Q. 25 - Precalculus Enhanced with Graphing Utilities

Question

Managing a Meat Market A meat market combines ground beef and ground pork in a single package for meat loaf. The ground beef is $75\%$ lean ( $75\%$ beef, $25\%$ fat ) and costs the market $\backslash (0.75$ per pound (lb). The ground pork is $60\%$ lean and costs the market $\backslash )0.45/lb$. The meat loaf must be at least $70\%$ lean. If the market wants to use at least $50/lb$ of its available pork, but no more than $200/lb$ of its available ground beef, how much ground beef should be mixed with ground pork so that the cost is minimized? 

Step-by-Step Solution

Answer

The minimum cost is $\$97.50$ and it is achieved with $100lb$ of ground beef and $50lb$ of ground pork. 

Step 1. Given information

Let x denote the number of pounds of ground beef and y denote the number of pounds of ground pork.
If C denotes the total cost of meat loaves to be produced, then C can be expressed as $C=0.75x+0.45y$. This expression is the objective function.

Step 2. Minimize the value of subject to the two constraints x and y .

As the number of pounds of ground beef should not exceed $200lb$, we get the first constraint as $x\le 200$.
Since the number of pounds of ground pork should be greater than $50lb$, we get the next constraint as $y\ge 50$.

It is given that $75\%$ of ground beef is lean and $60\%$ of ground pork is lean, which combine to give at least $70\%$ of meat loaf which is lean. So, we get the third inequality as

$0.75x+0.65y\ge 0.7(x+y)$

Step 3. Simplify the expression 0 . 75 x + 0 . 6 y ≥ 0 . 7 ( x + y ) by applying a distributive property in the right side of the inequality.

$0.75x+0.6y\ge \ge 0.7x+0.7y$

Subtract $0.7x+0.7y$ from both sides of the inequality.

$$\begin{gathered} 0.75x+0.6y-0.7x-0.7y\ge 0.7x+0.7y-0.7x-0.7y \\ 0.05x-0.1y\ge 0 \end{gathered}$$

Multiply both sides of the equation by 20 to clear the decimals 

$$\begin{gathered} 20(0.05x-0.1y)\ge 0\left(20\right) \\ x-2y\ge 0 \end{gathered}$$

The linear programming problem is to minimize $C=28x+52y$ subject to the constraints  $x\le 200,y\ge 50 and x-2y\ge 0$

Step 4. Graph the inequalities we obtained and label the corner points.

The shaded portion of the graph represents the set of feasible points. 

Step 5. Find the value of the object function at each corner point

Vertex	Value of cost $C=0.75x+0.45y$
$(100,50)$	$C=0.75\left(100\right)+0.45\left(50\right)=97.5$
$(200,50)$	$C=0.75\left(200\right)+0.45\left(50\right)=172.5$
$(200,100)$	$C=0.75\left(200\right)+0.45\left(100\right)=195$

Therefore, from the above table, we can see that the minimum cost is $\$97.5$ and it is achieved with $100lb$ of ground beef and $50lb$ of ground pork.