Question

A farmer plans to mix two types of food to make a mix of low cost feed for the animals in his farm. A bag of food $A$ costs 10$andcontains$40$unitsofproteins,$20$unitsofmineralsand$10$unitsofvitamins.Abagoffood$B$costs$ $12$ contains $30$ units of proteins, $20$ units of minerals and $30$ units of vitamins. How many bags of food $A$ and $B$ should the consumed by the animals each day in order to meet the minimum daily requirements of $150$ units of proteins, $90$ units of minerals and $60$ units of vitamins at a minimum cost?

Answer 1

Let $x$ be the number of bags of food $A$ and $y$ the number of bags of food $B$
Cost $C(x,y)=10x+12y$
The solution set of the system of inequalities above and the vertices of the feasible solution set obtained are shown below:
Vertices:
$A$ at intersection (x-axis) coordinates of $A:(6,0)$
$B$ at intersection of $10x+30y=60$ coordinates of $B:(15/4,4/3)$
$C$ at intersection of $40x+30y=150$ and $20x+20y=90$ coordinates of $D:(0,5)$
Evaluate the cost $c(x,y)=10x+12y$ at each one of the vertices $A(x,y),B(x,y),C(x,y)$ and $D(x,y)$
At $A(6,0):c(6,0)=10\left(6\right)+12\left(0\right)=60$
At $B(15/4,3/4):c(15/4,3/4)=10\left(15/4\right)+12\left(3/4\right)=46.5$
At $C(3/2,3):c(2,3)=10\left(3/2\right)+12\left(3\right)=51$
At $D(0,5):c(0,5)=10\left(0\right)+12\left(5\right)=60$
The cost $c(x,y)$ is minimum at the vertex $B(15/4,3/4)$ where $x=15/4=3.75$ and $y=3/4=0.75$
Hence $3.75$ bags of food $A$ and $0.75$ bags of food $B$ are needed to satisfy the minimum daily requirements in terms of proteins, minerals and vitamins at the lowest possible cost.