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Home  >>  CBSE XII  >>  Math  >>  Linear Programming
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Read the details below and Determine the number of bags of each brand which should be mixed in order to produce a mixture having a minimum cost per bag? What is the minimum cost of the mixture per bag? A farmer mixes two brands P and Q of cattle feed. Brand P, costing Rs 250 per bag, contains 3 units of nutritional element A, 2.5 units of element B and 2 units of element C. Brand Q costing Rs 200 per bag contains 1.5 units of nutritional element A, 11.25 units of element B, and 3 units of element C. The minimum requirements of nutrients A, B and C are 18 units, 45 units and 24 units respectively. Determine the number of bags of each brand which should be mixed in order to produce a mixture having a minimum cost per bag? What is the minimum cost of the mixture per bag?

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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 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 $1950$.}$
$\textbf{B) It takes 3 bags of brand P and 6 bags of brand Q to minimize the cost to Rs. 1950}$.
answered Apr 19, 2013 by balaji.thirumalai
edited Feb 11, 2014 by rvidyagovindarajan_1
 

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