Question

A farmer mixes two brands $P$ and $Q$ of cattle feed. Brand $P$, costing $Rs.250$ per bag, contains $3$ units of nutritional element $A,2.5$ units of element $B$ and $2$ units of element $C$. Brand $Q$ costing $Rs.200$ per bag contains $1.5$ units of nutritional element $A,11.25$ units of element $B$, and $3$ units of element $C$. The minimum requirements of nutrients $A,B$ and $C$ are $18units,45units$ and $24units$ respectively. Determine the number of bags of each brand which should be mixed in order to produce a mixture having a minimum cost per bag? What is the minimum cost of the mixture per bag?

Answer 1

Let's assume that mixture contains $Xbags$ of brands P and $Ybags$ of brands Q.

Cost of 1 bag of brands P $=250Rs$

Cost of 1 bag of brands Q $=200Rs$

So, Total cost (C) $=250X+200YRs...\left(1\right)$

Now, brands P contains $3units$ and brands Q contains $1.5units$ of nutritional element A.

So, Total nutritional element A $=3X+1.5Yunits$

Since, minimum requirement of nutritional element A is 18 units.

$\therefore 3X+1.5Y\ge 18$

$\Rightarrow 2X+Y\ge 12...\left(2\right)$ (After dividing by 1.5 both side)

Again, brands P contains $2.5units$ and brands Q contains $11.25units$ of nutritional element B.

So, Total nutritional element B $=2.5X+11.25Yunits$

Since, minimum requirement of nutritional element B is 45 units.

$\therefore 2.5X+11.25Y\ge 45$

$\Rightarrow 2X+9Y\ge 36...\left(3\right)$ (After dividing by 1.25 both side)

Also, brands P contains $2units$ and brands Q contains $3units$ of nutritional element C.

So, Total nutritional element C $=2X+3Yunits$

Since, minimum requirement of nutritional element C is 24 units.

$\therefore 2X+3Y\ge 24...\left(4\right)$

Since, amount of brands can never be negative.

So, $X\ge 0,Y\ge 0...\left(5\right)$

We have to minimise the cost of mixture given in equation (1) subject to the constraints given in (2),(3), (4) and (5).

After plotting all the constraints, we get the feasible region as shown in the image.

Corner points	Value of $Z=250X+200Y$
A (0,12)	2400
B (3,6)	1950 (minimum)
C (9,2)	2650
D (18,0)	4500

Now, since region is unbounded. Hence we need to confirm that minimum value obtained through corner points is true or not.

Now, plot the region $Z<1950$ to check if there exist some points in feasible region where value can be less than 1950.

$\Rightarrow 250X+200Y<1950$

$\Rightarrow 5X+4Y<39$.

Since there is no common points between the feasible region and the region which contains $Z<1950$ (See the image). So 1950 is the minimum cost.