $Z = 2x + y$ has 8 as minimum value subject to the given constraints.

Hence, the person should take 1 unit of X tablet and 6 units
of Y tablets to satisfy the given requirements and at the minimum cost of Rs.8.