This is a typical diet problem. Let the mixture contain x kg of Food ‘P’ and y kg of Food ‘Q’. Clearly, x, y ≥ 0. Let us construct the following table from the given data:

$\begin{array}{llllll} \textbf{Foods} & \textbf{Packets} & \textbf{Calcium} &\textbf{Iron} &\textbf{Cholesterol} &\textbf{Vitamin A} \\ P & x & 12x & 4x & 6x & 6x \\ Q & y & 3y & 20y & 4y & 3y\\ \text{Total} & x+y & 12+3y & 4x+20y & 6x+4y & 6x+3y\\ \textbf{Requirements}& & \text{Atleast 240}& \text{Atleast 460} & \text{Atmost 300} & \end{array}$

We need to maximize Z = 6x+3y given the following constraints:

$(1):\qquad$ 12x+3y $\geq$ 240 $\to$ 4x+y $\geq$ 80

$(2):\qquad$ 4x+20y $\geq$ 460 $\to$ x+5y $\geq$ 115

$(3):\qquad$ 6x+4y $\leq$ 300 $\to$ 3x+2y $\leq$ 150

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

$\textbf{Plotting the constraints}$

First draw the graph of the line 4x+y = 80.

If x = 0 $\to$ y = 80, and if y = 0 $\to$ 4x = 80 $\to$ x = 20.

At (0,0), in the inequality, we have 0 + 0 = 0 which is not ≥ 80. So the area associated with this inequality is unbounded and away from the origin

Next, draw the line x+5y = 115.

If x = 0 $\to$ 5y = 115 $\to$ y = 23, and if y = 0 $\to$ x = 115.

At (0,0), in the inequality, we have 0 + 0 = 0 which is not ≥ 115. So the area associated with this inequality is unbounded and away from the origin

Next, plot the line 3x+2y = 150.

If x = 0 $\to$ 2y = 150 $\to$ y = 75, and if y = 0, 3x = 150 $\to$ x = 50.

At (0,0), in the inequality, we have 0 + 0 = 0 which is $\leq$ 150. So the area associated with this inequality is bounded and towards 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 4x+y = 80 and x+5y = 115, we get:

Substituting for x = 115-5y $\to$ 4 $\times$ (115-5y) + y = 80 $\to$ 460 - 20y + y = 80 $\to$ 19y = 380 $\to$ y = 20.

Therefore, x = 115 - 5 $\times$ 20 = 115 - 100 = 15.

$\to$ Point G (15, 20)

On solving equations 3x+2y = 150 and x + 5y = 115, we get:

Substituting for x = 115 - 5y $\to$ 3 \times (115-5y) + 2y = 150 $\to$ 345 - 15y + 2y = 150 $\to$ 13y = 195 $\to$ y = 15.

Therefore, x = 115 - 5 $\times$ 15 = 115 - 75 = 40.

$\to$ Point H (40, 15).

On solving equations 3x+2y = 150 and 4x+y = 80, we get:

Substituting for y = 80-4x $\to$ 3x + 2 $\times$ (80-4x) = 150 $\to$ 3x + 160 - 8x = 150 $\to$ 5x = 10 $\to$ x = 2.

Therefore, y = 80 - 4 $\times$ 2 = 80 - 8 = 72.

$\to$ Point I (2, 72).

Therefore the feasible region is the area bounded by GIHG,

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

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

$\begin{array}{ll} \textbf{Corner Point} & \textbf{ Z = 6x+3y} \\ (15,20) & 150 \\ (40,15) & 285 \; \textbf{(Max Value)} \\ (2,72) & 228 \end{array}$

$\textbf{A) The maximum value of Z is 285 at (40,15).}$

$\textbf{B) To maximize the amount of Vitamin A, 40 packets of P and 15 packets of Q should be used.}$