Chapter 10, Problem 12P

Feeding Lab Rabbits A biologist wishes to feed laboratory rabbits a mixture of two types of foods. Type I contains 8 g of fat, 12 g of carbohydrate, and 2 g of protein per ounce. Type II contains 12 g of fat, 12 g of carbohydrate, and 1 g of protein per ounce. Type I costs $0.20 per ounce and type II costs $0.30 per ounce. Each rabbit receives a daily minimum of 24 g of fat 36 g of carbohydrate, and 4 g of protein, but get no more than 5 oz of food per day. How many ounces of each food type should be fed to each rabbit daily to satisfy the dietary requirements at minimum cost?

To determine

To find: The number of ounces of Type I and Type II should be given to each rabbit as per the dietary requirement at minimum cost.

Answer to Problem 12P

The number of ounces of Type I is 3 and Type II ounces is 0 as per the dietary requirements at minimum cost.

Explanation of Solution

Given:

The Type I contains $8\text{g}$ of fat, $12\text{g}$ of carbohydrate, $2\text{g}$ of protein per ounce.

The Type II contains $12\text{g}$ of fat, $12\text{g}$ of carbohydrate, and $1\text{g}$ of protein per ounce.

Type I cost $\$0.20$ per ounce and Type II cost $\$0.30$ per ounce.

The rat receives the minimum $24\text{g}$ of fat, $36\text{g}$ of carbohydrate, $4\text{g}$ of protein but no more than $5\text{oz}$ of food per day.

Calculation:

Suppose the number of ounce of Type I is x and the number of ounce of Type II is y.

Use the given information to make the inequalities and the objective function for the feasible region.

The required information is shown in the table below.

	Type I	Type II	Daily minimum fed
Fat	$8\text{g}$	$12\text{g}$	$24\text{g}$
Carbohydrates	$12\text{g}$	$12\text{g}$	$36\text{g}$
Protein	$2\text{g}$	$1\text{g}$	$4\text{g}$
Cost	$\$0.20$	$\$0.30$

The objective function is,

$N=\$0.20x+\$0.30y$,

The constraint to get the feasible region has shown below.

$\begin{array}{l}8x+12y\ge 24\\ 12x+12y\ge 36\\ 2x+y\ge 4\\ x+y\le 5\\ x\ge 0,y\ge 0\end{array}\}$

Simplify the above inequalities as,

$\begin{array}{l}2x+3y\ge 6\\ x+y\ge 3\\ 2x+y\ge 4\\ x+y\le 5\\ x\ge 0,y\ge 0\end{array}\}$

The equations are,

$2x+3y=6$ (1)

And,

$x+y=3$ (2)

Now, get the value of y in terms of x from the equation (1),

$\begin{array}{c}2x+3y=6\\ 3y=6-2x\\ y=\frac{6-2x}{3}\end{array}$

Substitute $\frac{6-2x}{3}$ for y in the equation (2)

$\begin{array}{c}x+\frac{6-2x}{3}=3\\ 3x+6-2x=9\\ x=9-6\\ x=3\end{array}$

Substitute 3 for x in equation (1),

$\begin{array}{c}y=\frac{6-2\cdot 3}{3}\\ =\frac{6-0}{3}\\ =0\end{array}$

The intersection point is $\left(x,y\right)=\left(3,0\right)$.

Now, draw the graph of the above equations,

Figure (1)

The vertices which lies in the feasible region is shown below.

$\left(0,4\right),\left(0,5\right),\left(3,0\right),\left(5,0\right),\left(1,2\right)$

Substitute the 0 for x and 4 for y in the objective function $N=\$0.20x+\$0.30y$,

$\begin{array}{c}N=\$0.20x+\$0.30y\\ =\$0.20\cdot 0+\$0.30\cdot 4\\ =\$1.20\end{array}$

Substitute the 0 for x and 5 for y in the objective function $N=\$0.20x+\$0.30y$,

$\begin{array}{c}N=\$0.20x+\$0.30y\\ =\$0.20\cdot 0+\$0.30\cdot 5\\ =\$1.50\end{array}$

Substitute the 3 for x and 0 for y in the objective function $N=\$0.20x+\$0.30y$,

$\begin{array}{c}N=\$0.20x+\$0.30y\\ =\$0.20\cdot 3+\$0.30\cdot 0\\ =\$0.60\end{array}$

Substitute the 5 for x and 0 for y in the objective function $N=\$0.20x+\$0.30y$,

$\begin{array}{c}N=\$0.20x+\$0.30y\\ =\$0.20\cdot 5+\$0.30\cdot 0\\ =\$1.00\end{array}$

Substitute the 1 for x and 2 for y in the objective function $N=\$0.20x+\$0.30y$,

$\begin{array}{c}N=\$0.20x+\$0.30y\\ =\$0.20\cdot 1+\$0.30\cdot 2\\ =\$0.20+\$0.60\\ =\$0.80\end{array}$

So, all the satisfies these vertices are shown in the table below.

Vertices	$N=\$0.20x+\$0.30y$
$\left(0,4\right)$	$\$1.20$0
$\left(0,5\right)$	$\$1.50$
$\left(3,0\right)$	$\$0.60$(Minimum)
$\left(5,0\right)$	$\$1.00$
$\left(1,2\right)$	$\$0.80$

Thus, The number of ounces of Type I is 3 and Type II ounces is 0 as per the dietary requirements at the minimum cost for the objective function $N=\$0.20x+\$0.30y$.