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Home  >>  CBSE XII  >>  Math  >>  Linear Programming
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A factory manufactures two types of screws, A and B. Each type of screw requires the use of two machines, an automatic and a hand operated. It takes 4 minutes on the automatic and 6 minutes on hand operated machines to manufacture a package of screws A, while it takes 6 minutes on automatic and 3 minutes on the hand operated machines to manufacture a package of screws B. Each machine is available for at the most 4 hours on any day. The manufacturer can sell a package of screws A at a profit of Rs 7 and screws B at a profit of Rs 10. Assuming that he can sell all the screws he manufactures, how many packages of each type should the factory owner produce in a day in order to maximise his profit? Determine the maximum profit.

$\begin{array}{1 1} 400 \\ 280 \\ 410 \\ 430 \end{array} $

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1 Answer

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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 x be the number of packages of screws A and y be the number of packages of screws B that we can make.
Clearly, x, y ≥ 0. Let us construct the following table from the given data
  Screw A (x) Screw B (x) # minutes
Machine A (h) 4 6 4h x 60 = 240 min
Machine B (h) 6 3 4h x 60 = 240 min
Profit (Rs.) 7 10  
Since we have only 240m of machine time, we have the following constraints:
4x + 6y $\leq$ 240 $\to$2x + 3y $\leq$ 120
6x + 3y $\leq$ 240 $\to$2x + y $\leq$ 80
The profits on Screw A is Rs. 7 and on Screw B is Rs. 10. We need to maximize the profits, i.e. maximize 7x + 10y, given the above constraints.
$\textbf{Plotting the constraints}$:
Plot the straight lines 2x + 3y = 120 and 2x + y = 80
First draw the graph of the line 2x + 3y = 120
If x = 0, y = 4 and if y = 0, x = 12. So, this is a straight line between (0,4) and (12,0).
If x = 0, y = 40 and if y = 0, x = 60. So, this is a straight line between (0,40) and (60,0). At (0,0), in the inequality, we have 0 + 0 = 0 which is $\leq$ 0. So the area associated with this inequality is bounded towards the origin.
Similarly, draw the graph of the line 2x + y = 80.
If x = 0, y = 80 and if y =0, x = 40. So, this is a straight line between (0,80) and (40,0).
At (0,0), in the inequality, we have 0 + 0 = 0 which is $\leq$ 0. So the area associated with this inequality is bounded towards 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 + 3y = 120 and 2x + y = 80, we get,
2x – 2x + 3y – y = 120 – 80 $\to$2y = 40 $\to$y = 20
If y = 20,2 x = 120 – 2x20 $\to$x = 30.
$\Rightarrow x = 30, y = 20 $
Therefore the feasible region has the corner points (0,0), (0,40), (30,20), (40,0) 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 = 7x + 10y

O (0,0)

0

A (0,40)

280

E (30,20)

410 (Max value)

D (40,0)

400

$\textbf{ The maximum profit we can make is Rs. 410, which involves making 30 packages of Screw A and 20 packages of Screw B.}$

 

answered Apr 18, 2013 by balaji.thirumalai
edited Apr 18, 2013 by balaji.thirumalai
 

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