Chapter 3.8, Problem 53SR

To determine

To calculate: The gallons of fat-free milk and whole milk to be produced by the dairy to maximize its profit.

Answer to Problem 53SR

The maximum profit is obtained by producing $179$ gallons of fat-free milk and 21 gallon of whole milk is $162.53$ .

Explanation of Solution

Given information:

The profit on a gallon of fat-free milk is $0.82 and gallon of whole milk is $ 0.75. Dairy produces 200 gallons of fat-free milk and whole milk each day, and regular customers require 15 gallons of fat-free milk and 21 gallons of whole milk each day.

Calculation:

Suppose that $x$ gallon of fat-free milk and $y$ gallon of whole.

Consider the case of Yoder Family diary that produces at most 200 gallons of whole and milk cream.

The inequality is:

  $x+y\le 200$

The customers require a minimum of 15 gallons fat-free milk and 21 gallons whole milk every day,

So the inequality is:

  $\begin{array}{l}x\ge 15\\ y\ge 21\end{array}$

Profit from fat-free milk is $ $0.82$ and profit from whole milk $ $0.75$ is Profit is maximized

Suppose that p is profit.

Then $P=0.82x+0.75y$

Consider the inequalities and profit:

  $\begin{array}{l}x+y\le 200\\ x\ge 15\\ y\ge 21\end{array}$

To maximize:

  $P=0.82x+0.75y$

To solve the equation:

  $\begin{array}{cc}x+y& =200\\ x& =15\\ y& =21\end{array}$

The points are (15,21)

Substitute $x=15$ into the equation

  $\begin{array}{l}x+y=200\\ 15+y=200\end{array}$

Simplify the above equation:

  $\begin{array}{l}y=200-15\\ y=185\end{array}$

The points are (15,185)

Substitute $y=21$ into the equation $x+y=200$

  $x+21=200$

Simplify the above equation:

  $\begin{array}{l}x=200-21\\ x=179\end{array}$

The points are (179,21)

Use maple to draw the inequality:

  

In the graph there is a common region $\text{ΔABC}$

Substitute all the points into the profit to check where the profit is maximized:

Substitute $\text{B}(15,21)$ into $\text{P}=0.82x+0.75y$

  $\begin{array}{cc}P& =0.82(15)+0.75(21)\\ & =28.05\end{array}$

Substitute $\text{A}(15,185)$ into $\text{P}=0.82x+0.75y$

  $\begin{array}{cc}P& =0.82(15)+0.75(185)\\ & =151.05\end{array}$

Substitute $\text{C}(179,21)$ into $\text{P}=0.82x+0.75y$

  $\begin{array}{cc}P& =0.82(179)+0.75(21)\\ & =162.53\end{array}$

So, the profit is maximizing at point (179,21)

Therefore, the maximum profit is obtained by producing $179$ gallons of skim milk and 21 gallon of whole milk is $162.53$ .