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Home  >>  CBSE XII  >>  Math  >>  Linear Programming
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A company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type A require 5 minutes each for cutting and 10 minutes each for assembling. Souvenirs of type B require 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours 20 minutes available for cutting and 4 hours for assembling. The profit is Rs 5 each for type A and Rs 6 each for type B souvenirs. How many souvenirs of each type should the company manufacture in order to maximise the profit?

$\begin{array}{1 1} 120 \\ 160 \\ 300 \\ 200 \end{array} $

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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 souveniers of type A and y be the number of souveniers of type B that we can make. Our problem is to maximize x and y and the maximum profit at a given capacity.
Clearly, x, y ≥ 0. Let us construct the following table from the given data
  Souvenier A (x) Souvenier B (x) Time
Cutting Machine (min) 5 8 3 x  60 + 20 = 200
Assembling (min) 10 8 4 x 60 = 240
Profits (Rs.) 5 6  
We have the following constraints: 5x+8y $\leq$ 200 and 10x+8y $\leq$ 240 $\to$5x + 4y $\leq$ 120
The profit on Souvenir A is Rs. 5 and on Souvenir B is Rs. 6. We need to maximize the profits, i.e. maximize 5x + 3y, given the above constraints.
$\textbf{Plotting the constraints}$:
Plot the straight lines 5x+8y = 200 and 5x+4y = 120
First draw the graph of the line 5x+8y = 200
If x = 0, y = 200/8 = 25 and if y = 0, x = 200/5 = 40. So, this is a straight line between (0,25) 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.
Similarly, draw the graph of the line 5x+4y = 120.
If x = 0, y = 120/4 = 30 and if y =0, x = 120/5 = 24. So, this is a straight line between (0,30) and (24,0).
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 5x+8y = 200 and 5x+4y = 120, we get,
8y – 4y = 200 – 120 $\to$4y = 80 $\to$y = 20.
If y = 20, x = (120-4x20)/5 = 8.
$\Rightarrow x = 8, y = 20 $
Therefore the feasible region has the corner points (0,0), (0,25), (8,20), (0,25) 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 = 5x+6y
(0,0) 0
(0,25) 150
(8,20) 160 (Max Value)
(24,0) 120
$\textbf{The maximum profit we can make is Rs. 160, which involves making 8 Type A Souvenirs and 20 Type B Souvenirs. }$

 

answered Apr 18, 2013 by balaji.thirumalai
 

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