Let the mixture contain x bas of brand ‘P’ and y bags of brand ‘Q’. Clearly, x, y ≥ 0. Let us construct the following table from the given data:

$\begin{array}{llllll} \textbf{Brands} & \textbf{Bags} & \textbf{Element A} &\textbf{Element B} &\textbf{Element C} &\textbf{Cost Rs.} \\ P & x & 3x & 2.5x & 2x & 250x \\ Q & y & 1.5y & 11.25y & 3y & 200y\\ \text{Total} & & 3x+1.5y & 2.5x+11.25y & 2x+3y & 250x+200y\\ \textbf{Requirements}& & \text{Min 18}& \text{Min 45} & \text{Min 24} & \end{array}$

We need to minimize the cost $Z =250x+200y$ given the following constraints:

$(1):\qquad$ $3x+1.5y \geq18 \:\Rightarrow\: 2x+y \geq 12$

$(2):\qquad$ $2.5x+11.25y \geq 45\: \Rightarrow\: 2x+9y \geq 36$

$(3):\qquad$ $ 2x+3y \geq 24$

$(4):\qquad$ $x \geq 0\:\: and\:\:$ $(5):\qquad$$ y \geq 0$.

$\textbf{Plotting the constraints}$

First draw the graph of the line $2x+y = 12.$

If $x = 0,\: y = 12,$ and if $y = 0, \: 2x = 12\Rightarrow\: x = 6$.

At (0,0), in the inequality, we have 0 + 0 = 0 which is not ≥ 12. So the area associated with this inequality is unbounded and away from the origin

Next, draw the line $2x+9y = 36$.

If $x = 0, \:9y = 36\Rightarrow\: y = 4$ and if $ y = 0, \:2x = 36\:\Rightarrow\: x = 18$.

At (0,0), in the inequality, we have 0 + 0 = 0 which is not ≥ 36. So the area associated with this inequality is unbounded and away from the origin

Next, plot the line $2x+3y = 24$.

If $x = 0, 3y = 24\:\Rightarrow\: y = 8$ and if $y = 0, 2x = 24 \:\Rightarrow\: x = 12$.

At (0,0), in the inequality, we have 0 + 0 = 0 which is not ≥ 24. So the area associated with this inequality is unbounded and away from the origin

$\textbf{Finding the feasible region}$

Since $x$ and $y$ are $\geq 0$, the feasible region is in the first quadrant.

On solving equations $2x+y = 12$ and $2x+9y = 36$, we get:

Substituting for $2x = 12 - y \:\Rightarrow\: 12 - y + 9y = 36\:\Rightarrow\: 8y = 24\: \Rightarrow\: y = 3$.

$\therefore\: 2x + 3 = 12\:\Rightarrow\: 2x = 9 \:\Rightarrow\: x = 4.5$.

$\Rightarrow\:$ Point G $ (4.5, 3)$

On solving equations $2x+y = 12$ and $2x+3y = 24$

Substituting for $2x + 12 - y \:\Rightarrow\: 12 - y + 3y = 24\:\Rightarrow\: 2y = 12\:\Rightarrow\: y = 6$

$\therefore 2x = 12 - 6 = 6 \:\Rightarrow\: x = 3$

$\Rightarrow\:$ Point $H (3,6)$

On solving equations $2x+9y = 36$ and $2x+3y = 24$

Substituting for $2x = 36-9y \:\Rightarrow\: 36-9y + 3y = 24\:\Rightarrow\: 6y = 12 \:\Rightarrow\: y = 2$.

$\therefore\: 2x = 36 - 9 \times 2 = 36 - 18 = 18 \:\Rightarrow\: x = 9$.

$\Rightarrow\:$ Point $I (9,2)$

Therefore the feasible region is the unbounded area with the core points EHID.

$\textbf{Solving the objective function using the corner point method}$

The values of $Z$ at the corner points are calculated as follows:

$\begin{array}{ll} \textbf{Corner Point} & \textbf{ Z = 250x+200y} \\ (0,12) & 2400 \\ (3,6) & 1950 \; \textbf{(Min Value)} \\ (9,2) & 2650 \\ (18,0) & 4500 \end{array}$

However, since the feasible region is unbounded, $1950$ may or may not be the minimum value. We therefore plot the inequality $ 250x + 200y < 1950$ to see if it has any points in common w the feasible region or not.

It can be seen that the inequality does not have any common points w the feasible region, hence $1950$ at $(3,6)$ must be the minimum value.

$\textbf{A) Minimum value of $Z$ is $1950$.}$

$\textbf{B) It takes 3 bags of brand P and 6 bags of brand Q to minimize the cost to Rs. 1950}$.