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 elements A, 11.25 units of element B, and 3 units of element C. The minimum requirements of nutrients A, B and C are 18 units, 45 units and 24 units 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 the mixture contain x bags of brand P and the mixture contain y bags of brand Q.

We have to tabulate the given data as,

Brands	Number of bags	Element A	Elements B	Element C
P	x	3units	2.5 units	2units
Q	y	1.5 units	11.25units	3units
Cholesterol		18 units	45 units	24units

The equation for element A is given as,

3x+1.5y≥18 2x+y≥12

The equation for element B is given as,

2.5x+11.25y≥45 2x+9y≥36

The equation for element C is given as,

2x+3y≥24

We need to minimize the cost so we can use function which Minimize Z.

Minimize Z=250x+200y(1)

All constraints are given as,

2x+y≥12 2x+9y≥36 2x+3y≥24 x≥0,y≥0

2x+y≥12

x	6	0
y	0	12

2x+9y≥36

x	18	0
y	0	4

2x+3y≥24

x	12	0
y	0	8

Plot the graph using equations of constraint.

Substitute the value of x and y in the equation (1) to check the max value.

Corner points	Value of Z
( 0,12 )	2400
( 3,6 )	1950(Minimum)
( 9,2 )	2650
( 18,0 )	4500

Where 1950 may or may not be the minimum value of Z so we have to check graph inequality of 250x+200y<1950.

There is no point common between feasible region and 250x+200y<1950.

Thus, the cost will be minimum if the mixture contains 3 bags of brand P and 6 bags of brand Q and the minimum cost is Rs. 1950.