question_answer

A-fliet for a sick person must contain at least 4000 units of vitamins, 50 units of minerals and 1400 calories. Two foods A and B are available at a cost of Rs. 4 and Rs. 3 per unit, respectively. 1 unit of food A contains 200 units of vitamins, 1 unit of minerals and 40 calories. Food B contains 100 units of vitamins, 2 units of minerals and 40 calories. Find what combination of food should be used to have the least cost. Why a proper diet is required for us?

Answer:

Let quantity of food A be x units and quantity of food B be y units. Then, the objective function is minimise \[Z=4x+3y.\] Subject to the constraints \[200x+100y\ge 4000\] \[\Rightarrow \] \[2x+y\ge 40\] [dividing both sides by 100]             \[x+2y\ge 50\]             \[40x+40y\ge 1400\] \[\Rightarrow \]   \[x+y\ge 35\]      [dividing both sides by 40] and       \[x,\,\,y\ge 0\]     Consider the constraints as equations, we get             \[2x+y=40\]                              ? (i)             \[x+2y=50\]                              ? (ii)             \[x+y=35\]                                ? (iii) and       \[x=0,\,\,y=0\]                           ? (iv) Table for line \[2x+y=40\] is

x	0	20
y	40	0

So, line \[2x+y=40\] passes through (0, 40) and (20, 0) On putting (0, 0) in the inequality \[2x+y\ge 40,\] we get \[2(0)+0\ge 40\] \[\Rightarrow \] \[0\ge 40\]             [false] So, the half plane is away from the origin.     Table for line \[x+2y=50\] is                    

x	0	50
y	25	0

So, line \[x+2y=50\] passes through (0, 25) and (50, 0). On putting (0, 0) in the inequality \[x+2y\ge 50,\] we get                                                     \[0+2(0)\ge 50\] \[\Rightarrow \] \[0\ge 50\]             [false]   So, the half plane is away from the origin,         Table for line \[x+y=35\] is

x	0	35
y	35	0

So, line \[x+y=35\] passes through (0, 35) and (35, 0) On putting (0, 0) in the inequality \[x+y\ge 35,\] we get \[0+0\ge 35\] \[\Rightarrow \] \[0\ge 35\]                [false] So, the half plane is away from the origin. The points of intersection of lines (i) and (ii), (ii) and (iii), (i) and (iii) are (10, 20), (20, 15)and (5, 30), respectively.            The graphical representation of these lines is given below The shaded region in the graph represents the feasible region which is unbounded and its corner points are A (50, 0), 8 (20, 15), C (5, 30) and D (0, 40). Now, the values of Z at corner points are

Corner points	\[\mathbf{Z=4x+3y}\]
A (50, 0)	\[Z=4(50)+3(0)=200\]
B (20, 15)	\[Z=4(20)+3(15)=125\]
C (5, 30)	\[Z=4(5)+3(30)=110\] (minimum)
D (0, 40)	\[Z=4(0)+3(40)=120\]

From the table, minimum value of Z is 110 at C (5, 3) but the feasible region is unbounded, therefore Z = 110 may or may not be minimum value of Z.          For this, we draw the graph of the inequality \[4x+3y<110\] and check whether the resulting half plane has points in common with the feasible region or not. It can be seen that the feasible region has no common point with \[4x+3y<110,\] therefore the minimum value of Z is 110. Hence, 5 units of food A and 30 units of food B should be used to have the least cost and least cost is Es. 110. Value A proper diet is required for proper functioning of the different body parts or to provide sufficient energy to do work.