**Toolbox:**

- First formulate the objective function and identify the constraints from the problem statement, To solve a Linear Programming problem graphically, first plot the constraints for the problem. This is done by plotting the boundary lines of the constraints and identifying the points that will satisfy all the constraints.
- One we graphically plot the area bounded by the constraints, it’s easy to see which points satisfy all constraints. This common region determined by all the constraints including non-negative constraints of a linear programming problem is called the $\textbf{Feasible Region (or solution region).}$
- Now, any point in the feasible region that gives the optimal value (maximum or minimum) of the objective function is called an $\textbf{Optimal Solution.}$ We see that every point in the feasible region satisfies all the constraints, and there are infinitely many points.
- Since we know from theory that the optimal value must occur at a corner point (vertex) of the feasible region, calculate the objective function values associated with the coordinates of all the extreme points. This method is called the $\textbf{Corner Point Method.}$
- If the feasible region is bounded (if it can be enclosed), the point with the best objective function value is the best optimal solution. If the feasible region is unbounded (means that the feasible region does extend indefinitely in any direction), the then a maximum or a minimum value of the objective function may not exist. However, if it exists, it must occur at a corner point of the feasible region, which can be calculated.

Let the mixture contain x kgs of food X and y kgs of food Y. Clearly, x, y ≥ 0. Let us construct the following table from the given data:

$\begin{array}{lcccccc} \textbf{Food} & \textbf{Amount} & \textbf{Vitamin A} &\textbf{Vitamin B} &\textbf{Vitamin C} &\textbf{Cost Rs.} \\ X & x & x & 2X & 3x & 16x \\ Y & y & 2y& 2y& y&20y\\ \text{Total} & x+y& x+2y&2x+2y&3x+y&16x+20y\\ \textbf{Requirements}& & \text{Atleast}\; 10&\text{Atleast}\;12&\text{Atleast}\;8&\\ \end{array}$

We need to minimize Z =16x+20y given the following constraints:

$(1):\qquad$x+2y$\;\geq\;$10

$(2):\qquad$2x+2y$\;\geq\;$12 $\to$ x+y $\geq$ 6

$(3):\qquad$3x+y$\;\geq\;$8

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

$\textbf{Plotting the constraints}$

First draw the graph of the line x+2y = 10

If x = 0 $\to$ 2y = 10 $\to$ y = 5, and if y = 0 $\to$ x=10.

At (0,0), in the inequality, we have 0 + 0 = 0 which is not ≥ 10. So the area associated with this inequality is unbounded and away from the origin

Next, draw the line x+y = 6.

If x = 0 $\to$ y=6 and vice versa.

At (0,0), in the inequality, we have 0 + 0 = 0 which is not ≥ 6. So the area associated with this inequality is unbounded and away from the origin

Next, plot the line 3x+y = 8.

If x = 0 $\to$ y =8 and if y = 0, 3x = 8 $\to$ x = $\large\frac{8}{3}$

At (0,0), in the inequality, we have 0 + 0 = 0 which is not ≥ 8. 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 x+2y = 10 and x+y = 6, we get:

Substituting for x=6-y $\to$ 6-y+2y=10 $\to$ y=4.

Therefore,x = 6-4 = 2.

$\to$ Point G (2,4)

On solving equations x+2y=10 and 3x+y = 8, we get:

Substituting for x = 10-2y $\to$ 3 $\times$ (10-2y) + y = 8 $\to$ 30 - 6y + y = 8 $\to$ 5y = 22$\to$ y = $\large\frac{22}{5}$

Therefore x = 10 - 2 $\times \large \frac{22}{5} = \frac{50-44}{5} = \frac{6}{5}$

$\to$ Point H ($\large\frac{6}{5},\frac{22}{5}$)

On solving equations x+y = 6 and 3x+y = 8, we get:

Substituting for y = 6-x $\to$ 3x + 6 -x = 8 $\to$ 2x = 2 $\to$ x = 1.

Therefore, y = 6-1 = 5.

$\to$ Point I (1,5)

Therefore the feasible region is the unbounded area w the corepoints EIGB.

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

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

$\begin{array}{cc} \textbf{Corner Point} & \textbf{ Z = 16x+20y} \\ (0,8) & 160\\ (1,5) & 116 \\ (2,4) & 112 \; \textbf{(Min Value)}\\ (0,8) & 160 \end{array}$

However, since the feasible region is unbounded, 112 may or may not be the minimum value. We therefore plot the inequality 16x+20y<112 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 112 at (2,4) must be the minimum value.

$\textbf{A) Minimum value of Z is 112.}$

$\textbf{B) It takes2 kgs of X and 4 kgs of Y to minimize the cost to Rs. 112}$.