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Home  >>  CBSE XII  >>  Math  >>  Linear Programming
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A manufacturer makes two types of toys A and B. Three machines are needed for this purpose and the time (in minutes) required for each toy on the machines is given below: \[\] \begin{array}{cccc} \textbf{Types} & \textbf{Machine I} & \textbf{Machine II}& \textbf{Machine III}\\ A&12&18&6 \\ B&6&0&9 \end{array} \[\] Each machine is available for a maximum of 6 hours per day. If the profit on each toy of type A is Rs 7.50 and that on each toy of type B is Rs 5, Formulate the L.P.P and solve to find the maximum profit.

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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 toys of type A and y toys of type B. Clearly, x, y ≥ 0. Let us construct the following table from the given data:
$\begin{array}{lcccccc} \textbf{Type of Toys} & \textbf{No. of Toys} & \textbf{Machine I time} &\textbf{Machine II time} &\textbf{Machine III time} &\textbf{Profit} \\ & & (\text{in min}) & (\text{in min})& (\text{in min}) & \text{Rs.}\\ A & x & 12x&18x&6x&7.5x\\ B&y&6y&0&9y&5y\\ \text{Total} & x+y& 12x+6y&18x&6x+9y&7.5x+5\\ \textbf{Requirements}& & \text{Max}\; 6\times60=360&\text{Max}\; 6\times60=360&\text{Max}\; 6\times60=360&\\ \end{array}$
We need to maximize Z =7.5x+5y given the following constraints:
$(1):\qquad$12x+6y$\;\leq\;$360 $\to$ 2x+y $\leq\;$ 60
$(2):\qquad$18x $\leq\;$ 360 $\to$ x $\leq\;$ 20
$(3):\qquad$6x+9y$\leq\;$ 360 $\to$ 2x+3y $\leq\;$ 120
$(4):\qquad$ x $\geq$ 0 and $(5):\qquad$ y $\geq$ 0.
$\textbf{Plotting the constraints}$
First draw the graph of the line 2x+y=60.
If x = 0 $\to$ y=60, and if y=0 $\to$ 2x = 60 $\to$ x = 30.
At (0,0), in the inequality, we have 0 + 0 = 0 which is $\leq$ 10. So the area associated with this inequality is bounded and towards the origin
Next, draw the line x = 20.
At (0,0), in the inequality, we have 0 + 0 = 0 which is $\leq$ 20. So the area associated with this inequality is bounded and towards the origin
Next, plot the line 2x+3y=120.
If x = 0 $\to$ 3y=120 $\to$ y = 40, and if y = 0 $\to$ 2x=120 $\to$ x = 60.
At (0,0), in the inequality, we have 0 + 0 = 0 which is $\leq$ 120. 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 2x+y = 60 and x = 20, we get:
Substituting for x=20, 2 $\times$ 20 + y = 60 $\to$ y = 60-40 = 20.
$\to$ Point G (20,20)
On solving equations 2x+y=60 and 2x+3y = 120, we get:
Substituting for 2x=60-y, we get 60-y+3y=120 $\to$ 2y = 60 $\to$ y = 30.
Therefore, 2x = 60 - 30 = 30 $\to$ x=15.
$\to$ Point H (15,30)
On solving equations x=20 and 2x+3y = 120, we get:
Substituting for x=20 $\to$ 2 $\times$ 20 + 3y=120 $\to$ 3y = 120-40 = 80 $\to$ y = $\large\frac{80}{3}$
$\to$ Point I (20,$\large\frac{80}{3}$).
Therefore the feasible region is the area bounded by the vertex corner points OEGHCO.
$\textbf{Solving the objective function using the corner point method}$
The values of Z at the corner points are calculated as follows:
$\begin{array}{cc} \textbf{Corner Point} & \textbf{ Z = 7.5x+5y} \\ (0,0) & 0\\ (20,0) & 150 \\ (20,20) & 250 \; \\ (15,30) & 712.5 \; \textbf{(Max Value)} \\ (0,40) & 200 \end{array}$
$\textbf{A) Maximum value of Z is 712.5.}$
It takes $15$ toys of type $A$ and $30$ toys of type $B$ to maximize the profits to $Rs. 712.50$.
answered Apr 19, 2013 by balaji.thirumalai
edited Feb 6, 2014 by rvidyagovindarajan_1
 

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