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| Meaties | Yummies | |
|---|---|---|
| Sales price per package | $2.80 | $2.00 |
| Raw materials per package | ||
| Cereal | 2.0lb. | 3.0lb. |
| Meat | 3.0lb. | 1.5lb. |
| Variable cost-blending and packing | $0.25 package | $0.20 package |
| Resources | ||
| quad Production capacity for Meaties | 90,000 packages per month | |
| Cereal available per month | 400,000lb. | |
| Meat available per month | 300,000lb. |
.
Solution:
The Decision Variables. We first identify those things over which we have control: the decision variables. In this problem we have direct control over two quantities: the number of packages of Meaties to make each month, and the number of packages of Yummies to make each month. Within the model these two quantities appear repeatedly, so we represent them in a simple fashion. We designate these variables by the symbols $M$ and $Y$.
$M=$ number of packages of Meaties to make each month
$Y=$ number of packages of Yummies to make each month
Note that the amount of meat used each month and the amount of cereal used each month are not good choices for the variables. First, we control these only indirectly through our choice of $M$ and $Y$. More important, using these as variables could lead to ambiguous production plans. Determining how much cereal and meat to use in production does not tell us how to use it-how much of each dog food to make. In contrast, after determining the values for $M$ and $Y$, we know what to produce and how much meat and cereal are needed.
| Grain | {:[” Minimum “],[” Daily “],[” Requirement “]:} | ||||
|---|---|---|---|---|---|
| 1 | 2 | 3 | (units) | ||
| Nutrient | A | 20 | 30 | 70 | 110 |
| Nutrient | B | 10 | 10 | 0 | 18 |
| Nutrient | C | 50 | 30 | 0 | 90 |
| Nutrient | D | 6 | 2.5 | 10 | 14 |
| Cost (c//lb) | 41 | 36 | 96 |
Solution
The quantities that the manager controls are the amounts of each grain to feed each sheep daily. We define
$x_j=$ number of pounds of grain $j(=1,2,3)$ to feed each sheep daily
Note that the units of measure are completely specified. In addition, the variables are expressed on a per sheep basis. If we minimize the cost per sheep, we minimize the cost for any group of sheep. The daily feed cost per sheep will be
(cost per lb of grain $j) \times(\mathrm{lb}$. of grain $j$ fed to each sheep daily)
That is, the objective function is to
$$
\text { Minimize } z=41 x_1+36 x_2+96 x_3
$$
Why can’t the manager simply make all the variables equal to zero? This keeps costs at zero, but the manager would have a flock of dead sheep, because there are minimum nutrient constraints that must be satisfied. The values of the variables must be chosen so that the number of units of nutrient A consumed daily by each sheep is equal to or greater than 110 . Expressing this in terms of the variables yields
$$
20 x_1+30 x_2+70 x_3 \geq 110
$$
The constraints for the other nutrients are
$$
\begin{array}{rr}
10 x_1+ & 10 x_2 \geq 18 \
50 x_1+ & 30 x_2 \geq 90 \
6 x_1+2.5 x_2+10 x_3 \geq 110
\end{array}
$$
and finally
$$
\text { all } x_j s \geq 0
$$
The optimal solution to this problem (obtained using a computer software package) is $x_1=0.595, x_2=2.008, x_3=0.541$, and $z=148.6$ cents.
e will shade the half-plane containing $(0,0)$.
To graph Line (2), we will find its intercepts.
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