Question

Certain animals in a rescue shelter must have at least 30 g of protein and at least 20 g of fat per feeding
period. These nutrients come from food A, which costs 18 cents per unit and supplies 2 g of protein
and 4 g of fat; and food B, which costs 12 cents per unit and has 6 g of protein and 2 g of fat. Food B
is bought under a contract requiring that at least 2 units of B be used per serving. How much of each
food must be bought to produces the minimum cost per serving? What is the lowest cost

Answer 1

Let $x$ denote the number units of food A;

Let $y$ denote the number units of food B;

Here food A costs 18 cents ($0.18) per unit and food B - 12 cents ($0.12);

The optimization function is: $Z = 0.18*x + 0.12*y$ ;

Food A has 2 g protein and 4 g of fat and food B has 6 g of protein and 2 g of fat;

Hence we have: $2*x + 6*y$ for protein and $4*x + 2*y$ for fat.;

Animals must have at least 30 g of protein and at least 20 g of fat ;

Therefore, the constrains are:

$2*x + 6*y >= 30$

$4*x+2*y>=20$

Food B is bought under a contract requiring that at least 2 units of B be used per serving, therefore $y >= 2$ ;

The linear programming problem is:

minimize $z = 0.18*x + 0.20*y$

subject to

$2*x+6*y>=30$

$4*x+2*y>=20$

$y >= 2$

$x>=0;$

The graphical method:

First equation:

$2*x+6*y=30$

$x + 3*y = 15$

take $y = 0$ to get x intercept: $x = 15$ , therefore, the point on the graph is (15,0),

take $x=0$ to get y intercept: $y = 5$ , therefore, the point on the graph is (0,5)

Second equation:

$4*x+2*y=20$

$2*x + y = 10$

take $y = 0$ to get x intercept: $x = 5$ , therefore, the point on the graph is (5,0),

take $x=0$ to get y intercept: $y = 10$ , therefore, the point on the graph is from the graph, the feasible points are (0.10)

To find the intersection of two lines($x + 3*y = 15$ and $2*x + y = 10$ ), multiply them to get $2*x+6*y=30$ ;

Then subtract $2*x + y = 10$ from $2*x+6*y=30$ , we get $5*y = 20$ , $y = 4$ ;

Now subtract $y = 4$ from $2*x+6*y=30$ , we get $x=3$ ;

Therefore, the intersection point is (3,4)

From the graph, the feasible points are: (0,10),(3,4),(9,2),(15,0);

To find optimization of the function:

minimum at (0,10):

$Z = 0.18x + 0.12y$

$Z = 0.18*0 + 0.12*10$

$Z = 1.2$

minimum at (3,4):

$Z=1.02$

minimum at (9,2):

$Z = 1.86$

minimum at (15,0):

$Z = 2.7$

Here 1.02 is the smallest value, therefore 3 units of food A and 4 units of food B are used to minimize the cost.

Answer:  the minimum cost per serving is $1.02.