Linear Programming — Class 12 Maths Solution

.: Let Reshma mix x kg of food P and y kg of food Q. Clearly, $x \ge 0,y \ge 0$.

Now, we make the following table from the given data :

Since the mixture must contain at least 8

units of vitamin A and 11 units of Vitamin B, we have the constraints :

$3x + 4y \ge 8,5x + 2y \ge 11$

Total cost Z of purchasing x kg of food P and y kg of food Q is $Z = 60x + 80y$

Hence, the mathematical formulation of the problem is

Minimize:
$Z = 60x + 80y$ ...(1)

subject to the constraints :
$3x + 4y \ge 8$ ….(2)

$5x + 2y \ge 11$ ….(3)

$x,y \ge 0$ …(4)

${l_1}:3x + 4y = 8;$ ${l_2}:5x + 2y = 11$

Let us graph the inequalities (2) to (4).

The feasible region determined by the system is shown in the graph.

Here, we again observe that the feasible region is unbounded.

Let us evaluate Z at the comer points $A\left( {\cfrac{8}{3},0} \right)$, $E\left( {2,\cfrac{1}{2}} \right)$

and $D\left( {0,\cfrac{{11}}{2}} \right)$

In the table, we find that smallest value of Z is 160 at points $E\left( {2,\cfrac{1}{2}} \right)$ and
$A\left( {\cfrac{8}{3},0} \right)$.

Now, that the feasible region is unbounded.

$\therefore$ We have to draw the graph of the inequality $60x + 80y < 160\;i.e.\;3x + 4y < 8$

to check whether the resulting open half plane has any points common with the feasible region.

$\therefore$ We see that it has no points in common.

$\therefore$ The minimum value of Z is Rs. 160 at all points lying on segment joining

$\left( {\cfrac{8}{3},0} \right)$ and $\left( {2,\cfrac{1}{2}} \right)$