Chapter 4.6, Problem 51STP

Solution

To calculate: the number of gallons of each type of milk the dairy produce to maximize profits.

179 gallon of skim milk and 21 gallon of whole milk.

Given information:

Diary produces at most 200 gallons of skim and whole milk. Regular customers require at least 15 gallons of skim and 21 gallons of whole milk each day. The profit on a gallon of skim milk is $0.82 and the profit on a gallon of whole milk is $0.75

Formula used:

Optimization with linear programming:-

Step 1. Define the variables.

Step 2. Write a system of inequalities.

Step 3. Graph the system of inequalities.

Step 4. Find the coordinates of the vertices of the feasible region.

Step 5. Write a linear function to be maximized or minimized.

Step 6. Substitute the coordinates of the vertices into the function.

Step 7. Select the greatest or least result. Answer the problem.

Calculation:

Consider,

Let the x be the amount of skim produced.

Let the y be the amount of whole milk produced.

According to the question,

  $x+y\le 200\text{ (i)}$

Requirement of customers

  $\begin{array}{l}=x\ge 15\text{ (ii)}\\ =y\ge 21\text{ (iii)}\end{array}$

The profit is to be maximized. This implies :-

  $\text{Maximize function}=0.82x+0.75y$

  

The following shades represent the graph as respective: $15\le x$ (red shade), $21\le y$ (blue shade), $x+y\le 200$ (black shade). The common region comes out the triangle with vertices $(15,21),(179,21)\text{ and }(15,185)$ .

$(x,y)$	$0.82x+0.75y$	$f(x,y)$
$(15,185)$	$(0.82\times 15)+(0.75\times 185)$	$151.05$
$(15,21)$	$(0.82\times 15)+(0.75\times 21)$	$28.05$
$(179,21)$	$(0.82\times 179)+(0.75\times 21)$	$162.53$

This implies the maximum profit is at point $(179,21)$

Therefore, 179 gallon of skim milk and 21 gallon of whole milk should be produced to maximize profits.