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# There are two types of fertilisers $$F_{1}$$ and $$F_{2}$$. $$F_{1}$$ consists of 10% nitrogen and 6% phosphoric acid and $$F_{2}$$ consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, a farmer finds that she needs atleast 14 kg of nitrogen and 14 kg of phosphoric acid for her crop. If $$F_{1}$$ costs Rs 6/kg and $$F_{2}$$ costs Rs 5/kg, determine how much of each type of fertiliser should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?

$\begin{array}{1 1}1000 \\ 1400 \\ 1200 \\ 1100 \end{array}$

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A)
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 x be the weight of fertilizer F1 and y be the weight for fertilizer F2. We can construct the following table with the information given in the problem:
Clearly, x, y ≥ 0. Let us construct the following table from the given data
 F1 (x) kg F2 (y) kg Requirements Nitrogen 10% 5% At least 14kg Phosphoric Acid 6% 10% At least 14kg Cost Rs. 6 5
We have the following constraints:
10% x + 5% y ≥ 14 $\to$10x + 5y ≥ 1400 $\to$2x + y ≥ 280
6%x + 10% y ≥ 14 $\to$6x + 10y ≥ 1400 $\to$3x + 5y ≥ 700
We need to minimize the cost while the nutrient requirements are met, i.e minimize Z = 6x + 5y.
$\textbf{Plotting the constraints}$:
Plot the straight lines 2x + y = 280 and 3x + 5y = 700
First draw the graph of the line 2x + y = 280.
If x = 0, y = 280 and if y = 0, x = 140. So, this is a straight line between (0,280) and (140,0). At (0,0), in the inequality, we have 0 + 0 = 0 which is not ≥ 0.
At (0,0), in the inequality, we have 0 + 0 = 0 which is not ≥ 0. So the area associated with this inequality is unbounded and away from the origin.
Similarly, draw the graph of the line 3x + 5y = 700
If x = 0, y = 700/5 = 140 and if y =0, x = 700/3. So, this is a straight line between (0,140) and (700/3,0).
At (0,0), in the inequality, we have 0 + 0 = 0 which is not ≥ 0. So the area associated with this inequality is unbounded and away from the origin.
$\textbf{Finding the feasible region}$:
We can see that the feasible region is bounded and in the first quadrant.
On solving the equations 2x + y = 280 and 3x + 5y = 700, we get,
3x + 5(280-2x) = 700 $\to$3x + 1400 – 10x = 700 $\to$-7x = -700 $\to$x = 100.
If x = 100, y = 280-200 = 80.
$\Rightarrow x = 100, y = 80$
Therefore the feasible region has the corner points (0,280), (100,80) and (700/3, 0) as shown in the figure.
$\textbf{Solving the objective function using the corner point method}$
The values of Z at the corner points are calculated as follows:
 Corner Point Z = 6x+5y (0, 280) 1400 (100, 80) 1000 (Minimum Value) (700/3, 0) 1400
However, we still don’t know if 1000 is indeed the minimum value for Z. We have to check through the inequality Z $\leq$ 1000, i.e, 6x + 5y $\leq$ 1000
If we graph this, we can see that there is no common point w/ the feasible region.
Therefore the minimum cost has to be 1000.
$\textbf{The minimum feasible value for the mixture is 1000.}$