Chapter 8.5, Problem 37E

To determine

To Find:The combination of beef and pork for the minimum cost. What is the minimum cost per hundred pounds?

Answer to Problem 37E

The minimum cost is $\$76$ in which combination of beef and pork is $30\%and20\%.$

Explanation of Solution

Given:A meat packers makes a kind of sausage using the beef, pork and cereal, fat, water and spices.

Using the given data in the question the inequality equations become.

  $\begin{array}{l}b\ge 0.3\\ p\ge 0.2\\ b+p\le 0.66\\ b\ge p\end{array}$

Now consider the graph.

Now from the graph the vertices are $(0.2,0.3),(0.3,0.3),(0.33,0.33),(0.2,0.46).$

Now the cost function is

  $\begin{array}{l}C(p,b)=90p+140b+32(1-p-b)\\ =32+58p+108b\\ now\\ plottingpo\mathrm{ints}\\ C(0.2,0.3)=32+58(0.2)+108(0.3)=76\\ C(0.3,0.3)=32+58(0.3)+108(0.3)=81.8\\ C(0.33,0.33)=32+58(0.33)+108(0.33)=86.8\\ C(0.2,0.46)=32+58(0.2)+108(0.46)=93.3\end{array}$

Thus the minimum cost is $\$76$ in which combination of beef and pork is $30\%and20\%.$