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Home  >>  CBSE XII  >>  Math  >>  Linear Programming
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A merchant plans to sell two types of personal computers, a desktop model and a portable model that will cost Rs 25000 and Rs 40000 respectively. He estimates that the total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computers which the merchant would stock to get maximum profit if he does not want to invest more than Rs 70 lakhs and if his profit on the desktop model is Rs 4500 and on portable model is Rs 5000.

This question has appeared in model paper 2012

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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 is 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 desktop computers and y be the number of portable computers that we can stock. Our problem is to maximize x and y and the maximum profit.
Clearly, x, y ? 0. Let us construct the following table from the given data
Desktop (x) Desktop (y) Investment 25000 40000 Profit 4500 5000 The profit on desktop models is 4500 and on the portables is 5000, so we need to maximize 4500x + 5000y, given the following constraints:
Monthly demand is 250 computers. Therefore, x + y $\leq$ 250.
The maximum he can invest is Rs. 70L. Therefore, 25000x + 40000y $\leq$ 700000, i.e., 5x + 8y $\leq$ 1400
$\textbf{Plotting the constraints}$:
Plot the straight lines x+y = 250 and 5x+8y = 1400.
First draw the graph of the line x+y = 250
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).
If x = 0, y = 250 and vice versa. So, this is a straight line between (0,250) and (250,0). 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+8y = 1400.
If x = 0, y = 1400/8 = 175 and if y =0, x = 1400/5 = 280 So, this is a straight line between (0,175) and (280,0).
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 x+y = 250 and 5x+8y = 1400, we get,
5 (250-y) + 8y = 1400 $\to$1250 – 5y + 8y = 1400 $\to$ 3y = 150 $\to$y = 50.
If y = 50, x = 250 – 50 = 200.
$\Rightarrow x = 200, y = 50 $
Therefore the feasible region has the corner points (0,0), (0,175), (200,50), (0,250) 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 = 4500x+5000y (0,0) 0 (0,175) 875000 (200,50) 1150000 (250,0) 1125000
$\textbf{The maximum profit we can make is Rs. 11.5l, which involves stocking 250 desktops and 50 portable computers.}$
answered Dec 13, 2013 by balaji.thirumalai
 

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