Question

A house wife wishes to mix together two kinds of food, I and II, in such a way that the mixture contains at least $10$ units of vitamin A, $12$ units of vitamin B and $8$ units of vitamin C. The vitamin contents of one kg of food is given below:

	Vitamin A	Vitamin B	Vitamin C
Food I	1	2	3
Food II	2	2	1

One Kg of food I costs $E6$ and one Kg of food II costs $E10$. Formulate the above problem as a linear programming problem and find the least cost of the mixture which will product the diet.

Answer 1

Given

	Vitamin A	Vitamin B	Vitamin C	Cost Rs
Food i	1	2	3	16
Food ii	2	2	1	10
Requirement	Atleast 10	Atleast 12	Atleast 8

Solution:We need to maximise $z=6x+10y$ for the following contraints

i)$x+2y\ge 10$

ii)$2x+2y\ge 12\Rightarrow$ $x+y\ge 6$

iii)$3x+y\ge 8$

iv)$x\ge 0,y\ge 0$

Finding the feasible region

Since $x$ and $y$ are$\ge ,0$,,the fasible region is in first quadrant

i)On solving equation $x+y=6$ and $x+2y=10$

We get

$x=6-y$ $\Rightarrow 6-y+2y=10$ $y=4$

$x=6-y$ $\Rightarrow =6-4$ $x=2$

ii)On standing equation $x+2y=10$and$3x+y =8$ we get

$x+2y=10$ $\underset{–}{3x+y=8}$ $x+2y=10$ $\underset{–}{6x+2y=16}$ $-5x=-6$ $x=\frac{6}{5}$

$x+2y=10$ $\frac{6}{5}+2y=10$ $y=\frac{22}{5}$

Point H$(\frac{6}{5},\frac{22}{5})$

iii)On solving equation $x+y=6$ and $3x+y=8$ we get

$x+y=6$ $\underset{–}{(-)3x+y=8}$ $-2x=-2$ $x=1$

$x+y=6$ $1+y=6$ $y=5$

Point$I(1,5)$

i)$x+2y\ge 10$

$x$	0	10
$y$	5	0

Point A(0,5) B(10,0)

ii)$x+y\ge$ 6

$x$	0	6
$y$	6	0

Point C(0,6) D(6,0)

iii)$3x+y\ge 8$

$x$	0	$8/3$
$y$	8	0

Point E(0,8) F$(8/3,0)$

Corner point	cost $z=6x+10y$
(0,8)	80
(135)	$56\to$ MInimum
(2,4)	66
(0,8)	80

The least mixture cost is 56 for point,

$x=1$and $y=5$

Least cost=56