Question

Formulate this problem an LLP and solve it graphically.
A Diet for a sick person must contain at least $4000$ units of vitamins, $50$ units of minerals and $1400$ units of calories. Two foods $A$ and $B$ are available at a cost of Rs $5$ ans Rs $4$ per unit respectively. one unit of food $A$ contains $200$ units of vitamins, $1$ units of minerals and $40$ units of calories, while one unit of the food $B$ contains $100$ unit of vitamins, $2$ unit of minerals and $40$ units of calories. Find what combination of the foods $A$ and $B$ should be used to have least cost, but it must satisfy the requirements of the sick person.. form the question as LLP and solve it graphically.

Answer 1

We have,

Let $x$ units of food A

$y$ units of food B

Where $x\ge 0,y\ge 0$

Then table

We are to minimize

$z=5x+4y$

Subject to constraints

$200x+100y\ge 4000$

$2x+y\ge 40......\left(1\right)$

$x+2y\ge 50......\left(2\right)$

$40x+40y\ge 1400$

$x+y\ge 35.......\left(3\right)$

$\text{where}x\ge 0,y\ge 0$

From equation (1) to and we get,

$2x+y=40$

If $x=0$ then $y=40$ OY in $L(0,40)$

If $y=0$ then $x=20$ OX in $A(20,0)$

From equation (2) to and we get,

$x+2y=50$

If $x=0$ then $y=25$ OY in $M(0,25)$

If $y=0$ then $x=50$ OX in $B(50,0)$

From equation (3) to and we get,

$x+y=35$

If $x=0$ then $y=35$ OY in $N(0,35)$

If $y=0$ then $x=35$ OX in $C(35,0)$

Shaded region is the feasible region, which is unbounded.

Then,

Corner point is

$B(50,0)$

$D(20,15)$

$E(5,30)$

$L(0,40)$

At point $B(50,0)\Rightarrow Z=5\left(50\right)+4\left(0\right)=250$

At point $D(20,15),Z=5\left(20\right)+4\left(15\right)=100+60=160$

At point $E(5,30),Z=5\left(5\right)+4\left(30\right)=25+120=145$

At point $L(0,40),Z=5\left(0\right)+4\left(40\right)=0+160=160$

Then at least cost $=Rs145at(5,30)$

Hence, this is the answer.