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Home  >>  CBSE XII  >>  Math  >>  Linear Programming
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Reshma wishes to mix two types of Food P and Q in such a way that the vitamin contents of the mixture contain at least 8 units of vitamin A and 11 units of vitamin B. Food P costs Rs 60/kg and Food Q costs Rs 80/kg. Food P contains 3 units/kg of Vitamin A and 5 units / kg of Vitamin B while Food Q contains 4 units/kg of Vitamin A and 2 units/kg of vitamin B. Determine the minimum cost of the mixture

$\begin{array}{1 1} 160 \\ 80 \\ 100 \\ 440 \end{array} $

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Toolbox:
  • 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.
This is a typical diet problem. Let the mixture contain x kg of Food ‘P’ and y kg of Food ‘Q’. Clearly, x, y ≥ 0. Let us construct the following table from the given data:
-
Resources P Q Requirements
Vitamin A 3 4 8
Vitamin B 5 2 11
Cost (Rs./Kg) 60 80  

 

Since the mixture must contain at least 8 units of Vitamin A and 10 units of Vitamin B, we have the following constraints: 3x + 4y ≥ 8. and 5x + 2y ≥ 11.
The total cost of purchasing x kg of food P and y kg of food Q is Z = 60x + 80y.
Hence we need to Minimize Z = 50x + 60y, subject to the constraints
$(1) \quad 3x+4y \geq 8$
$(2) \quad 5x+2y \geq 11$
$\textbf{Plotting the constraints}$
Plot the straight lines 3x+4y = 8 and 4x+2y = 11.
First draw the graph of the line 3x+4y = 8.
If x = 0, y = 2 and if y = 0, x = 8/3.
At (0,0), in the inequality, we have 0 + 0 = 0 which is not ≥ 8. So the area associated with this inequality is unbounded and away from the origin.
Similarly, draw the graph of the line 5x + 2y = 11.
If x = 0, y = 11/2 and if y =0, x = 11/5.
At (0,0), in the inequality, we have 0 + 0 = 0 which is not ≥ 11. So the area associated with this inequality is unbounded and away from the origin.
$\textbf{Finding the feasible region}$
We can see that the feasible region is unbounded.
On solving the equations 3x + 4y = 8 and 5x + 2y = 11, we get,
3x + (22-10x) = 8 $\rightarrow$ -7x = -14 $\rightarrow$ x = 2.
If x = 2, then y = (11 – 10) /2 = 1/2 .
x = 2 and y = $\frac{1}{2}$
Therefore the feasible region has the corner points $(\frac{8}{3}, 0), (2,\frac{1}{2}), (0, \frac{11}{2}).$
$\textbf{Solving using Corner Point Method}$
The values of Z at the corner points are summarized as follows:
Corner Point Z = 60x+80y
(8/3, 0) 160 (Min Value)
(2, 1/2) 160 (Min Value)
(0, 11/2) 440
 
In the table we find that the smallest value of Z is 160. Can we say that this is the minimum value? Remember that the feasible region is unbounded.
Therefore we have to draw the graph of the inequality 60x + 80y < 160, i.e, 3x + 4y < 8 to check if the resulting half plane has any common point with the feasible region.
From the figure and table we can see that it has no points in common.
$\textbf{Thus, the minimum value of Z is 160 attained at two of the points}$
$\textbf{The optimum strategy would be to mix 2 kgs of P and ½ kg of Q or just use 8/3 kgs of P, at a cost of Rs. 160.}$

 

answered Apr 16, 2013 by balaji.thirumalai
edited Apr 17, 2013 by balaji.thirumalai
 

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